Free math worksheets

- A base-10 logarithm returns the exponent on \(10\). - Keep track of the negative sign in the pH definition.

\(\text{pH}=-\log_{10}(10^{-4})=-(-4)=4\).

\(4\)

- First divide the sound-level equation by \(10\). - Rewrite the base-10 logarithmic statement in exponential form.

\(20=10\log_{10}\left(\frac{I}{I_0}\right)\), so \(2=\log_{10}\left(\frac{I}{I_0}\right)\). Therefore, \(\frac{I}{I_0}=10^2=100\).

\(\frac{I}{I_0}=100\)

- Read the numerical labels from left to right. - Compare consecutive differences and consecutive ratios. - Ask which comparison stays constant across equal visual intervals.

1. The equally spaced labeled positions represent \(1\), \(10\), \(100\), and \(1000\). 2. Their differences are not equal, so the axis is not linear in the displayed values. 3. Each value is \(10\) times the preceding value. Equal spatial intervals therefore represent equal multiplicative factors. 4. The axis is logarithmic, and one marked interval represents multiplication by \(10\).

The axis is logarithmic; one marked interval represents multiplication by \(10\).

- Relate the number of digits to consecutive powers of \(10\). - Evaluate common logarithms of powers such as \(10^3\) and \(10^4\). - Determine which consecutive powers of \(10\) bound a number with a given number of digits.

1. A positive integer with \(k\) digits satisfies \(10^{k-1} \le n < 10^k\). 2. Taking common logarithms gives \(k - 1 \le \log(n) < k\). 3. Therefore, all positive integers with the same number of digits have common logarithms with the same integer part: \(\lfloor \log(n) \rfloor = k - 1\). 4. A six-digit positive integer satisfies \(10^5 \le x < 10^6\). Thus, \(5 \le \log(x) < 6\), so \(\log(x) \in [5, 6)\).

a) Their common logarithms have the same integer part. b) \([5, 6)\)

- Compare both the additive changes and the multiplicative ratios of the inputs. - Evaluate the index at powers of \(10\). - For \(r=1000\), continue the pattern formed by successive tenfold changes.

1. \(S(1)=10\log_{10}(1)=0\), \(S(10)=10\), and \(S(100)=20\). Each tenfold increase in \(r\) raises \(S\) by \(10\). 2. On a linear horizontal axis, the first input change is \(10-1=9\), while the second is \(100-10=90\). 3. Even though the absolute input change is ten times larger in the second interval, both changes multiply the input by \(10\), so both produce the same additive increase of \(10\) in \(S\). This is logarithmic compression. 4. Since \(1000\) is another factor of \(10\) above \(100\), \(S(1000)=30\).

a) \(10\) units in each case b) The linear-axis changes are \(9\) and \(90\), but each is a tenfold multiplication of the input, so each raises the logarithmic index by \(10\). Large multiplicative changes are compressed into equal additive steps. c) \(S(1000)=30\)

- Read the two labeled values immediately to the left and right of \(p\). - Compare the position of \(p\) with the exponents of \(10\) represented by those labels. - A midpoint on a logarithmic axis is based on ratios rather than equal numerical differences.

1. From the image, \(p\) is halfway between the adjacent labeled values \(10\) and \(100\) on the logarithmic axis. 2. Their base-10 logarithms are \(1\) and \(2\), so the halfway logarithmic position is \(1.5\). 3. Therefore, \(p=10^{1.5}=10\sqrt{10}\approx31.6\). 4. The arithmetic midpoint \(55\) is halfway on a linear scale. A logarithmic midpoint corresponds to the geometric mean, not the arithmetic mean.

a) \(p=10\sqrt{10}\approx31.6\) b) A logarithmic midpoint represents equal multiplicative spacing, so it is the geometric mean rather than the arithmetic mean of the adjacent labeled values.

- Start with the difference between the two magnitudes. - Use the quotient property for a difference of logarithms. - Isolate the logarithm of the energy ratio, then exponentiate with base \(10\).

1. The magnitude difference is \(7.5-5.5=2\). 2. Subtract the two magnitude equations: \(\frac{2}{3}\left[\log_{10}\!\left(\frac{E_2}{E_0}\right)-\log_{10}\!\left(\frac{E_1}{E_0}\right)\right]=2\). 3. Apply the quotient property: \(\frac{2}{3}\log_{10}\!\left(\frac{E_2}{E_1}\right)=2\). 4. Therefore, \(\log_{10}\!\left(\frac{E_2}{E_1}\right)=3\), so \(\frac{E_2}{E_1}=10^3=1000\).

The magnitude \(7.5\) earthquake releases \(1000\) times as much energy.

- Substitute the given values and solve for \(k\). - Write the change as \(S(2I)-S(I)\) and use the quotient property. - A change independent of the starting value should contain no \(I\) after simplification. - Use the same difference method for multiplication by \(10\).

