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Half-life and doubling time models

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Difficulty: 1 – 5

55149212

A sample contains \(80\,\text{mg}\) of a substance. Its half-life is \(6\) hours. How much remains after exactly one half-life?

Hints

- A half-life is the time required for an amount to become half as large. - Only one half-life has passed.

Solution

One half-life reduces the amount to one-half of its starting value. Half of \(80\,\text{mg}\) is \(40\,\text{mg}\).

Answer

\(40\,\text{mg}\)

55149312

The graph shows the amount \(A(t)\) of a substance over time. The marked values are \(A(0)=160\), \(A(5)=80\), and \(A(10)=40\). What is the half-life?

Figure for problem 551493

Hints

- Compare two marked values where one amount is half the other. - Measure the horizontal time difference between those two points.

Solution

The amount falls from \(160\) to \(80\) in \(5\) time units, and from \(80\) to \(40\) in another \(5\) time units. Therefore, the half-life is \(5\) time units.

Answer

\(5\) time units

52645512

A decay process is modeled by \(h(t)=250(0.75)^t\), where \(t\) is measured in seconds. 1. Rewrite the function in the form \(h(t)=h(0)e^{kt}\). Round \(k\) to four decimal places. 2. Find the half-life \(T_H\) of the process. 3. Find the percent decrease per second.

Hints

- Use \(e^k=0.75\) to find \(k\). - At the half-life, the exponential factor equals \(0.5\). - Interpret the factor \(0.75\) as the percent that remains after one second.

Solution

1. The initial value is \(h(0)=250\). Since \(e^k=0.75\), \(k=\ln(0.75)\approx -0.2877\). Thus, \(h(t)=250e^{\ln(0.75)t}\approx250e^{-0.2877t}\). 2. At the half-life, \((0.75)^{T_H}=0.5\). Therefore, \(T_H=\frac{\ln(0.5)}{\ln(0.75)}\approx 2.41\,\text{s}\). 3. The factor \(0.75\) means that \(75\%\) remains each second, so the decrease is \(25\%\) per second.

Answer

1. \(h(t)=250e^{\ln(0.75)t}\approx250e^{-0.2877t}\) 2. \(T_H\approx 2.41\,\text{s}\) 3. \(25\%\) per second

52833512

The Rule of 70 estimates the doubling time \(d\) in years for an investment earning \(p\%\) annually: \(pd\approx70\). Use the rule to estimate the doubling time for each annual rate. 1. \(1.4\%\) 2. \(2.5\%\) 3. \(3.5\%\) 4. \(5.0\%\)

Hints

- Rearrange the formula to isolate the doubling time. - Substitute each percent rate for \(p\). - The result is measured in years.

Solution

1. Solve for doubling time: \(d\approx\frac{70}{p}\). 2. For \(1.4\%\), \(d\approx\frac{70}{1.4}=50\) years. 3. For \(2.5\%\), \(d\approx\frac{70}{2.5}=28\) years. 4. For \(3.5\%\), \(d\approx\frac{70}{3.5}=20\) years. 5. For \(5.0\%\), \(d\approx\frac{70}{5.0}=14\) years.

Answer

1. \(50\) years 2. \(28\) years 3. \(20\) years 4. \(14\) years

52833612

A biologist uses the Rule of 70, \(pd\approx70\), to estimate the relationship between a culture's hourly growth rate \(p\), as a percent, and its doubling time \(d\), in hours. Estimate the hourly growth rate for each culture. 1. Culture A: \(d=20\,\text{hours}\) 2. Culture B: \(d=35\,\text{hours}\) 3. Culture C: \(d=50\,\text{hours}\)

Hints

- Identify the unknown in the Rule of 70. - Consider how a longer doubling time affects the growth rate. - Substitute each doubling time into the formula.

Solution

1. Rearrange the Rule of 70: \(p\approx\frac{70}{d}\). 2. Culture A: \(p\approx\frac{70}{20}=3.5\), so the hourly growth rate is about \(3.5\%\). 3. Culture B: \(p\approx\frac{70}{35}=2\), so the hourly growth rate is about \(2\%\). 4. Culture C: \(p\approx\frac{70}{50}=1.4\), so the hourly growth rate is about \(1.4\%\).

Answer

1. About \(3.5\%\) per hour 2. About \(2\%\) per hour 3. About \(1.4\%\) per hour

52841712

A forestry company estimates that the timber volume in a forest grows by \(3.5\%\) each year. The initial volume is \(2000\,\text{m}^3\). After how many years will the volume double? Round to the nearest tenth.

Hints

- Convert the annual percent increase to a growth factor. - A doubled amount is twice the initial value. - Use logarithms to solve for an unknown exponent. - Notice that the initial volume cancels.

Solution

1. The growth factor is \(1.035\). Doubling gives \(2000\cdot1.035^t=4000\). 2. Divide by \(2000\): \(1.035^t=2\). 3. Therefore, \(t=\frac{\log(2)}{\log(1.035)}\approx20.15\), which rounds to \(20.1\) years.

Answer

About \(20.1\) years

52848812

A radioactive sample has a half-life of \(15\) minutes. At the beginning of an experiment, the sample has a mass of \(120\,\text{mg}\). a) Write a function \(f(t)=cb^t\) that gives the mass, in milligrams, after \(t\) minutes. b) Determine how many minutes it takes for the mass to decrease to \(15\,\text{mg}\).

Hints

- A half-life tells you the factor over a \(15\)-minute interval. - Express the decay using a power with base \(0.5\). - Determine how many halvings reduce \(120\,\text{mg}\) to \(15\,\text{mg}\). - Multiply the number of half-life intervals by \(15\) minutes.

Solution

1. The initial value is \(c=120\). 2. The minute-by-minute decay factor satisfies \(b^{15}=0.5\), so \(b=0.5^{1/15}\approx0.9548\). Thus \(f(t)=120\left(0.5^{1/15}\right)^t\), or equivalently \(f(t)=120(0.5)^{t/15}\). 3. Set the mass equal to \(15\): \(120(0.5)^{t/15}=15\), so \((0.5)^{t/15}=\frac{1}{8}=(0.5)^3\). 4. Therefore, \(\frac{t}{15}=3\), giving \(t=45\) minutes.

Answer

a) \(f(t)=120\left(0.5^{1/15}\right)^t\), or equivalently \(f(t)=120(0.5)^{t/15}\) b) \(45\) minutes

52849212

A medication is eliminated from the body so that its concentration in the blood decreases by \(14\%\) each hour. Find the medication's half-life in hours. Round to the nearest hundredth.

Hints

- What fraction of the concentration remains after one hour? - Write the exponential decay factor. - At the half-life, the remaining fraction is \(0.5\). - Use logarithms to solve for time.

Solution

1. The hourly decay factor is \(0.86\). 2. At the half-life, \(0.86^t=0.5\). 3. Therefore, \(t=\frac{\ln(0.5)}{\ln(0.86)}\approx4.59577\). 4. Rounded to the nearest hundredth, the half-life is \(4.60\) hours.

Answer

About \(4.60\) hours

53449212

Two types of algae grow in a test pond. The graphs show the area, in square meters, covered by each type after \(t\) days. The curve labeled \(p\) represents type \(1\), and the curve labeled \(q\) represents type \(2\). a) Use the graph to find the doubling time for each type. b) The growth of type \(1\) is modeled by \(f(t)=5(2)^{t/4}\). Find the area it covers after \(10\) days.

Figure for problem 534492

Hints

- Identify the initial area, then locate twice that value on the vertical axis. - Read the time at which each graph reaches the doubled area. - Substitute \(t=10\) into the given model and check that the result agrees with the graph.

Solution

1. Both types start at \(5\,\text{m}^2\). Type \(1\), shown by curve \(p\), reaches \(10\,\text{m}^2\) after \(4\) days, and type \(2\), shown by curve \(q\), reaches \(10\,\text{m}^2\) after \(6\) days. Their doubling times are \(4\) days and \(6\) days, respectively. 2. \(f(10)=5\cdot2^{10/4}=5\cdot2^{2.5}\approx28.28\,\text{m}^2\).

Answer

a) Type \(1\): \(4\) days; type \(2\): \(6\) days b) About \(28.28\,\text{m}^2\)

51008912

Hydrogen peroxide (\(\text{H}_2\text{O}_2\)) decays exponentially in the presence of a catalyst. During an experiment, half of the hydrogen peroxide decays in the first \(10\) minutes. How many minutes after the experiment begins will \(90\%\) of the hydrogen peroxide have decayed? a) \(18\,\text{min}\) b) \(30\,\text{min}\) c) \(32.1\,\text{min}\) d) \(33.2\,\text{min}\)

Hints

- Use the half-life to write an exponential decay model. - If \(90\%\) has decayed, what percent remains? - Set the remaining fraction equal to the decay model, then use logarithms to solve for time.

Solution

1. With a half-life of \(10\) minutes, the amount remaining is \(N(t)=N_0\cdot0.5^{\frac{t}{10}}\). 2. If \(90\%\) has decayed, then \(10\%\) remains, so \(0.1N_0=N_0\cdot0.5^{\frac{t}{10}}\). 3. Divide by \(N_0\): \(0.1=0.5^{\frac{t}{10}}\). 4. Take natural logarithms: \(\ln(0.1)=\frac{t}{10}\ln(0.5)\). 5. Therefore, \(t=10\cdot\frac{\ln(0.1)}{\ln(0.5)}\approx33.2\,\text{min}\).

Answer

d) \(33.2\,\text{min}\)

52641612

A medication is eliminated from the body over time. Immediately after the dose is taken (\(t=0\)), its concentration in the blood is \(80\,\text{mg/L}\). After \(4\) hours, the concentration has decreased to \(50\,\text{mg/L}\). 1. Find the hourly decay factor \(a\) for a model of the form \(c(t)=c_0a^t\), where \(t\) is measured in hours. 2. Rewrite the model in the form \(c(t)=c_0e^{kt}\). 3. Find the half-life of the medication in this person’s body.

Hints

- Substitute the two given concentrations into the exponential model. - In a decay model, should \(k\) be positive or negative? - Half-life is the time required for the concentration to reach half its initial value. - Set the model equal to one-half of the initial concentration and solve for time.

Solution

1. Since \(c(0)=80\), the initial concentration is \(c_0=80\). Using \(c(4)=50\) gives \(80a^4=50\), so \(a^4=\frac{5}{8}\). Therefore, \(a=\left(\frac{5}{8}\right)^{1/4}\approx 0.8891\). 2. Because \(a=e^k\), \(k=\ln(a)=\frac{1}{4}\ln\left(\frac{5}{8}\right)\approx -0.1175\). Thus, \(c(t)=80e^{\frac{1}{4}\ln\left(\frac{5}{8}\right)t}\approx80e^{-0.1175t}\). 3. At the half-life \(T\), \(e^{kT}=0.5\). Therefore, \(T=\frac{\ln(0.5)}{k}\approx 5.90\). The half-life is about \(5.9\) hours.

