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Logarithmic scales

Learning goals

  • Recognizes quantities modeled on logarithmic scales.
  • Interprets equal additive changes on the scale as multiplicative changes in the underlying quantity.
  • Converts between scale values and underlying ratios when formulas are given.
  • Compares quantities using the logarithmic model.

Click problems to add them to your worksheet.

Logarithmic scales 1
The pH of a solution is modeled by \(\text{pH}=-\log_{10}([H^+])\), where \([H^+]\) means the numerical hydrogen-ion concentration in \(\text{mol}/\text{L}\). Find the pH when \([H^+]=10^{-4}\).

Hints

- What exponent on ten gives the stated numerical concentration? - How does the minus sign in the pH formula change that exponent?

Solution

1. A base-ten logarithm gives the exponent on \(10\), so \(\log_{10}(10^{-4})=-4\). 2. Therefore, \(\text{pH}=-(-4)=4\).

Answer

\(4\)
Logarithmic scales 2
Sound level is modeled by \(L=10\log_{10}\left(\frac{I}{I_0}\right)\), where \(I_0\) is a reference intensity. If \(L=20\,\text{dB}\), what is the ratio \(\frac{I}{I_0}\)?

Hints

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

Solution

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

Answer

\(\frac{I}{I_0}=100\)
Logarithmic scales 3
Is this measurement axis linear or logarithmic? If it is logarithmic, what factor does one equal interval represent? Explain.
Figure for problem 555628

Hints

- Read the numerical labels from left to right. - Compare the differences and ratios between consecutive labels. - Which of those comparisons stays constant across equal visual intervals?

Solution

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 represent equal multiplicative factors. 4. Therefore, the axis is logarithmic, and one marked interval means multiplication by \(10\).

Answer

The axis is logarithmic; one marked interval represents multiplication by \(10\).
Logarithmic scales 4
The vertical axis uses a logarithmic scale. Find the value at \(P\), and tell how much the value is multiplied by when you move up one tick.
Figure for problem 561617

Hints

- Compare the two lowest labeled ticks. - On a logarithmic axis, what stays the same between equally spaced ticks: the difference or the factor? - Use that pattern to fill the missing tick.

Solution

1. Equal steps represent the same multiplication factor. Moving from \(1\) to \(10\) is a factor of \(10\). 2. The next tick is therefore \(10\cdot10=100\). 3. This agrees with the tick above it: \(100\cdot10=1000\).

Answer

\(P=100\); each step multiplies the value by \(10\).
Logarithmic scales 5
Consider positive integers written in base \(10\). a) What do \(\log(n)\) and \(\log(m)\) have in common when \(n\) and \(m\) have the same number of digits? Here \(\log\) denotes the common logarithm. b) In what interval does \(\log(x)\) lie when \(x\) is any six-digit positive integer? Write the interval in the form \([a, b)\).

Hints

- Between which consecutive powers of ten does a number with a given number of digits fall? - What is the common logarithm of an exact power of ten? - How does taking a logarithm change those power-of-ten bounds?

Solution

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. All positive integers with the same number of digits therefore have common logarithms with the same integer part, \(k-1\). 4. A six-digit positive integer satisfies \(10^5\le x<10^6\), so \(5\le\log(x)<6\), or \(\log(x)\in[5,6)\).

Answer

a) Their common logarithms have the same integer part. b) \([5,6)\)
Logarithmic scales 6
A logarithmic index is defined by \(S(r)=10\log_{10}(r)\) for a positive ratio \(r\). a) How much does \(S\) increase when \(r\) changes from \(1\) to \(10\)? How much does it increase from \(10\) to \(100\)? b) Suppose \(r\) is plotted on an ordinary linear horizontal axis. Compare the input changes \(1\to10\) and \(10\to100\). Explain how the logarithmic index compresses multiplicative changes. c) Predict \(S(1000)\) without evaluating a logarithm on a calculator.

Hints

- 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.

Solution

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 \(100\), \(S(1000)=30\).

Answer

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) \(S(1000)=30\)
Logarithmic scales 7
On this base-10 logarithmic axis: a) What value does \(p\) represent? Give an exact expression and a decimal approximation to the nearest tenth. b) Explain why averaging the two neighboring labeled values does not give \(p\).
Figure for problem 555629

Hints

- Which labeled values are nearest to the marker on either side? - How would you describe the midpoint in terms of powers of ten? - Do equal physical distances represent equal differences or equal ratios here?