1. Substitute the data: \(50=k\ln\!\left(\frac{10^{-7}}{10^{-12}}\right)=k\ln(10^5)=5k\ln(10)\). Thus, \(k=\frac{10}{\ln(10)}\approx4.343\). 2. The change from \(I\) to \(2I\) is \(S(2I)-S(I)=k\left[\ln\!\left(\frac{2I}{I_0}\right)-\ln\!\left(\frac{I}{I_0}\right)\right]=k\ln(2)\). This expression does not depend on \(I\). 3. Multiplying intensity by \(10\) changes the perception strength by \(k\ln(10)\). Substituting \(k=\frac{10}{\ln(10)}\) gives a change of \(10\).

1. \(k=\frac{10}{\ln(10)}\approx4.343\) 2. The change is \(k\ln(2)\), independent of the starting intensity. 3. The change is \(10\).

- Compare how the sound level changes as the intensity increases from one table entry to another. - Substitute two ordered pairs from the table into the proposed model. - Use \(\log(10^x)=x\) to simplify. - To isolate the input of a common logarithm, apply a power of \(10\).

1. a) Multiplying the intensity by \(100\) increases the sound level by \(20\,\text{dB}\), so multiplying the intensity by \(10\) increases the level by \(10\,\text{dB}\). 2. b) Using \((10^{-12},0)\) gives \(0=a\log(10^{-12})+b=-12a+b\), so \(b=12a\). 3. Using \((10^{-10},20)\) gives \(20=-10a+12a=2a\), so \(a=10\) and \(b=120\). Thus, \(L=10\log(I)+120\). 4. Check with \(I=10^{-6}\): \(L=10\log(10^{-6})+120=10\cdot(-6)+120=60\), which matches the table. 5. c) Solve for \(I\): \(L-120=10\log(I)\), so \(\log(I)=\frac{L-120}{10}\) and \(I=10^{\frac{L-120}{10}}\). 6. For \(L=50\), \(I=10^{\frac{50-120}{10}}=10^{-7}\,\text{W}/\text{m}^2\).

a) \(10\,\text{dB}\) b) \(a=10\), \(b=120\), so \(L=10\log(I)+120\). For example, \(I=10^{-6}\) gives \(L=60\,\text{dB}\), matching the table. c) \(I=10^{\frac{L-120}{10}}\); at \(50\,\text{dB}\), \(I=10^{-7}\,\text{W}/\text{m}^2\).

- Look for the constant change in voltage each time the illuminance is multiplied by \(10\). - Use \(\log(1)=0\) to find one parameter immediately. - A common logarithm is inverted by raising \(10\) to a power.

1. a) Each time \(x\) is multiplied by \(10\), \(U\) increases by \(20\,\text{mV}\). 2. b) Substitute \(x=1\): \(5=a\log(1)+b\). Since \(\log(1)=0\), \(b=5\). 3. Substitute \(x=10\): \(25=a\log(10)+5\). Since \(\log(10)=1\), \(a=20\). Therefore, \(U=20\log(x)+5\). 4. c) Set \(U=100\): \(100=20\log(x)+5\). Then \(\log(x)=4.75\), so \(x=10^{4.75}\approx56{,}234.13\,\text{lx}\).

a) \(U\) increases by \(20\,\text{mV}\). b) \(U=20\log(x)+5\) c) \(x=10^{4.75}\,\text{lx}\approx56{,}234.13\,\text{lx}\)

- Substitute the given multiple of \(I_0\) into the sound-level formula and simplify the ratio. - For part b), compare the two decibel levels before solving for an intensity ratio. - In part c), combine the difference of two logarithms into one logarithm. - Here, \(\log\) denotes the common logarithm.

1. \(L=10\log(5\times10^6)=10(\log(5)+6)\approx66.99\,\text{dB}\). 2. The change is \(84-66=18\,\text{dB}\). Thus, \(18=10\log\left(\frac{I_1}{I_2}\right)\), so \(\frac{I_1}{I_2}=10^{1.8}\approx63.10\). The intensity is reduced by a factor of about \(63.10\). 3. Suppose \(L_1-L_2=10\). Then \(10\log\left(\frac{I_1}{I_0}\right)-10\log\left(\frac{I_2}{I_0}\right)=10\). 4. Divide by \(10\) and use the quotient property: \(\log\left(\frac{I_1}{I_2}\right)=1\). Therefore, \(\frac{I_1}{I_2}=10\), so \(I_2=\frac{1}{10}I_1\).

a) \(L\approx66.99\,\text{dB}\) b) The intensity is reduced by a factor of \(10^{1.8}\approx63.10\). c) A \(10\,\text{dB}\) decrease gives \(I_2=\frac{1}{10}I_1\).

- In part a), isolate the intensity ratio before converting from logarithmic to exponential form. - In part b), use a logarithm property to separate the factor \(2\). - For \(n\) identical speakers, the total intensity is \(nI\). - The increase in sound level depends on the multiplicative change in intensity.