Answer

1. \(a=\left(\frac{5}{8}\right)^{1/4}\approx 0.8891\) 2. \(c(t)=80e^{\frac{1}{4}\ln\left(\frac{5}{8}\right)t}\approx80e^{-0.1175t}\) 3. About \(5.9\,\text{hours}\)

52645612

A growth process is modeled by \(p(t)=15(1.04)^{0.5t}\), where \(t\) is measured in seconds. 1. Rewrite the function in the form \(p(t)=p(0)e^{kt}\). Round \(k\) to four decimal places. 2. Find the doubling time \(T_D\). 3. Find the percent increase per second.

Hints

- Use exponent rules to rewrite \((1.04)^{0.5t}\) with \(t\) as the entire variable exponent. - Use \(a^t=e^{\ln(a)t}\). - At the doubling time, the exponential factor equals \(2\). - Find the growth factor for exactly one second before converting it to a percent.

Solution

1. The initial value is \(p(0)=15\). Since \((1.04)^{0.5t}=e^{0.5\ln(1.04)t}\), \(k=0.5\ln(1.04)\approx 0.0196\). Therefore, \(p(t)=15e^{0.5\ln(1.04)t}\approx15e^{0.0196t}\). 2. At the doubling time, \((1.04)^{0.5T_D}=2\). Thus, \(T_D=\frac{2\ln 2}{\ln(1.04)}\approx 35.35\,\text{s}\). 3. The one-second growth factor is \((1.04)^{0.5}\approx 1.0198\), so the increase is about \(1.98\%\) per second.

Answer

1. \(p(t)=15e^{0.5\ln(1.04)t}\approx15e^{0.0196t}\) 2. \(T_D\approx 35.35\,\text{s}\) 3. About \(1.98\%\) per second

52646112

Find the doubling time \(T_D\) or half-life \(T_H\) for each function. Round each answer to two decimal places. a) \(f(t)=40(1.15)^t\) b) \(g(t)=100e^{-0.5t}\) c) \(h(t)=15(0.98)^t\)

Hints

- First decide whether each function represents growth or decay. - At a doubling time, the exponential factor equals \(2\); at a half-life, it equals \(0.5\). - The initial coefficient does not affect the doubling time or half-life.

Solution

1. For \(f\), solve \((1.15)^{T_D}=2\). Thus, \(T_D=\frac{\ln 2}{\ln(1.15)}\approx 4.96\). 2. For \(g\), solve \(e^{-0.5T_H}=0.5\). Thus, \(T_H=\frac{\ln(0.5)}{-0.5}\approx 1.39\). 3. For \(h\), solve \((0.98)^{T_H}=0.5\). Thus, \(T_H=\frac{\ln(0.5)}{\ln(0.98)}\approx 34.31\).

Answer

a) \(T_D\approx 4.96\) b) \(T_H\approx 1.39\) c) \(T_H\approx 34.31\)

52646212

A quantity is modeled by \(f(t)=f(0)e^{kt}\), where \(t\) is measured in hours. a) One quantity doubles every \(12\) hours. Find its growth coefficient \(k\). b) Another quantity has a half-life of \(25\) hours. Find the time when only \(10\%\) of its initial amount remains.

Hints

- Relate \(k\) to the given doubling time. - Use the half-life to determine the decay coefficient for part b). - When \(10\%\) remains, the ratio \(f(t)/f(0)\) equals \(0.10\).

Solution

1. For a doubling time of \(12\) hours, \(e^{12k}=2\). Therefore, \(k=\frac{\ln 2}{12}\approx 0.0578\). 2. A half-life of \(25\) hours gives \(e^{25k}=0.5\), so \(k=\frac{\ln(0.5)}{25}\). 3. To find when \(10\%\) remains, solve \(e^{kt}=0.10\). Thus, \(t=\frac{\ln(0.10)}{k}=\frac{25\ln(0.10)}{\ln(0.5)}\approx 83.05\) hours.

Answer

a) \(k=\frac{\ln 2}{12}\approx 0.0578\) b) \(t\approx 83.05\,\text{hours}\)

52646412

A new app has \(5000\) users at \(t=0\), where \(t\) is measured in months. After \(3\) months, the app has \(8000\) users. One analyst claims that the number of users grows linearly. A second analyst claims that it grows exponentially and doubles every \(5\) months. After \(6\) months, the app has \(12{,}800\) users. a) Determine which analyst chose the correct type of growth model. b) Use the data to evaluate the second analyst's claim about the doubling time.

Hints

- Compare differences and ratios over the two equal \(3\)-month intervals. - A doubling time satisfies an equation in which the growth factor equals \(2\). - Use logarithms to solve for the unknown time.

Solution

1. In the first \(3\)-month interval, the increase is \(8000-5000=3000\). In the second interval, the increase is \(12{,}800-8000=4800\). The differences are not equal, so a linear model does not fit. 2. The growth factors are \(\frac{8000}{5000}=1.6\) and \(\frac{12{,}800}{8000}=1.6\). Since the factors are equal over equal time intervals, an exponential model fits. 3. If \(T\) is the doubling time, then \((1.6)^{T/3}=2\). Therefore, \(T=\frac{3\ln 2}{\ln(1.6)}\approx 4.42\) months. 4. The second analyst is correct about the model type but incorrect about the claimed \(5\)-month doubling time.

Answer

a) The second analyst chose the correct model type: exponential growth b) The doubling time is about \(4.42\) months, not \(5\) months

52647312

A medication is eliminated from the body. At \(t=0\), the body contains \(250\,\text{mg}\) of the medication. The amount decreases by \(15\%\) each hour, and \(t\) is measured in hours. a) Write a model of the form \(m(t)=ab^t\). b) Rewrite the model in the form \(m(t)=ae^{kt}\). Round \(k\) to four decimal places. c) Find the half-life of the medication. d) Find when only \(10\,\text{mg}\) remains.

Hints

- Convert the percent decrease to a multiplication factor. - Use \(e^k=b\) to rewrite the model with base \(e\). - At the half-life, the ratio of the remaining amount to the initial amount is \(0.5\). - Use logarithms when the unknown is in the exponent.

Solution

1. A \(15\%\) hourly decrease gives the hourly factor \(b=1-0.15=0.85\). Thus, \(m(t)=250(0.85)^t\). 2. Since \(e^k=0.85\), \(k=\ln(0.85)\approx -0.1625\). Therefore, \(m(t)=250e^{\ln(0.85)t}\approx250e^{-0.1625t}\). 3. For the half-life \(T\), solve \((0.85)^T=0.5\). Thus, \(T=\frac{\ln(0.5)}{\ln(0.85)}\approx 4.27\) hours. 4. To find when \(10\,\text{mg}\) remains, solve \(250(0.85)^t=10\). Then \((0.85)^t=0.04\), so \(t=\frac{\ln(0.04)}{\ln(0.85)}\approx 19.81\) hours.

Answer

a) \(m(t)=250(0.85)^t\) b) \(m(t)=250e^{\ln(0.85)t}\approx250e^{-0.1625t}\) c) About \(4.27\,\text{hours}\) d) About \(19.81\,\text{hours}\)

52647412

The population of a city is modeled by \(P(t)=45{,}000e^{0.028t}\), where \(t\) is the number of years since 2020. a) Give the population at the beginning of 2020. b) Find the annual growth factor and the annual percent growth rate. c) Determine during which year the city's population first exceeds \(75{,}000\). d) Find the amount of time required for the population to triple.

Hints

- Interpret the coefficient in front of the exponential expression. - The one-year factor is \(e^{0.028}\). - Use logarithms to solve for a time in an exponential equation. - The initial value cancels when finding a tripling time.

Solution

1. At \(t=0\), \(P(0)=45{,}000\). 2. The annual growth factor is \(b=e^{0.028}\approx 1.0284\). Therefore, the annual growth rate is about \(2.84\%\). 3. Solve \(45{,}000e^{0.028t}=75{,}000\). This gives \(e^{0.028t}=\frac{5}{3}\), so \(t=\frac{\ln(5/3)}{0.028}\approx 18.24\). The population exceeds \(75{,}000\) during 2038. 4. For the tripling time \(T\), solve \(e^{0.028T}=3\). Thus, \(T=\frac{\ln 3}{0.028}\approx 39.24\) years.

Answer

a) \(45{,}000\) people b) Growth factor: about \(1.0284\); growth rate: about \(2.84\%\) c) During 2038 d) About \(39.24\,\text{years}\)

52649812

A forestry district tracks two types of timber. The current spruce volume is \(8000\,\text{m}^3\) and decreases by \(2.8\%\) per year. The current beech volume is \(2500\,\text{m}^3\) and increases by \(1.5\%\) per year because of reforestation. a) Write an exponential model of the form \(V(t)=V_0e^{kt}\) for each type of timber, where \(t\) is measured in years. b) Find the half-life of the spruce volume. c) Find when the beech volume reaches \(4000\,\text{m}^3\). d) Find when the two timber volumes are equal.

Hints

- Convert each annual percent change to a yearly factor and use \(k=\ln(q)\). - The initial value cancels when finding a half-life. - Use logarithms to solve equations with time in the exponent. - Set the two models equal to find when the timber volumes match.

Solution

1. For spruce, the yearly factor is \(0.972\), so \(k_S=\ln(0.972)\approx -0.0284\). Thus, \(V_S(t)=8000e^{\ln(0.972)t}\). 2. For beech, the yearly factor is \(1.015\), so \(k_B=\ln(1.015)\approx 0.0149\). Thus, \(V_B(t)=2500e^{\ln(1.015)t}\). 3. For the spruce half-life, solve \(e^{k_ST}=0.5\). Then \(T=\frac{\ln(0.5)}{\ln(0.972)}\approx 24.41\) years. 4. For beech to reach \(4000\,\text{m}^3\), solve \(2500e^{k_Bt}=4000\). Then \(t=\frac{\ln(1.6)}{\ln(1.015)}\approx 31.57\) years. 5. Set the volumes equal: \(8000e^{k_St}=2500e^{k_Bt}\). Thus, \(t=\frac{\ln(8000/2500)}{k_B-k_S}\approx 26.87\) years.