Solution

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

Answer

a) \(p=10\sqrt{10}\approx31.6\) b) Equal distances on a logarithmic axis mean equal ratios, so the midpoint is the geometric mean of the adjacent labeled values.
Logarithmic scales 8
An earthquake-amplitude scale uses \(M=\log_{10}\!\left(\frac{A}{A_0}\right)\), where \(A\) is the measured amplitude and \(A_0\) is a reference amplitude. One quake has \(M=2\) and another has \(M=4\). Find each amplitude as a multiple of \(A_0\), and find the ratio of their amplitudes.

Hints

- How can you rewrite an equation involving a common logarithm as a power of ten? - How does a difference between the two scale readings relate to the ratio of their amplitudes?

Solution

1. The first quake has \(\log_{10}(A_1/A_0)=2\), so \(A_1=10^2A_0=100A_0\). 2. The second has \(\log_{10}(A_2/A_0)=4\), so \(A_2=10^4A_0=10{,}000A_0\). 3. The amplitude ratio is \(A_2/A_1=10^{4-2}=100\).

Answer

First amplitude: \(100A_0\); second amplitude: \(10{,}000A_0\); ratio: \(100\).
Logarithmic scales 9
The pH of a liquid is given by \(\text{pH}=-\log_{10}(c)\), where \(c\) is the numerical hydrogen-ion concentration in moles per liter. Find \(c\) for liquids with pH \(5\) and pH \(6\). How many times as concentrated is the first liquid?

Hints

- Which operation undoes a common logarithm? - Keep track of the negative sign in the pH formula. - Compare the two resulting powers of ten using a ratio.

Solution

1. From \(5=-\log_{10}(c_1)\), we get \(c_1=10^{-5}\). 2. From \(6=-\log_{10}(c_2)\), we get \(c_2=10^{-6}\). 3. The concentration ratio is \(c_1/c_2=10^{-5}/10^{-6}=10\).

Answer

\(c_1=10^{-5}\,\text{mol/L}\); \(c_2=10^{-6}\,\text{mol/L}\). The first is \(10\) times as concentrated.
Logarithmic scales 10
A simplified brightness scale for stars is \(m=-2.5\log_{10}\left(\frac{F}{F_0}\right)\), where \(F\) is a star's measured brightness and \(F_0\) is a fixed reference. Star A has \(F=100F_0\), and star B has \(F=F_0\). Find the magnitude of each star. Which appears brighter on this scale?

Hints

- Start with the brightness ratio inside the logarithm for each star. - What are the common logarithms of \(100\) and \(1\)? - Check the direction of the scale using the actual brightnesses, not just the magnitude signs.

Solution

1. For star A, \(\frac{F}{F_0}=100\), so \(m_A=-2.5\log_{10}(100)=-5\). 2. For star B, \(\frac{F}{F_0}=1\), so \(m_B=-2.5\log_{10}(1)=0\). 3. Star A is brighter because its measured brightness is \(100\) times star B's. On this scale, the brighter star has the smaller magnitude number.

Answer

Star A: \(m_A=-5\). Star B: \(m_B=0\). Star A appears brighter.
Logarithmic scales 11
In a simplified earthquake-energy model, \(\log_{10}\left(\frac{E}{E_0}\right)=1.5M\), where \(E\) is released energy, \(E_0\) is a fixed reference, and \(M\) is magnitude. How many times as much energy does a magnitude-\(5\) earthquake release as a magnitude-\(3\) earthquake according to this model?

Hints

- How does the difference in magnitudes affect the difference between the two logarithms? - Can you combine the logarithms into one logarithm of an energy ratio? - What ratio has that common logarithm?

Solution

1. Subtract the two logarithmic equations: \(\log_{10}\left(\frac{E_5}{E_0}\right)-\log_{10}\left(\frac{E_3}{E_0}\right)=1.5(5-3)=3\). 2. The logarithm quotient rule gives \(\log_{10}\left(\frac{E_5}{E_3}\right)=3\). 3. Therefore \(\frac{E_5}{E_3}=10^3=1000\).

Answer

\(1000\) times as much energy.
Logarithmic scales 12
The vertical logarithmic scale has the same multiplication factor between each pair of neighboring ticks. Find the value at \(A\), then find the ratio of the values at \(B\) and \(A\).
Figure for problem 561622

Hints

- Count how many equal steps separate the two numbered readings. - If the factor per step is \(q\), what repeated multiplication connects the endpoint readings? - Compare the values at the two marked positions only after finding that factor.

Solution

1. The value rises from \(4\) to \(108\) in three equal steps, so the factor \(q\) satisfies \(4q^3=108\). 2. Thus \(q^3=27\) and \(q=3\). 3. Point \(A\) is one step above \(4\), so it represents \(4\cdot3=12\). 4. Point \(B\) represents \(108\), so \(\frac{B}{A}=\frac{108}{12}=9\).