1. \(75=10\log\left(\frac{I}{I_0}\right)\), so \(\log\left(\frac{I}{I_0}\right)=7.5\). Therefore, \(\frac{I}{I_0}=10^{7.5}\). 2. For two identical speakers, \(L_{\text{new}}=10\log\left(\frac{2I}{I_0}\right)=10\log(2)+10\log\left(\frac{I}{I_0}\right)\). 3. Thus, \(L_{\text{new}}=10\log(2)+75\approx78.01\,\text{dB}\). 4. For \(n\) identical speakers, \(85=10\log\left(\frac{nI}{I_0}\right)=10\log(n)+75\). 5. Then \(10=10\log(n)\), so \(\log(n)=1\) and \(n=10\).

a) \(\frac{I}{I_0}=10^{7.5}\) b) \(L_{\text{new}}\approx78.01\,\text{dB}\) c) \(10\) speakers

- Keep track of the negative sign in the pH definition. - To solve for concentration, rewrite the logarithmic equation in exponential form. - A pH difference corresponds to a multiplicative concentration ratio. - When comparing the two liquids, simplify the quotient of powers of \(10\).

1. For the stomach acid, \(\text{pH}=-\log(0.032)\approx1.49\). 2. For the soap solution, \(\text{pH}=-\log(10^{-10})=10\). 3. From \(\text{pH}=-\log(c)\), solve for \(c\): \(c=10^{-\text{pH}}\). Thus, \(c=10^{-5.5}\approx3.16\times10^{-6}\,\text{mol}/\text{L}\). 4. For the final comparison, the concentrations are \(c_1=10^{-2.4}\) and \(c_2=10^{-2.9}\). 5. Their ratio is \(\frac{c_1}{c_2}=10^{-2.4-(-2.9)}=10^{0.5}\approx3.16\), so lemon juice has about \(3.16\) times the hydronium-ion concentration of vinegar.

a) (1) \(\text{pH}\approx1.49\); (2) \(\text{pH}=10\) b) \(c(H_3O^+)\approx3.16\times10^{-6}\,\text{mol}/\text{L}\) c) Lemon juice has about \(3.16\) times the hydronium-ion concentration of vinegar.

- Simplify the intensity ratio inside the logarithm before evaluating it. - For part b), isolate the logarithmic expression before rewriting it in exponential form. - For part c), compare the level for \(I\) with the level for \(2I\). - A logarithm turns a multiplicative intensity change into an additive level change.

1. For the conversation-level sound, \(L=10\log\left(\frac{10^{-6}}{10^{-12}}\right)=10\log(10^6)=60\,\text{dB}\). 2. For \(120\,\text{dB}\), \(120=10\log\left(\frac{I}{10^{-12}}\right)\), so \(12=\log\left(\frac{I}{10^{-12}}\right)\). 3. Therefore, \(10^{12}=\frac{I}{10^{-12}}\), giving \(I=1\,\text{W}/\text{m}^2\). 4. If intensity doubles, the level increase is \(10\log\left(\frac{2I}{I_0}\right)-10\log\left(\frac{I}{I_0}\right)=10\log(2)\approx3.01\,\text{dB}\).

a) \(60\,\text{dB}\) b) \(1\,\text{W}/\text{m}^2\) c) About \(3.01\,\text{dB}\)

- Inspect the actual numeric labels on each horizontal axis rather than the panel letters. - On a linear scale, equal distances correspond to equal differences. - On a logarithmic scale, equal distances correspond to equal ratios. - Compare how clearly each panel separates the smaller measurements.

1. In panel a), the horizontal axis is numbered by equal additive increments, and the smaller measurements \(p\) and \(q\) are compressed near the origin while \(s\) lies far to the right. This is the linear panel. 2. In panel b), the displayed values \(1\), \(10\), \(100\), and \(1000\) occupy equal horizontal intervals. Their consecutive ratios are all \(10\), so panel b) is logarithmic. 3. On the linear panel, equal horizontal distance means equal numerical difference, so successive tenfold changes occupy increasingly large gaps. On the logarithmic panel, equal tenfold ratios occupy equal gaps. 4. Panel b) is more informative for repeated multiplicative change because equal multiplicative factors receive equal visual spacing across orders of magnitude.

a) Panel b) is logarithmic. b) The linear panel compresses the smaller measurements near the origin and gives increasingly large gaps; the logarithmic panel gives equal spacing to the successive tenfold values. c) Panel b), because equal multiplicative changes are represented by equal spatial intervals.

- Doubling the number of identical sources doubles the intensity, not the decibel value. - Use logarithm properties to separate a multiplicative intensity factor. - For several identical sources, represent the total intensity as a multiple of one source's intensity. - In part b), the number of machines must be a whole number that actually reaches the threshold.

1. Two identical sources double the intensity. The increase in level is \(\frac{10}{\ln(10)}\ln(2)\approx3.01\,\text{dB}\). 2. Therefore, the total level is approximately \(85+3.01=88.01\,\text{dB}\). 3. With \(n\) identical sources, the increase over one source is \(\frac{10}{\ln(10)}\ln(n)\). To reach \(100\,\text{dB}\), require \(\frac{10}{\ln(10)}\ln(n)\ge15\). 4. This gives \(\ln(n)\ge1.5\ln(10)\), so \(n\ge e^{1.5\ln(10)}=10^{1.5}\approx31.62\). 5. The number of jackhammers must be a whole number, so the minimum is \(32\).

a) Approximately \(88.01\,\text{dB}\) b) \(32\) jackhammers