Answer

a) \(V_S(t)=8000e^{\ln(0.972)t}\); \(V_B(t)=2500e^{\ln(1.015)t}\) b) About \(24.41\,\text{years}\) c) About \(31.57\,\text{years}\) d) About \(26.87\,\text{years}\)

52670212

A microorganism population grows continuously according to \(N(t)=N_0e^{kt}\), where \(t\) is measured in hours. At \(t=0\), the population is \(1200\). After \(3\) hours, it is \(1850\). a) Find \(k\), rounded to four decimal places, and write the population model. b) Find the percent increase during the first hour. c) Find the doubling time of the initial population.

Hints

- Substitute the value at \(t=3\) into the model and use a logarithm. - The one-hour growth factor is \(e^k\). - At the doubling time, the exponential factor equals \(2\).

Solution

1. Using \(N(3)=1850\), \(1850=1200e^{3k}\). Therefore, \(k=\frac{1}{3}\ln\left(\frac{1850}{1200}\right)\approx 0.1443\), so \(N(t)=1200e^{\frac{1}{3}\ln\left(\frac{1850}{1200}\right)t}\approx1200e^{0.1443t}\). 2. The one-hour growth factor is \(e^k\approx 1.1552\), so the population increases by about \(15.52\%\) during the first hour. 3. For the doubling time \(T\), solve \(e^{kT}=2\). Thus, \(T=\frac{\ln 2}{k}\approx 4.80\) hours.

Answer

a) \(k\approx 0.1443\); \(N(t)=1200e^{\frac{1}{3}\ln\left(\frac{1850}{1200}\right)t}\approx1200e^{0.1443t}\) b) About \(15.52\%\) c) About \(4.80\,\text{hours}\)

52824512

An investment grows by a fixed percentage each year and is modeled by \(K(t)=K_0q^t\), where \(t\) is time in years and \(q\) is the growth factor. The time needed for the initial investment \(K_0\) to double is the doubling time \(t_D\). a) Show that \(t_D=\frac{\log(2)}{\log(q)}\). b) Find the doubling time for an investment earning \(4.5\%\) interest per year.

Hints

- What value must the investment have at the doubling time? - How can you solve an equation when the unknown appears in an exponent? - Convert the annual percent increase to a growth factor. - First divide both sides by the common factor \(K_0\).

Solution

1. a) At the doubling time, \(K(t_D)=2K_0\), so \(2K_0=K_0q^{t_D}\). 2. Divide by \(K_0\): \(2=q^{t_D}\). 3. Take logarithms: \(\log(2)=t_D\log(q)\). Therefore, \(t_D=\frac{\log(2)}{\log(q)}\). 4. b) A yearly interest rate of \(4.5\%\) gives \(q=1+0.045=1.045\). 5. Thus, \(t_D=\frac{\log(2)}{\log(1.045)}\approx15.75\) years.

Answer

a) \(t_D=\frac{\log(2)}{\log(q)}\) b) About \(15.75\) years

52824612

Two radioactive isotopes decay at different rates. Isotope A is modeled by \(N_A(t)=N_0\cdot0.92^t\), where \(t\) is measured in days. Isotope B has a half-life of exactly \(5\) days. a) Find the half-life of Isotope A to the nearest hundredth of a day. b) Find the daily decay factor \(b\) for Isotope B if its decay is modeled by \(N_B(t)=N_0b^t\).

Hints

- At the half-life, what fraction of the initial amount remains? - How can you solve when the unknown appears in an exponent? - How can you solve \(b^5=0.5\) when the unknown is the base? - Keep track of which information belongs to each isotope.

Solution

1. a) At the half-life, \(0.5N_0=N_0\cdot0.92^{t_H}\), so \(0.5=0.92^{t_H}\). 2. Take logarithms: \(\log(0.5)=t_H\log(0.92)\). Therefore, \(t_H=\frac{\log(0.5)}{\log(0.92)}\approx8.31\) days. 3. b) For Isotope B, \(0.5=b^5\). 4. Taking the fifth root gives \(b=0.5^{\frac{1}{5}}\approx0.871\).

Answer

a) About \(8.31\) days b) \(b\approx0.871\)

52831312

The Rule of 70 estimates the doubling time \(d\), in years, for an investment growing at a constant annual rate of \(p\%\): \(d\approx\frac{70}{p}\). a) Use the Rule of 70 to estimate the doubling time for annual rates of \(2\%\), \(5\%\), and \(10\%\). b) Find the exact doubling time for each rate using logarithms. Round to the nearest hundredth. c) For which rate does the Rule of 70 have the smallest error?

Hints

- A doubling means the growth factor raised to the unknown time equals \(2\). - Use logarithms to solve for an exponent. - Convert each percent rate to a growth factor. - Compare the absolute differences between each estimate and exact value.

Solution

1. a) Using \(d\approx\frac{70}{p}\): for \(2\%\), \(d\approx35\) years; for \(5\%\), \(d\approx14\) years; for \(10\%\), \(d\approx7\) years. 2. b) The exact doubling time satisfies \((1+\frac{p}{100})^d=2\), so \(d=\frac{\ln(2)}{\ln(1+\frac{p}{100})}\). 3. For \(2\%\), \(d\approx\frac{\ln(2)}{\ln(1.02)}\approx35.00\) years. 4. For \(5\%\), \(d\approx\frac{\ln(2)}{\ln(1.05)}\approx14.21\) years. 5. For \(10\%\), \(d\approx\frac{\ln(2)}{\ln(1.10)}\approx7.27\) years. 6. c) The absolute errors are about \(0.00\), \(0.21\), and \(0.27\) years, respectively. The smallest error occurs at \(2\%\).

Answer

a) \(35\) years, \(14\) years, and \(7\) years b) \(35.00\) years, \(14.21\) years, and \(7.27\) years c) \(2\%\)

52831412

Under ideal conditions, a bacteria culture doubles every \(14\) hours. a) Use the Rule of 70, \(pd\approx70\), to estimate the hourly growth rate \(p\) as a percent. b) Find the exact hourly growth factor \(b\) and the corresponding growth rate. Round the factor to four decimal places and the rate to two decimal places. c) The culture begins with \(500\) bacteria. Use the exact factor to predict the population after exactly \(3\) days. Round to the nearest whole bacterium.

Hints

- Convert days to hours. - A doubling time of \(14\) hours means the \(14\)-hour growth factor is \(2\). - Rearrange the Rule of 70 to estimate the rate. - Round the population only at the end.

Solution

1. Using \(14p\approx70\), \(p\approx\frac{70}{14}=5\), so the estimated hourly rate is \(5\%\). 2. The exact factor satisfies \(b^{14}=2\), so \(b=2^{1/14}\approx1.0507566\). Thus the hourly rate is \((b-1)\cdot100\%\approx5.08\%\). 3. Three days is \(72\) hours. The population is \(500b^{72}=500\cdot2^{72/14}\approx17{,}665.43\), or about \(17{,}665\) bacteria.

Answer

a) About \(5\%\) b) \(b\approx1.0508\); growth rate about \(5.08\%\) c) About \(17{,}665\) bacteria

52832512

An amount of \(\$4000.00\) is invested at a fixed annual interest rate of \(1.5\%\), compounded annually. a) Find the balance after \(8\) years. b) Determine when the balance first reaches \(\$5000.00\). c) Find the doubling time in two ways: exactly using logarithms and approximately using the Rule of 70, \(n\approx\frac{70}{p}\).

Hints

- Convert the percent increase to a growth factor. - Use logarithms to isolate a variable in an exponent. - Why does the starting balance cancel when finding doubling time? - In the Rule of 70, use the numerical percent rate without the percent sign.

Solution

1. The growth factor is \(1.015\), so \(K(n)=4000\cdot1.015^n\). 2. a) \(K(8)=4000\cdot1.015^8\approx4505.97\). The balance is about \(\$4505.97\). 3. b) Solve \(4000\cdot1.015^n=5000\). Then \(1.015^n=1.25\), so \(n=\frac{\ln(1.25)}{\ln(1.015)}\approx14.99\). The balance reaches \(\$5000.00\) after about \(15\) years. 4. c) For exact doubling time, solve \(1.015^n=2\): \(n=\frac{\ln(2)}{\ln(1.015)}\approx46.56\) years. 5. The Rule of 70 gives \(n\approx\frac{70}{1.5}\approx46.67\) years.

Answer

a) About \(\$4505.97\) b) About \(15\) years c) Exact: about \(46.56\) years; Rule of 70: about \(46.67\) years

52832612

A bacterial culture begins with \(250\) bacteria and grows by \(8\%\) each hour. a) State the hourly growth factor \(q\) and write a function \(B(t)\) for the population after \(t\) hours. b) After how many hours will the population have quadrupled? c) Find the exact doubling time and compare it with the Rule of 70 estimate, \(t\approx\frac{70}{p}\).

Hints

- Convert the hourly percent increase to a growth factor. - A quadrupled population is four times the initial population. - The starting value cancels when finding doubling time. - Compare the exact result and the Rule of 70 estimate by subtracting.

Solution

1. a) The growth factor is \(q=1+0.08=1.08\), so \(B(t)=250\cdot1.08^t\). 2. b) For a quadrupled population, solve \(250\cdot1.08^t=1000\), so \(1.08^t=4\). 3. Therefore, \(t=\frac{\ln(4)}{\ln(1.08)}\approx18.01\) hours. 4. c) The exact doubling time satisfies \(1.08^t=2\), so \(t=\frac{\ln(2)}{\ln(1.08)}\approx9.01\) hours. 5. The Rule of 70 gives \(t\approx\frac{70}{8}=8.75\) hours. The estimate is about \(0.26\) hour, or about \(15\) minutes, shorter than the exact value.

Answer

a) \(q=1.08\); \(B(t)=250\cdot1.08^t\) b) About \(18.01\) hours c) Exact: about \(9.01\) hours; Rule of 70: \(8.75\) hours

52832912

A rare wild-bee population in a protected area grows by \(6.5\%\) each year. a) After how many years will the population double? b) The habitat can support at most four times the current population. After how many years will the population reach that capacity if the growth rate remains constant?

Hints

- Convert the percent increase to a yearly growth factor. - Write an equation in which the growth factor raised to time equals \(2\). - Use logarithms to solve for the exponent. - How is quadrupling related to two consecutive doublings?

Solution

1. The yearly growth factor is \(1.065\). 2. a) For doubling, solve \(1.065^t=2\). Thus, \(t=\frac{\log(2)}{\log(1.065)}\approx11.01\) years. 3. b) For quadrupling, solve \(1.065^t=4\). Thus, \(t=\frac{\log(4)}{\log(1.065)}\approx22.01\) years. 4. Since \(4=2^2\), the quadrupling time is twice the doubling time.

Answer

a) About \(11.0\) years b) About \(22.0\) years

52833012

A manufacturing company increases its annual production by \(5.2\%\) each year. At the same time, its material cost per unit increases by \(2.8\%\) each year. Find the doubling time, in years, for each quantity, assuming the rates remain constant.