Answer

\(A=12\); the ratio \(\frac{B}{A}=9\).
Logarithmic scales 13
Determine the number of digits in the decimal representation of \(6^{60}\). Use \(\log(2)\approx0.3010\) and \(\log(3)\approx0.4771\).

Hints

- How is the number of digits of a positive integer related to its common logarithm? - Which logarithm properties let you rewrite \(\log(6^{60})\)? - What does the integer part of a common logarithm tell you about the size of a number?

Solution

1. Use the power rule: \(\log(6^{60})=60\log(6)\). 2. Use the product rule: \(\log(6)=\log(2\cdot3)=\log(2)+\log(3)\). 3. Substitute the given values: \(\log(6)\approx0.3010+0.4771=0.7781\). 4. Thus, \(\log(6^{60})\approx60\cdot0.7781=46.686\). 5. Since the logarithm is between \(46\) and \(47\), the number lies between \(10^{46}\) and \(10^{47}\). Therefore, \(6^{60}\) has \(47\) digits.

Answer

\(6^{60}\) has \(47\) digits.
Logarithmic scales 14
A computer program can store whole numbers with at most \(100\) decimal digits. Find the greatest natural number \(n\) for which \(2^n\) can still be stored. Use \(\log(2)\approx0.30103\).

Hints

- What condition must the common logarithm satisfy for a number to have at most \(100\) digits? - Write an inequality involving \(n\). - Use the power property of logarithms.

Solution

1. A positive integer has at most \(100\) digits when its common logarithm is less than \(100\), so \(\log(2^n)<100\). 2. Apply the power property: \(n\log(2)<100\). 3. Then \(n<\frac{100}{\log(2)}\approx\frac{100}{0.30103}\approx332.19\). 4. Since \(n\) is a natural number, the greatest possible value is \(n=332\). 5. Check: \(\log(2^{332})\approx99.94\), so \(2^{332}\) has \(100\) digits, while \(\log(2^{333})\approx100.24\), so \(2^{333}\) has \(101\) digits.

Answer

\(n=332\)
Logarithmic scales 15
The magnitude \(M\) of an earthquake can be modeled from its released seismic energy \(E\), in joules, by \(M=\frac{2}{3}\log_{10}\!\left(\frac{E}{E_0}\right)\), where \(E_0=10^{4.8}\,\text{J}\). How many times as much energy is released by an earthquake of magnitude \(7.5\) as by an earthquake of magnitude \(5.5\)?

Hints

- 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 \(10\).

Solution

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 \(\frac{E_2}{E_1}=10^3=1000\).

Answer

The magnitude \(7.5\) earthquake releases \(1000\) times as much energy.
Logarithmic scales 16
The Weber-Fechner model for perceived stimulus strength is \(S(I)=k\ln\!\left(\frac{I}{I_0}\right)\), where \(I\) is physical intensity, \(I_0\) is the threshold intensity, and \(k\) is a constant. 1. For sound perception, suppose \(I_0=10^{-12}\,\text{W/m}^2\). At \(I_1=10^{-7}\,\text{W/m}^2\), the measured perception strength is \(S_1=50\). Find \(k\). 2. Show algebraically that doubling the intensity always produces the same absolute change in perceived strength, regardless of the starting intensity. 3. Using the value of \(k\) from part 1, find the absolute change in perceived strength when intensity is multiplied by \(10\).

Hints

- 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 \(10\).

Solution

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 \(10\).

Answer

1. \(k=\frac{10}{\ln(10)}\approx4.343\) 2. The change is \(k\ln(2)\), independent of the starting intensity. 3. The change is \(10\).
Logarithmic scales 17
In acoustics, sound intensity and sound level are related logarithmically. The table shows several values. <table> <tr> <td>Sound intensity \(I\) (in \(\text{W}/\text{m}^2\))</td> <td>\(10^{-12}\)</td> <td>\(10^{-10}\)</td> <td>\(10^{-8}\)</td> <td>\(10^{-6}\)</td> </tr> <tr> <td>Sound level \(L\) (in \(\text{dB}\))</td> <td>\(0\)</td> <td>\(20\)</td> <td>\(40\)</td> <td>\(60\)</td> </tr> </table> In the formula below, \(I\) means the numerical intensity reading in \(\text{W}/\text{m}^2\), and \(L\) is the numerical sound-level reading in decibels. a) By how many decibels does \(L\) increase when \(I\) is multiplied by \(10\)? b) The relationship can be modeled by \(L=a\log(I)+b\). Find \(a\) and \(b\), and verify the model using another pair from the table. c) Find the inverse formula for intensity in \(\text{W}/\text{m}^2\) as a function of \(L\). What intensity corresponds to \(50\,\text{dB}\)?