Hints

- Convert each percent increase to a growth factor. - Set the growth factor raised to time equal to \(2\). - The initial values cancel when finding doubling time.

Solution

1. The growth factors are \(1.052\) for production and \(1.028\) for material cost. 2. For production, solve \(1.052^t=2\): \(t=\frac{\log(2)}{\log(1.052)}\approx13.67\) years. 3. For material cost, solve \(1.028^t=2\): \(t=\frac{\log(2)}{\log(1.028)}\approx25.10\) years.

Answer

Production doubles after about \(13.7\) years. Material cost per unit doubles after about \(25.1\) years.

52833212

The rule of thumb \(p\cdot T_H\approx70\) can estimate the half-life \(T_H\) of an exponential decay process with a decrease rate of \(p\%\) per time unit. a) A radioactive sample loses \(10\%\) of its activity each hour. Estimate its half-life using the rule of thumb. b) Find the exact half-life by solving \(\left(1-\frac{p}{100}\right)^{T_H}=0.5\). c) Find the percent error of the estimate relative to the exact value.

Hints

- A percent decrease produces a decay factor less than \(1\). - At the half-life, one half of the initial amount remains. - Use the exact value in the denominator when calculating percent error.

Solution

1. a) Using \(p=10\), \(T_H\approx\frac{70}{10}=7\) hours. 2. b) Solve \(0.9^{T_H}=0.5\): \(T_H=\frac{\ln(0.5)}{\ln(0.9)}\approx6.58\) hours. 3. c) The percent error is \(\frac{|7-6.5788|}{6.5788}\cdot100\%\approx6.40\%\).

Answer

a) \(7\) hours b) About \(6.58\) hours c) About \(6.4\%\)

52833412

An investment earns a fixed annual interest rate and doubles in exactly \(14\) years. a) Find the annual interest rate \(p\). b) Find how long it takes for the original investment to grow to ten times its initial value.

Hints

- Write an equation relating the growth factor to the \(14\)-year doubling time. - Convert the growth factor to a percent rate. - The initial investment cancels when working with a growth multiple. - For tenfold growth, set the growth factor raised to time equal to \(10\). - Use logarithm properties to simplify the expression for time.

Solution

1. a) If \(q\) is the annual growth factor, then \(q^{14}=2\), so \(q=2^{\frac{1}{14}}\approx1.05076\). 2. Therefore, \(p=(q-1)\cdot100\%\approx5.08\%\). 3. b) For tenfold growth, \(q^t=10\). 4. Thus, \(t=\frac{\ln(10)}{\ln(q)}=14\cdot\frac{\ln(10)}{\ln(2)}\approx46.51\) years.

Answer

a) About \(5.08\%\) b) About \(46.5\) years

52833712

Two radioactive isotopes decay exponentially. Isotope A has a half-life of \(8\) hours, and isotope B has a half-life of \(14\) hours. Find the hourly percent decrease for each isotope. Round to the nearest hundredth of a percent.

Hints

- After one half-life, one-half of the original amount remains. - Write an equation in which the hourly decay factor is unknown. - Relate the decay factor to the percent decrease. - Use a root or a fractional exponent to solve for the factor.

Solution

1. If \(q\) is the hourly decay factor and \(T\) is the half-life, then \(q^T=0.5\), so \(q=0.5^{1/T}\). 2. For isotope A, \(q=0.5^{1/8}\approx0.9170\). The hourly percent decrease is \((1-q)\cdot100\%\approx8.30\%\). 3. For isotope B, \(q=0.5^{1/14}\approx0.9517\). The hourly percent decrease is \((1-q)\cdot100\%\approx4.83\%\).

Answer

Isotope A: about \(8.30\%\) per hour Isotope B: about \(4.83\%\) per hour

52833812

The area of a lake covered by algae grows exponentially. During a cool summer, the covered area doubles every \(10\) days. During a very warm summer, it doubles every \(6\) days. Find the daily percent growth rate for each summer. Round to the nearest hundredth of a percent.

Hints

- A doubling means the total growth factor over the given time is \(2\). - Write an equation of the form \(q^T=2\). - Use a root or a fractional exponent to solve for \(q\). - Convert the growth factor to a percent growth rate.

Solution

1. If \(q\) is the daily growth factor and \(T\) is the doubling time, then \(q^T=2\), so \(q=2^{1/T}\). 2. For the cool summer, \(q=2^{1/10}\approx1.0718\). The daily growth rate is \((q-1)\cdot100\%\approx7.18\%\). 3. For the very warm summer, \(q=2^{1/6}\approx1.1225\). The daily growth rate is \((q-1)\cdot100\%\approx12.25\%\).

Answer

Cool summer: about \(7.18\%\) per day Very warm summer: about \(12.25\%\) per day

52834312

A pain medication has a half-life of \(3\) hours in the human body. At time \(t=0\), a patient has \(600\,\text{mg}\) of the medication in the bloodstream. a) Find the amount remaining after \(6\), \(9\), and \(12\) hours. b) After how many hours will the amount decrease to \(5\,\text{mg}\)? c) What percent of the initial amount remains after \(10\) hours?

Hints

- Count how many half-life intervals fit into each given time. - Write a general exponential decay model using the half-life. - The half-life determines the exponent \(\frac{t}{3}\). - Use logarithms when solving for a time in the exponent. - To find a percent of the initial amount, use the remaining fraction from the model.

Solution

1. The decay model is \(N(t)=600\cdot0.5^{\frac{t}{3}}\). 2. a) \(N(6)=600\cdot0.5^2=150\,\text{mg}\), \(N(9)=600\cdot0.5^3=75\,\text{mg}\), and \(N(12)=600\cdot0.5^4=37.5\,\text{mg}\). 3. b) Solve \(600\cdot0.5^{\frac{t}{3}}=5\). Then \(0.5^{\frac{t}{3}}=\frac{1}{120}\). 4. Thus, \(t=3\cdot\frac{\ln\left(\frac{1}{120}\right)}{\ln(0.5)}\approx20.72\) hours. 5. c) The remaining fraction after \(10\) hours is \(0.5^{\frac{10}{3}}\approx0.0992\), or about \(9.92\%\).

Answer

a) After \(6\) hours: \(150\,\text{mg}\); after \(9\) hours: \(75\,\text{mg}\); after \(12\) hours: \(37.5\,\text{mg}\) b) About \(20.72\) hours c) About \(9.92\%\)

52834412

Radon-222 has a half-life of \(3.8\) days. a) What percent of an initial amount remains after \(14\) days? b) After how many days will the amount decrease to \(10\%\) of its initial value? c) After how many half-lives, and after how many days, will only \(\frac{1}{64}\) of the original amount remain?

Hints

- Use \(100\%\) or \(1\) as the initial amount. - Express elapsed time as a number of half-life intervals. - Relate fractions such as \(\frac{1}{2}\), \(\frac{1}{4}\), and \(\frac{1}{8}\) to powers of \(\frac{1}{2}\). - Multiply the number of half-lives by \(3.8\) days.

Solution

1. Model the percent remaining by \(N(t)=100\cdot0.5^{\frac{t}{3.8}}\). 2. a) \(N(14)=100\cdot0.5^{\frac{14}{3.8}}\approx7.78\%\). 3. b) Solve \(0.5^{\frac{t}{3.8}}=0.10\). Then \(t=3.8\cdot\frac{\ln(0.1)}{\ln(0.5)}\approx12.62\) days. 4. c) Since \(\frac{1}{64}=\left(\frac{1}{2}\right)^6\), the amount reaches this level after \(6\) half-lives. 5. The corresponding time is \(6\cdot3.8=22.8\) days.

Answer

a) About \(7.78\%\) b) About \(12.62\) days c) \(6\) half-lives, or \(22.8\) days

52834612

A pollutant in a treatment tank decays exponentially. Measurements show that its concentration decreases by \(12\%\) each hour. a) Find the hourly decay factor \(a\). b) What percent of the original amount remains after \(5\) hours? c) Find the half-life of the pollutant concentration. d) What hourly percent decrease \(p\) would produce a half-life of exactly \(4\) hours?

Hints

- Convert the percent decrease to a multiplier. - Raise the hourly factor to the number of hours. - At the half-life, \(50\%\) remains. - In part d, find the factor whose fourth power is \(0.5\).

Solution

1. a) The hourly decay factor is \(a=1-0.12=0.88\). 2. b) After \(5\) hours, the remaining fraction is \(0.88^5\approx0.5277\), or about \(52.8\%\). 3. c) Solve \(0.88^{T_H}=0.5\): \(T_H=\frac{\ln(0.5)}{\ln(0.88)}\approx5.42\) hours. 4. d) A \(4\)-hour half-life requires \(a^4=0.5\), so \(a=0.5^{\frac{1}{4}}\approx0.8409\). 5. Therefore, \(p=(1-0.8409)\cdot100\%\approx15.9\%\).

Answer

a) \(a=0.88\) b) About \(52.8\%\) c) About \(5.42\) hours d) About \(15.9\%\) per hour

52836112

A radioactive sample has a half-life of \(5\) days. Its initial mass is \(640\,\text{mg}\). a) Find the mass remaining after \(2\), \(3\), and \(5\) half-life intervals. Make a table and give each mass in both milligrams and grams. b) Estimate the mass remaining after \(10\) half-life intervals, or \(50\) days. Use \(\left(\frac{1}{2}\right)^{10}\approx\frac{1}{1000}\).

Hints

- Halving during each interval means repeatedly multiplying by the same factor. - Convert milligrams to grams by dividing by \(1000\). - Write a formula for the mass after \(n\) half-life intervals. - For the estimate, use the approximation supplied in the problem.

Solution

1. The mass is multiplied by \(0.5\) during each half-life interval, so \(M(n)=640(0.5)^n\) milligrams. 2. After \(2\) intervals, \(M(2)=640\cdot(0.5)^2=160\,\text{mg}=0.16\,\text{g}\). 3. After \(3\) intervals, \(M(3)=640\cdot(0.5)^3=80\,\text{mg}=0.08\,\text{g}\). 4. After \(5\) intervals, \(M(5)=640\cdot(0.5)^5=20\,\text{mg}=0.02\,\text{g}\). 5. Using the given estimate, \(M(10)\approx640\cdot\frac{1}{1000}=0.64\,\text{mg}\).

Answer

a) <table> <tr> <th>Half-life intervals \(n\)</th> <th>Mass in mg</th> <th>Mass in g</th> </tr> <tr> <td>2</td> <td>\(160\)</td> <td>\(0.16\)</td> </tr> <tr> <td>3</td> <td>\(80\)</td> <td>\(0.08\)</td> </tr> <tr> <td>5</td> <td>\(20\)</td> <td>\(0.02\)</td> </tr> </table> b) About \(0.64\,\text{mg}\)

52848712

A laboratory begins with \(80\,\text{mg}\) of iodine-131. After exactly \(16\) days, \(20\,\text{mg}\) remains. a) Write a function of the form \(m(t)=m_0a^t\) that gives the mass \(m\), in milligrams, after \(t\) days. b) Find the mass remaining after \(24\) days.

Hints

- Identify the value at \(t=0\). - Use the known mass after \(16\) days to solve for the daily factor. - Notice how many halvings reduce \(80\) to \(20\). - Substitute \(t=24\) into the completed model.

Solution

1. The initial value is \(m_0=80\). 2. Use the point \((16,20)\): \(80a^{16}=20\), so \(a^{16}=0.25\) and \(a=0.25^{1/16}=0.5^{1/8}\approx0.9170\). 3. An exact model is \(m(t)=80(0.5)^{t/8}\). In the requested form, this is \(m(t)=80\left(0.5^{1/8}\right)^t\), or approximately \(m(t)=80(0.9170)^t\). 4. \(m(24)=80\cdot(0.5)^{24/8}=80\cdot(0.5)^3=10\,\text{mg}\).

Answer

a) \(m(t)=80\left(0.5^{1/8}\right)^t\), or approximately \(m(t)=80(0.9170)^t\) b) \(10\,\text{mg}\)

52849012

A forest currently contains \(25{,}000\,\text{m}^3\) of timber and grows by \(2.8\%\) each year. a) Find the timber volume after \(10\) years and after \(20\) years. b) Find the doubling time. c) After how many years will the original timber volume have tripled?

Hints

- Convert the annual percent increase to a growth factor. - Doubling means the model is twice its initial value. - The initial value cancels when finding a growth multiple time. - Use logarithms to solve for the time in the exponent.

Solution

1. The growth factor is \(1.028\), so \(W(t)=25{,}000\cdot1.028^t\). 2. a) \(W(10)\approx32{,}951.19\,\text{m}^3\) and \(W(20)\approx43{,}431.25\,\text{m}^3\). 3. b) For doubling, solve \(1.028^t=2\): \(t=\frac{\ln(2)}{\ln(1.028)}\approx25.10\) years. 4. c) For tripling, solve \(1.028^t=3\): \(t=\frac{\ln(3)}{\ln(1.028)}\approx39.78\) years.

Answer

a) After \(10\) years: about \(32{,}951.19\,\text{m}^3\); after \(20\) years: about \(43{,}431.25\,\text{m}^3\) b) About \(25.10\) years c) About \(39.78\) years

52852912

A patient takes a tablet containing \(400\,\text{mg}\) of medication. The amount in the bloodstream decreases by \(15\%\) each hour. a) Create a table showing the amount remaining after \(0\), \(1\), \(2\), and \(3\) hours. b) Find the medication's half-life. c) The medication is considered mostly eliminated when less than \(50\,\text{mg}\) remains. After how many complete hours does this first occur?

Hints

- Convert the hourly percent decrease to a decay factor. - Use the same factor repeatedly to fill the table. - At the half-life, one half of the initial amount remains. - Use logarithms to find the boundary time, then check whole hours.

Solution

1. The decay model is \(N(t)=400(0.85)^t\). 2. a) \(N(0)=400\), \(N(1)=340\), \(N(2)=289\), and \(N(3)=245.65\), all in milligrams. 3. b) For the half-life, solve \(0.85^t=0.5\): \(t=\frac{\ln(0.5)}{\ln(0.85)}\approx4.27\) hours. 4. c) Find the boundary by solving \(400(0.85)^t=50\), or \(0.85^t=0.125\). Then \(t\approx12.80\) hours. 5. After \(12\) hours, about \(56.90\,\text{mg}\) remains; after \(13\) hours, about \(48.36\,\text{mg}\) remains. Therefore, the amount first falls below \(50\,\text{mg}\) after \(13\) complete hours.

Answer

a) <table> <tr><td>Time (hours)</td><td>0</td><td>1</td><td>2</td><td>3</td></tr> <tr><td>Amount (mg)</td><td>400</td><td>340</td><td>289</td><td>245.65</td></tr> </table> b) About \(4.27\) hours c) After \(13\) complete hours

52853012

An algae population in a pond grows by \(18\%\) per day under ideal conditions. At the start of the observation, the algae cover \(2500\,\text{cm}^2\). a) Find the doubling time of the covered area. b) What area is covered after \(12\) days? c) After how many days will the covered area exceed \(2\,\text{m}^2\)?

Hints

- Convert the daily percent increase to a growth factor. - Doubling means the area is twice its initial value. - Convert square meters to square centimeters before comparing areas. - Use logarithms to solve for time.

Solution

1. The growth factor is \(1.18\), so \(A(t)=2500\cdot1.18^t\), where \(A\) is measured in \(\text{cm}^2\). 2. a) Solve \(1.18^t=2\): \(t=\frac{\ln(2)}{\ln(1.18)}\approx4.19\) days. 3. b) \(A(12)=2500\cdot1.18^{12}\approx18{,}218.98\,\text{cm}^2\). 4. c) Since \(2\,\text{m}^2=20{,}000\,\text{cm}^2\), solve \(2500\cdot1.18^t=20{,}000\). Then \(1.18^t=8\), so \(t=\frac{\ln(8)}{\ln(1.18)}\approx12.56\) days. 5. The covered area exceeds \(2\,\text{m}^2\) for times greater than about \(12.56\) days.

Answer

a) About \(4.19\) days b) About \(18{,}218.98\,\text{cm}^2\) c) After about \(12.6\) days

52853112

Cesium-137, \(^{137}\text{Cs}\), has a half-life of \(30.17\) years. a) Find the annual decay factor \(b\) and the annual percent decrease. Round the factor to four decimal places and the percent to the nearest hundredth. b) A soil sample has an initial activity of \(600\,\text{Bq}\). Find its activity after \(50\) years. Round to the nearest hundredth of a becquerel.

Hints

- A half-life gives the total decay factor over \(30.17\) years. - Relate the annual decay factor to the annual percent decrease. - Use an exponential decay model. - Express \(50\) years as a fraction of one half-life.

Solution

1. The annual decay factor satisfies \(b^{30.17}=0.5\), so \(b=0.5^{1/30.17}\approx0.977287\). 2. The annual percent decrease is \((1-b)\cdot100\%\approx2.27\%\). 3. The activity after \(50\) years is \(A(50)=600\cdot(0.5)^{50/30.17}\approx190.22\,\text{Bq}\).

Answer

a) \(b\approx0.9773\); annual decrease about \(2.27\%\) b) About \(190.22\,\text{Bq}\)

52853212

Iodine-131 (\(^{131}\text{I}\)) has a half-life of \(8.02\) days. a) Find its daily percent decay rate. b) After how many days will only \(10\%\) of the original amount remain? Round to the nearest tenth.

Hints

- Relate the daily decay factor to what happens after one half-life. - Express \(10\%\) as a decimal fraction of the initial amount. - Use logarithms to solve for an unknown time. - You may write the model as \(N(t)=N_0b^t\).

Solution

1. a) If \(b\) is the daily decay factor, then \(b^{8.02}=0.5\). 2. Thus, \(b=0.5^{\frac{1}{8.02}}\approx0.9172\). 3. The daily percent decay rate is \((1-0.9172)\cdot100\%\approx8.28\%\). 4. b) Solve \(0.5^{\frac{t}{8.02}}=0.1\). 5. Therefore, \(t=8.02\cdot\frac{\log(0.1)}{\log(0.5)}\approx26.64\) days, which rounds to \(26.6\) days.

Answer

a) About \(8.28\%\) per day b) About \(26.6\) days

52853712

A laboratory studies a \(400\,\text{mg}\) sample of strontium-90 that decreases by \(2.4\%\) each year. a) Find the mass after \(15\) years. b) By what percent does the mass decrease over those \(15\) years? c) Find the half-life.

Hints

- Convert the annual percent decrease to a decay factor. - Write an exponential decay model. - Compare the remaining fraction with \(1\) to find the percent decrease. - At the half-life, one half of the original mass remains. - Use logarithms to solve for time.

Solution

1. The yearly decay factor is \(0.976\), so \(N(t)=400\cdot0.976^t\). 2. a) \(N(15)=400\cdot0.976^{15}\approx277.85\,\text{mg}\). 3. b) The fraction lost is \(1-0.976^{15}\approx0.30538\), so the total decrease is about \(30.54\%\). 4. c) Solve \(0.976^T=0.5\): \(T=\frac{\ln(0.5)}{\ln(0.976)}\approx28.53\) years.

Answer

a) About \(277.85\,\text{mg}\) b) About \(30.54\%\) c) About \(28.5\) years

52853812

Carbon-14 has a half-life of \(5730\) years and can be used to date archaeological samples. In the same measurement setup, an excavated bone sample has carbon-14 activity of \(3.5\) relative units for every \(12.0\) relative units measured in comparable modern material. a) Find the annual decay factor \(b\). b) Estimate the age of the bone sample.

Hints

- Relate the annual decay factor to the half-life. - Compare the sample's activity with the modern-material activity as a ratio. - Use logarithms to isolate the time exponent. - Keep enough decimal places in the decay factor, or use its exact expression, to avoid a large rounding error.

Solution

1. The annual decay factor satisfies \(b^{5730}=0.5\). 2. Therefore, \(b=0.5^{\frac{1}{5730}}\approx0.999879\). 3. The activity ratio is \(\frac{3.5}{12.0}\), so \(\frac{3.5}{12.0}=b^t\). 4. To avoid rounding error, substitute \(b=0.5^{\frac{1}{5730}}\): \(t=5730\cdot\frac{\ln\left(\frac{3.5}{12.0}\right)}{\ln(0.5)}\approx10{,}185.69\). 5. The bone sample is about \(10{,}186\) years old.

Answer

a) \(b\approx0.999879\) b) About \(10{,}186\) years old

52982612

A radioactive isotope used in medical imaging has an activity of \(400\,\text{MBq}\) at the time of injection (\(t=0\)). After exactly \(4\) hours, its activity is \(240\,\text{MBq}\). Assume an exponential decay model \(A(t)=A_0e^{kt}\), where \(t\) is measured in hours. a) Find \(k\). b) Find the half-life of the isotope. c) The procedure may end when the activity has fallen to \(5\%\) of its initial value. Find how many hours after injection this occurs.

Hints

- Use the activity after \(4\) hours to solve for \(k\). - At the half-life, the exponential factor equals \(0.5\). - For part c), set the ratio \(A(t)/A_0\) equal to \(0.05\).

Solution

1. Using \(A(4)=240\), \(240=400e^{4k}\), so \(e^{4k}=0.6\). Therefore, \(k=\frac{\ln(0.6)}{4}\approx -0.1277\). 2. For the half-life \(T_H\), solve \(e^{kT_H}=0.5\). Thus, \(T_H=\frac{\ln(0.5)}{k}\approx 5.43\) hours. 3. For \(5\%\) remaining, solve \(e^{kt}=0.05\). Therefore, \(t=\frac{\ln(0.05)}{k}\approx 23.46\) hours.

Answer

a) \(k=\frac{\ln(0.6)}{4}\approx -0.1277\) b) About \(5.43\,\text{hours}\) c) About \(23.46\,\text{hours}\)

52982812

A medical contrast agent has a half-life of \(3\) hours in a patient's bloodstream. The patient receives \(120\,\text{mg}\) of the agent. 1. Write a function \(M(t)\) for the amount remaining after \(t\) hours. 2. Find the amount remaining after \(10\) hours. 3. Find when only \(10\%\) of the initial amount remains.

Hints

- Express the number of half-lives as \(t/3\). - Ten percent remaining means the ratio \(M(t)/M(0)\) is \(0.10\). - Use logarithms to solve for time.

Solution

1. With a half-life of \(3\) hours, the model is \(M(t)=120(0.5)^{t/3}\). 2. \(M(10)=120(0.5)^{10/3}\approx 11.91\,\text{mg}\). 3. Set \((0.5)^{t/3}=0.10\). Taking logarithms gives \(t=\frac{3\ln(0.10)}{\ln(0.5)}\approx 9.97\) hours.

Answer

1. \(M(t)=120(0.5)^{t/3}\) 2. About \(11.91\,\text{mg}\) 3. About \(9.97\,\text{hours}\)

52983412

An invasive aquatic plant spreads across a lake. a) The covered area doubles every \(5\) days and is initially \(1\,\text{m}^2\). Write a function \(A(t)\) for the area after \(t\) days. b) Use \(2^{10}\approx 10^3\) to estimate the area predicted by the unlimited model after \(50\) days and after \(100\) days. c) The lake has an area of \(1\,\text{km}^2\). Estimate when half the lake, or \(500{,}000\,\text{m}^2\), is covered. Use \(500\approx 2^9\). d) Use logarithms to find the exact model time when half the lake is covered, and compare it with the estimate from part c).

Hints

- Express the number of doubling periods as \(t/5\). - Convert square kilometers to square meters. - Decompose \(500{,}000\) using the given power-of-two approximations. - Use logarithms to solve \(2^x=y\).

Solution

1. Since the area doubles every \(5\) days, \(A(t)=2^{t/5}\). 2. \(A(50)=2^{10}\approx 1000\,\text{m}^2\). Also, \(A(100)=2^{20}=(2^{10})^2\approx 1{,}000{,}000\,\text{m}^2\). The second value is only a prediction of the unlimited model. 3. \(500{,}000=500\cdot 1000\approx 2^9\cdot 2^{10}=2^{19}\). Thus, \(t/5\approx 19\), so \(t\approx 95\) days. 4. Exactly, \(2^{t/5}=500{,}000\), so \(t=\frac{5\ln(500000)}{\ln 2}\approx 94.66\) days. The estimate is about \(0.34\) day greater than the logarithmic result.

Answer

a) \(A(t)=2^{t/5}\) b) \(A(50)\approx 1000\,\text{m}^2\); \(A(100)\approx 1{,}000{,}000\,\text{m}^2\) c) About \(95\,\text{days}\) d) About \(94.66\,\text{days}\); difference: about \(0.34\) day

52983912

A bacterial culture doubles every \(3\) hours. Its population is modeled by \(N(t)=N_0b^t\), where \(t\) is measured in hours. a) Find \(b\), rounded to four decimal places. b) By what factor does the population change in exactly one hour? c) By what factor does it change in \(4.5\) hours? d) Find the percent increase during \(30\) minutes.

Hints

- Translate the doubling time into an equation for \(b\). - Use time in hours as the exponent. - Convert a growth factor to a percent increase by subtracting \(1\).

Solution

1. The doubling condition gives \(b^3=2\), so \(b=2^{1/3}\approx1.2599\). 2. The one-hour factor is \(b\approx1.2599\). 3. The \(4.5\)-hour factor is \(b^{4.5}=2^{4.5/3}=2^{1.5}=\sqrt8\approx2.8284\). 4. Thirty minutes is \(0.5\) hour. The factor is \(b^{0.5}=2^{1/6}\approx1.1225\), so the percent increase is about \(12.25\%\).

Answer

a) \(b\approx1.2599\) b) About \(1.2599\) c) About \(2.8284\) d) About \(12.25\%\)

52984612

The concentration of a medication in the blood is modeled by an exponential function. The table gives the concentration \(y\), in \(\text{mg/L}\), after \(t\) hours. <table> <tr><td>\(t\)</td><td>0</td><td>2</td><td>5</td></tr> <tr><td>\(y\)</td><td>\(12.0\)</td><td>\(7.68\)</td><td>\(3.93216\)</td></tr> </table> a) Find a function of the form \(f(t)=ca^t\). b) Find the percent decrease per hour. c) Find the half-life of the medication.

Hints

- Use the value at \(t=0\) to identify \(c\). - Use another table value to solve for \(a\). - Interpret \(a\) as the percent remaining each hour. - At the half-life, the exponential factor equals \(0.5\).

Solution

1. Since \(f(0)=12.0\), \(c=12\). Using \(f(2)=7.68\), \(12a^2=7.68\), so \(a^2=0.64\) and \(a=0.8\). Thus, \(f(t)=12(0.8)^t\). The value at \(t=5\) confirms the model. 2. The factor \(0.8\) means \(80\%\) remains each hour, so the concentration decreases by \(20\%\) per hour. 3. For the half-life \(T\), solve \((0.8)^T=0.5\). Therefore, \(T=\frac{\ln(0.5)}{\ln(0.8)}\approx 3.11\) hours.

Answer

a) \(f(t)=12(0.8)^t\) b) \(20\%\) per hour c) About \(3.11\,\text{hours}\)

52993912

The concentration of a medication in the blood is modeled by \(C(t)=C_0e^{-kt}\), where \(C_0=250\,\text{mg/L}\), \(k=0.085\,\text{h}^{-1}\), and \(t\) is measured in hours. a) Find when the concentration decreases to \(100\,\text{mg/L}\). b) Find the half-life of the medication. c) Find the percent decrease per hour. Compare this percent with the numerical value of \(k\).

Hints

- Use logarithms to solve for time. - At the half-life, the concentration ratio is \(0.5\). - The one-hour remaining factor is \(e^{-k}\). - Distinguish a continuous rate constant from a discrete percent change.

Solution

1. Set \(C(t)=100\): \(100=250e^{-0.085t}\). Thus, \(e^{-0.085t}=0.4\), so \(t=\frac{\ln(0.4)}{-0.085}\approx 10.78\) hours. 2. The half-life satisfies \(e^{-0.085T_H}=0.5\). Therefore, \(T_H=\frac{\ln(0.5)}{-0.085}\approx 8.15\) hours. 3. The one-hour factor is \(e^{-0.085}\approx 0.9185\). Therefore, the hourly decrease is about \(8.15\%\). This is slightly less than \(8.5\%\); the constant \(k=0.085\,\text{h}^{-1}\) is a continuous decay rate, not the discrete hourly percent decrease.

Answer

a) About \(10.78\,\text{hours}\) b) About \(8.15\,\text{hours}\) c) About \(8.15\%\) per hour, slightly less than \(8.5\%\)

52994012

A threatened animal population in a reserve is modeled by \(B(t)=B_0e^{a(t-t_0)}\). In 2015, the reserve had \(1200\) animals, and the continuous growth rate is \(a=0.045\,\text{yr}^{-1}\). a) Find the model's predicted population in 2030. b) Determine during which year the population reaches \(2400\). c) Find the effective annual percent growth rate and explain why it is greater than \(4.5\%\).

Hints

- Find the elapsed time from 2015 to the target year. - A doubled population makes the exponential factor equal to \(2\). - The effective one-year growth factor is \(e^a\).

Solution

1. From 2015 to 2030 is \(15\) years. Thus, \(B(2030)=1200e^{0.045(15)}\approx 2356.84\), or about \(2357\) animals. 2. To double, solve \(e^{0.045\Delta t}=2\). Then \(\Delta t=\frac{\ln 2}{0.045}\approx 15.40\) years, so the population reaches \(2400\) during 2030. 3. The one-year factor is \(e^{0.045}\approx 1.0460\), giving an effective annual growth rate of about \(4.60\%\). It exceeds \(4.5\%\) because \(e^x>1+x\) for \(x>0\).

Answer

a) About \(2357\) animals b) During 2030 c) About \(4.60\%\) per year

52994412

Tritium is a radioactive isotope of hydrogen with a half-life of \(12.32\) years. Because small amounts occur in rainwater, tritium can be used to estimate the age of wine. a) Write a model of the form \(N(t)=N_0a^t\) for the amount remaining after \(t\) years. Round \(a\) to four decimal places. b) Find the annual percent decrease in tritium. c) A wine sample contains \(22\%\) of its original tritium. Estimate the wine's age to the nearest year.

Hints

- Use the half-life to find the one-year remaining factor. - Convert the remaining factor to a percent decrease. - Write \(22\%\) as \(0.22\) and solve with logarithms.

Solution

1. The half-life gives \(a^{12.32}=0.5\). Therefore, \(a=(0.5)^{1/12.32}\approx 0.9453\), so \(N(t)=N_0(0.9453)^t\). 2. The annual decrease is \(1-0.9453\approx 0.0547\), or about \(5.47\%\). 3. Solve \((0.9453)^t=0.22\). Then \(t=\frac{\ln(0.22)}{\ln(0.9453)}\approx 26.91\), so the wine is about \(27\) years old.

Answer

a) \(N(t)=N_0(0.9453)^t\) b) About \(5.47\%\) per year c) About \(27\,\text{years}\)

53002512

A radioactive substance decays exponentially. After \(12\) days, exactly \(75\%\) of the initial mass remains. 1. Find the daily decay factor \(q\) and the decay constant \(\lambda\) for \(M(t)=M_0e^{-\lambda t}\), where \(t\) is measured in days. 2. Find the half-life of the substance. 3. Find when only \(10\%\) of the initial mass remains.

Hints

- Find the daily factor whose twelfth power is \(0.75\). - Relate the daily factor to the base-\(e\) form. - At the half-life, the remaining ratio is \(0.5\). - Use logarithms to solve for time.

Solution

1. Since \(q^{12}=0.75\), \(q=(0.75)^{1/12}\approx 0.9763\). Also, \(e^{-12\lambda}=0.75\), so \(\lambda=-\frac{\ln(0.75)}{12}\approx 0.02397\,\text{day}^{-1}\). 2. The half-life is \(T_H=\frac{\ln 2}{\lambda}\approx 28.91\) days. 3. For \(10\%\) remaining, solve \(e^{-\lambda t}=0.10\). Thus, \(t=\frac{-\ln(0.10)}{\lambda}\approx 96.05\) days.

Answer

1. \(q\approx 0.9763\); \(\lambda\approx 0.02397\,\text{day}^{-1}\) 2. \(T_H\approx 28.91\,\text{days}\) 3. About \(96.05\,\text{days}\)

53002612

Two bacterial cultures are being studied in a laboratory. Culture A doubles every \(5\) hours. Culture B doubles every \(8\) hours but begins with four times as many bacteria as Culture A. Find when the two cultures will have the same number of bacteria.

Hints

- Write an exponential model for each culture. - Use a variable for Culture A's unknown initial population. - Rewrite \(4\) as a power of \(2\). - Once both sides have the same base, compare the exponents.

Solution

1. Let Culture A begin with \(N_0\) bacteria. Then \(N_A(t)=N_0 2^{t/5}\) and \(N_B(t)=4N_0 2^{t/8}\). 2. Set the populations equal: \(N_0 2^{t/5}=4N_0 2^{t/8}\). Dividing by \(N_0\) gives \(2^{t/5}=4\cdot 2^{t/8}\). 3. Since \(4=2^2\), \(2^{t/5}=2^{2+t/8}\), so \(\frac{t}{5}=2+\frac{t}{8}\). 4. Solving gives \(\frac{3t}{40}=2\), so \(t=\frac{80}{3}=26\frac{2}{3}\) hours.

Answer

The cultures have the same population after \(26\frac{2}{3}\) hours, or about \(26.67\) hours.

53002912

Two animal populations are being monitored in a wildlife reserve. At the start of the study, Species A has about \(15{,}000\) animals and grows by \(2.1\%\) each year. Species B has \(8000\) animals and grows by \(4.5\%\) each year. a) Find the doubling time for Species B. b) After how many years will the population of Species A be only \(1.5\) times the population of Species B?

Hints

- Convert each percent increase to a growth factor. - A doubling time makes the final-to-initial ratio equal to \(2\). - For part b, translate “\(1.5\) times” into an equation relating the two models. - Use logarithms to solve an equation with the variable in an exponent.

Solution

1. The population models are \(A(t)=15{,}000(1.021)^t\) and \(B(t)=8000(1.045)^t\). 2. For Species B to double, solve \((1.045)^t=2\). Thus, \(t=\frac{\ln 2}{\ln(1.045)}\approx 15.75\) years. 3. For part b, set \(A(t)=1.5B(t)\): \(15{,}000(1.021)^t=1.5\cdot 8000(1.045)^t\). 4. This simplifies to \(1.25=\left(\frac{1.045}{1.021}\right)^t\). Therefore, \(t=\frac{\ln(1.25)}{\ln(1.045)-\ln(1.021)}\approx 9.60\) years.

Answer

a) About \(15.75\) years b) About \(9.60\) years

53003212

A technology company tracks registered users of a new app. <table style="width: 100%; border-collapse: collapse;"> <thead> <tr style="border-bottom: 1px solid black;"> <th style="text-align: left; padding: 5px;">Year</th> <th style="text-align: left; padding: 5px;">Registered users</th> </tr> </thead> <tbody> <tr><td style="padding: 5px;">2016</td><td style="padding: 5px;">\(120{,}000\)</td></tr> <tr><td style="padding: 5px;">2018</td><td style="padding: 5px;">\(240{,}000\)</td></tr> <tr><td style="padding: 5px;">2021</td><td style="padding: 5px;">\(480{,}000\)</td></tr> <tr><td style="padding: 5px;">2025</td><td style="padding: 5px;">\(960{,}000\)</td></tr> </tbody> </table> a) Find each doubling interval. b) Find the average annual growth rate during each doubling interval. c) Interpret how the changing doubling intervals reflect the app's growth dynamics.

Hints

- Compare the years associated with consecutive doubled values. - For doubling time \(T\), the annual factor satisfies \(q^T=2\). - Longer doubling times correspond to smaller annual percent growth rates.

Solution

1. The user count doubles from 2016 to 2018 in \(2\) years, from 2018 to 2021 in \(3\) years, and from 2021 to 2025 in \(4\) years. 2. The yearly factors are \(2^{1/2}\approx1.4142\), \(2^{1/3}\approx1.2599\), and \(2^{1/4}\approx1.1892\). The corresponding annual growth rates are about \(41.42\%\), \(25.99\%\), and \(18.92\%\). 3. The doubling intervals are getting longer, so the relative annual growth rate is decreasing even though the total number of users continues to increase.

Answer

a) \(2\) years, \(3\) years, and \(4\) years b) About \(41.42\%\), \(25.99\%\), and \(18.92\%\) per year c) The relative growth rate is decreasing because each doubling takes longer

53003412

A threatened animal population in a wildlife reserve is assumed to grow exponentially. At the start of the study, there are \(200\) animals. After \(5\) years, there are \(350\). 1. Find the annual growth factor \(q\) and annual percent growth rate. 2. Find how long it takes the initial population to quadruple. 3. An equivalent continuous model has the form \(B(t)=B_0e^{kt}\). Find \(k\), and use the model to predict the population after \(10\) years.

Hints

- Use the two population values to find the annual factor. - Quadrupling means the final-to-initial ratio is \(4\). - Use logarithms to solve for time and for the continuous growth constant. - Relate the five-year factor to the ten-year factor.

Solution

1. From \(350=200q^5\), \(q=(1.75)^{1/5}\approx1.1184\). The annual growth rate is about \(11.84\%\). 2. For quadrupling, solve \(q^t=4\). Thus, \(t=\frac{\ln4}{\ln q}\approx12.39\) years. 3. From \(200e^{5k}=350\), \(k=\frac{\ln1.75}{5}\approx0.1119\,\text{yr}^{-1}\). Then \(B(10)=200e^{10k}=200(1.75)^2=612.5\), or about \(613\) animals.

Answer

1. \(q\approx1.1184\); about \(11.84\%\) per year 2. About \(12.39\) years 3. \(k\approx0.1119\,\text{yr}^{-1}\); about \(613\) animals after \(10\) years

53003512

The amount of a medication in a patient's blood is modeled by \(M(t)=150(0.82)^t\), where \(M(t)\) is measured in milligrams and \(t\) is measured in hours. a) Rewrite the model as \(M(t)=150e^{kt}\). Round \(k\) to four decimal places. b) Find the medication's half-life. c) A second medication has an initial dose of \(300\,\text{mg}\) and decay constant \(k=-0.12\,\text{h}^{-1}\). Write its model in the form \(g(t)=be^{kt}\), and compare its graph with the graph of \(M\).

Hints

- Use \(k=\ln a\) to convert the base. - Half-life means the remaining factor is \(0.5\). - Compare both the initial values and the magnitudes of the negative decay constants.

Solution

1. Since \(e^k=0.82\), \(k=\ln(0.82)\approx-0.1985\,\text{h}^{-1}\). Thus, \(M(t)=150e^{\ln(0.82)t}\approx150e^{-0.1985t}\). 2. Solve \((0.82)^{t_H}=0.5\). Then \(t_H=\frac{\ln(0.5)}{\ln(0.82)}\approx3.49\) hours. 3. The second model is \(g(t)=300e^{-0.12t}\). It starts at twice the value of \(M\). Because its decay constant has smaller magnitude, it decreases more slowly as a percentage of its current amount. Also, \(\frac{g(t)}{M(t)}=2e^{(-0.12-\ln(0.82))t}>2\) for \(t>0\), so its graph remains above the graph of \(M\).

Answer

a) \(k\approx-0.1985\,\text{h}^{-1}\) b) About \(3.49\) hours c) \(g(t)=300e^{-0.12t}\); it starts higher, remains above \(M\), and decreases more slowly proportionally

53218412

After a patient takes a medication, the amount of the active ingredient in the body decreases continuously. The graph shows the remaining amount \(W(t)\), in milligrams, after \(t\) hours. a) Use the graph to determine when the initial amount has been reduced to one-half, one-fourth, and one-eighth. What pattern do you notice? b) State the medication's half-life in the body. c) Find the percent decrease per hour. Round to the nearest tenth of a percent. d) Calculate the amount of medication remaining after exactly \(15\) hours.

Figure for problem 532184

Hints

- Read the initial value at \(t=0\). - Locate one-half, one-fourth, and one-eighth of the initial value on the vertical axis. - Compare the time intervals between successive halvings. - Relate the hourly factor to a half-life of \(3\) hours. - Determine how many half-lives pass in \(15\) hours.

Solution

1. The graph starts at \(80\,\text{mg}\). It reaches \(40\,\text{mg}\) after \(3\) hours, \(20\,\text{mg}\) after \(6\) hours, and \(10\,\text{mg}\) after \(9\) hours. Each additional halving takes \(3\) hours. 2. Therefore, the half-life is \(3\) hours. 3. If \(b\) is the hourly decay factor, then \(b^3=0.5\). Thus \(b=0.5^{1/3}\approx0.7937\), so the hourly percent decrease is \((1-b)\cdot100\%\approx20.6\%\). 4. Fifteen hours is \(5\) half-lives. Therefore, \(W(15)=80\cdot(0.5)^5=2.5\,\text{mg}\).

Answer

a) One-half after \(3\) hours, one-fourth after \(6\) hours, and one-eighth after \(9\) hours; each halving takes \(3\) hours. b) \(3\) hours c) About \(20.6\%\) per hour d) \(2.5\,\text{mg}\)

53245712

The graph shows the concentration \(C(t)\) of a medication in a patient's blood, in milligrams per liter, \(\text{mg/L}\). Time \(t\) is measured in hours after the dose. a) Use the marked points to find an exponential model of the form \(C(t)=C_0a^t\). Round \(a\) to four decimal places. Also state the medication's half-life. b) The patient may take another dose only after the concentration falls below \(0.1\,\text{mg/L}\). Find when this occurs. Round to the nearest tenth of an hour.

Figure for problem 532457

Hints

- Read the initial value at \(t=0\). - Compare the marked concentrations at equal time intervals. - Use one marked point to solve for the base \(a\). - Use logarithms to solve for the time in part b.

Solution

1. The point \((0, 8)\) gives \(C_0=8\). 2. The marked values are halved every \(3\) hours: \(8\) at \(t=0\), \(4\) at \(t=3\), and \(2\) at \(t=6\). Therefore, the half-life is \(3\) hours. 3. Using \((3, 4)\), \(4=8a^3\), so \(a^3=0.5\) and \(a=(0.5)^{1/3}\approx 0.7937\). Thus, \(C(t)=8(0.7937)^t\). 4. Solve \(8(0.7937)^t=0.1\). Then \(t=\frac{\ln(0.0125)}{\ln(0.7937)}\approx 19.0\) hours. The threshold occurs at about \(19.0\) hours, so the concentration is below the limit just after that time.

Answer

a) \(C(t)=8(0.7937)^t\); half-life: \(3\) hours b) Just after about \(19.0\) hours

53262612

The graph shows an exponential growth process \(f\) and an exponential decay process \(g\). Time \(t\) is measured in hours. a) Find the initial amount for each process. b) Use the graph to find the doubling time of \(f\) and the half-life of \(g\). c) Write an equation of the form \(N(t)=N_0b^t\) for each process. d) Find each amount after \(15\) hours. Round the value for \(g\) to the nearest tenth.

Figure for problem 532626

Hints

- The initial amount is the value when \(t=0\). - Locate twice the initial value of \(f\) and one-half the initial value of \(g\). - Use the initial value and the characteristic time to determine each hourly factor. - Substitute \(t=15\) into both equations.

Solution

1. The y-intercepts give \(f(0)=10\) and \(g(0)=80\). 2. The graph shows that \(f\) grows from \(10\) to \(20\) in \(3\) hours, so its doubling time is \(3\) hours. It shows that \(g\) decreases from \(80\) to \(40\) in \(4\) hours, so its half-life is \(4\) hours. 3. For \(f\), the hourly factor is \(2^{1/3}\), so \(f(t)=10\left(2^{1/3}\right)^t=10(2)^{t/3}\). 4. For \(g\), the hourly factor is \(0.5^{1/4}\), so \(g(t)=80\left(0.5^{1/4}\right)^t=80(0.5)^{t/4}\). 5. \(f(15)=10\cdot2^{15/3}=320\), and \(g(15)=80\cdot(0.5)^{15/4}\approx5.9\).

Answer

a) \(f(0)=10\); \(g(0)=80\) b) Doubling time of \(f\): \(3\) hours; half-life of \(g\): \(4\) hours c) \(f(t)=10(2)^{t/3}\); \(g(t)=80(0.5)^{t/4}\) d) \(f(15)=320\); \(g(15)\approx5.9\)

53402812

Iodine-131 is used in some thyroid treatments. The graph shows the decay of a sample with an initial mass of \(10\,\text{mg}\). a) Determine the half-life \(T_{1/2}\) from the graph. b) Write the model in the form \(m(t)=m_0\cdot 0.5^{t/T_{1/2}}\). c) Calculate the mass remaining after \(20\) days. Round to the nearest hundredth of a milligram.

Figure for problem 534028

Hints

- Find the time when the graph reaches half the initial mass. - Use the units shown on the axes. - Substitute \(t=20\) into the model for part c).

Solution

1. Half the initial mass is \(5\,\text{mg}\). The graph reaches \(5\,\text{mg}\) at \(t=8\) days, so \(T_{1/2}=8\) days. 2. Substituting \(m_0=10\) and \(T_{1/2}=8\) gives \(m(t)=10\cdot 0.5^{t/8}\). 3. After \(20\) days, \(m(20)=10\cdot 0.5^{20/8}\approx 1.77\,\text{mg}\).

Answer

a) \(T_{1/2}=8\) days b) \(m(t)=10\cdot 0.5^{t/8}\) c) Approximately \(1.77\,\text{mg}\)

52594812

A biologist observes two bacterial cultures. Culture A starts with \(100\) bacteria and doubles every \(20\) minutes. Culture B starts with \(100\) bacteria and triples every \(30\) minutes. Determine which culture reaches \(1{,}000{,}000\) bacteria first. Find the difference, in minutes, between the two times. Round to the nearest hundredth.

Hints

- Write an exponential growth model for each culture. - Express the number of doubling or tripling periods as a fraction of the unknown time. - Set each model equal to the target population and solve for time. - Use logarithms to isolate a variable in an exponent.

Solution

1. Let \(t\) be time in minutes. The models are \(N_A(t)=100\cdot2^{\frac{t}{20}}\) and \(N_B(t)=100\cdot3^{\frac{t}{30}}\). 2. For Culture A, solve \(100\cdot2^{\frac{t}{20}}=1{,}000{,}000\). Then \(2^{\frac{t}{20}}=10{,}000\), so \(t_A=20\cdot\frac{\ln(10{,}000)}{\ln(2)}\approx265.75\) minutes. 3. For Culture B, solve \(100\cdot3^{\frac{t}{30}}=1{,}000{,}000\). Then \(3^{\frac{t}{30}}=10{,}000\), so \(t_B=30\cdot\frac{\ln(10{,}000)}{\ln(3)}\approx251.51\) minutes. 4. Culture B reaches the target first. The time difference is \(265.75-251.51\approx14.25\) minutes.

Answer

Culture B reaches \(1{,}000{,}000\) bacteria first, about \(14.25\) minutes before Culture A.

52834512

A radioactive isotope decays so that after \(500\) years, \(25\%\) of the original mass has decayed. a) Find the decay factor \(q\) for a \(500\)-year period. b) Find the annual decay factor \(a\) to four decimal places. c) Find the half-life to the nearest year. d) After how many years will only \(1\%\) of the original mass remain?

Hints

- First determine what percent of the mass remains after \(500\) years. - Relate the long-term decay factor to the annual factor using a power. - At the half-life, one half of the initial mass remains. - Use logarithms to solve when the unknown is in the exponent.

Solution

1. a) After \(500\) years, \(75\%\) remains, so \(q=0.75\). 2. b) The annual factor satisfies \(a^{500}=0.75\), so \(a=0.75^{\frac{1}{500}}\approx0.9994\). 3. c) Solve \(0.75^{\frac{T_H}{500}}=0.5\). Then \(T_H=500\cdot\frac{\ln(0.5)}{\ln(0.75)}\approx1204.71\), so the half-life is about \(1205\) years. 4. d) Solve \(0.75^{\frac{t}{500}}=0.01\). Then \(t=500\cdot\frac{\ln(0.01)}{\ln(0.75)}\approx8003.92\), or about \(8004\) years.

Answer

a) \(q=0.75\) b) \(a\approx0.9994\) c) About \(1205\) years d) About \(8004\) years

52835812

In a lake, light intensity decreases exponentially with water depth. The intensity decreases by \(22\%\) for each meter of depth. Let the intensity at the surface be \(100\%\). a) What percent of the surface light intensity remains at depths of \(2\,\text{m}\), \(5\,\text{m}\), and \(10\,\text{m}\)? b) At what depth is the light intensity one-half of the surface intensity? c) The aphotic zone begins where the light intensity falls below \(1\%\) of the surface intensity. Find the depth at which the intensity is \(1\%\).

Hints

- Convert the percent decrease to a decay factor. - Use an exponential model with depth as the input. - To find a depth for a specified intensity, solve an exponential equation. - You may model the surface intensity as either \(100\) or \(1\), provided you use the corresponding target value.

Solution

1. The decay factor per meter is \(1-0.22=0.78\), so the model is \(I(d)=100(0.78)^d\), where \(d\) is depth in meters and \(I(d)\) is a percent. 2. \(I(2)=100\cdot(0.78)^2=60.84\%\), \(I(5)=100\cdot(0.78)^5\approx28.87\%\), and \(I(10)=100\cdot(0.78)^{10}\approx8.34\%\). 3. For half the surface intensity, solve \(0.78^d=0.5\): \(d=\frac{\ln(0.5)}{\ln(0.78)}\approx2.79\,\text{m}\). 4. For \(1\%\) intensity, solve \(0.78^d=0.01\): \(d=\frac{\ln(0.01)}{\ln(0.78)}\approx18.53\,\text{m}\).

Answer

a) At \(2\,\text{m}\): \(60.84\%\); at \(5\,\text{m}\): about \(28.87\%\); at \(10\,\text{m}\): about \(8.34\%\) b) About \(2.79\,\text{m}\) c) About \(18.53\,\text{m}\)

53003312

The Rule of 72 is a common estimate for the number of years \(n\) required for an investment to double at an annual interest rate of \(p\%\): \(n\approx \frac{72}{p}\). 1. An initial investment of \(\$5000\) earns \(3\%\) interest compounded annually. Find the exact doubling time and compare it with the Rule of 72 estimate. 2. For \(p=12\), find the percent error of the Rule of 72 estimate relative to the exact doubling time. 3. Show that the exact doubling time is \(n=\frac{\ln 2}{\ln\left(1+\frac{p}{100}\right)}\). Then use \(\ln(1+x)\approx x\) for small \(x\) to explain why \(np\approx 69.3\) provides a theoretical basis for doubling-time rules.

Hints

- Write the compound-interest model and set the final balance equal to twice the initial balance. - Use logarithms to solve for the exponent. - Percent error compares the absolute difference with the exact value. - For the approximation, substitute \(x=\frac{p}{100}\) into \(\ln(1+x)\approx x\).

Solution

1. The doubling equation is \(5000(1.03)^n=10{,}000\), so \((1.03)^n=2\). Therefore, \(n=\frac{\ln 2}{\ln(1.03)}\approx 23.45\) years. The Rule of 72 gives \(n\approx \frac{72}{3}=24\) years, a difference of about \(0.55\) year. 2. For \(p=12\), the exact doubling time is \(n=\frac{\ln 2}{\ln(1.12)}\approx 6.116\) years, while the estimate is \(\frac{72}{12}=6\) years. The percent error is \(\frac{|6-6.116|}{6.116}\cdot 100\%\approx 1.90\%\). 3. Starting with \(\left(1+\frac{p}{100}\right)^n=2\), take natural logarithms: \(n\ln\left(1+\frac{p}{100}\right)=\ln 2\). Solving for \(n\) gives \(n=\frac{\ln 2}{\ln\left(1+\frac{p}{100}\right)}\). For small \(p\), \(\ln\left(1+\frac{p}{100}\right)\approx \frac{p}{100}\), so \(n\approx \frac{100\ln 2}{p}\). Since \(100\ln 2\approx 69.3\), \(np\approx 69.3\).

Answer

1. Exact doubling time: about \(23.45\) years; Rule of 72 estimate: \(24\) years 2. About \(1.90\%\) 3. \(n=\frac{\ln 2}{\ln\left(1+\frac{p}{100}\right)}\), and for small \(p\), \(np\approx 69.3\)

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