Hints

- What happens to the sound level when intensity rises by two powers of ten? - How does the common logarithm change under multiplication by ten? - Use two table columns to find the coefficients. - To invert, what operation undoes a common logarithm?

Solution

1. The table shows that multiplying intensity by \(100\) increases the level by \(20\,\text{dB}\), so a factor of \(10\) increases it by \(10\,\text{dB}\). 2. For \(I=10^{-12}\), \(0=-12a+b\), so \(b=12a\). 3. For \(I=10^{-10}\), \(20=-10a+12a\), so \(a=10\), \(b=120\), and \(L=10\log(I)+120\). 4. Check \(I=10^{-6}\): \(L=10(-6)+120=60\), matching the table. 5. Solve \(L-120=10\log(I)\) for the numerical intensity: \(I=10^{(L-120)/10}\). Thus, the physical intensity is \(10^{(L-120)/10}\,\text{W}/\text{m}^2\). 6. At \(L=50\), the intensity is \(10^{-7}\,\text{W}/\text{m}^2\).

Answer

a) \(10\,\text{dB}\) b) \(a=10\), \(b=120\), so \(L=10\log(I)+120\); the \(10^{-6}\) reading gives \(60\,\text{dB}\). c) Physical intensity \(=10^{(L-120)/10}\,\text{W}/\text{m}^2\); at \(50\,\text{dB}\), intensity \(=10^{-7}\,\text{W}/\text{m}^2\).
Logarithmic scales 18
A digital light sensor measures illuminance in lux (\(\text{lx}\)) and produces a voltage in millivolts (\(\text{mV}\)). Let \(x\) be the numerical illuminance reading in lux and \(U\) the numerical voltage reading in millivolts. The sensor has a logarithmic response. <table> <tr> <td>Illuminance \(x\) (in \(\text{lx}\))</td> <td>\(1\)</td> <td>\(10\)</td> <td>\(100\)</td> <td>\(1000\)</td> </tr> <tr> <td>Voltage \(U\) (in \(\text{mV}\))</td> <td>\(5\)</td> <td>\(25\)</td> <td>\(45\)</td> <td>\(65\)</td> </tr> </table> a) How does \(U\) change when \(x\) is multiplied by \(10\)? b) Find a model of the form \(U=a\log(x)+b\). c) The sensor produces \(55\,\text{mV}\). Find the corresponding illuminance, rounded to the nearest hundredth of a lux. Assume the model applies between the table values.

Hints

- What stays the same as the illuminance changes by a factor of ten each time? - Which logarithm has input \(1\), making one coefficient easy to find? - How do you undo \(\log\) when the desired illuminance is unknown?

Solution

1. Each time \(x\) is multiplied by \(10\), \(U\) increases by \(20\,\text{mV}\). 2. At \(x=1\), \(5=a\log(1)+b=b\). At \(x=10\), \(25=a+b\), giving \(a=20\). The model is \(U=20\log(x)+5\). 3. Set \(U=55\): \(55=20\log(x)+5\), so \(\log(x)=2.5\). Then \(x=10^{2.5}\approx316.23\), corresponding to about \(316.23\,\text{lx}\).

Answer

a) \(U\) increases by \(20\,\text{mV}\). b) \(U=20\log(x)+5\), where \(x\) is the numerical lux reading c) \(x=10^{2.5}\approx316.23\), or \(316.23\,\text{lx}\)
Logarithmic scales 19
In acoustics, the sound level \(L\), in decibels (\(\text{dB}\)), of a sound with intensity \(I\) is modeled by \(L=10\log\left(\frac{I}{I_0}\right)\), where \(I_0\) is the threshold-of-hearing intensity. a) A passing car has intensity \(I=5\times10^6I_0\). Find its sound level. Round to the nearest hundredth of a decibel. b) A sound barrier reduces traffic noise from \(84\,\text{dB}\) to \(66\,\text{dB}\). By what factor is the sound intensity reduced? Round the factor to the nearest hundredth. c) Use logarithm properties to show that a decrease of exactly \(10\,\text{dB}\) always reduces the intensity to one tenth of its original value.

Hints

- 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.

Solution

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 \(I_2=\frac{1}{10}I_1\).

Answer

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 \(I_2=\frac{1}{10}I_1\).
Logarithmic scales 20
Sound level \(L\), in decibels (\(\text{dB}\)), is modeled by \(L=10\log\left(\frac{I}{I_0}\right)\). a) A speaker produces a sound level of \(75\,\text{dB}\). What multiple of the threshold intensity \(I_0\) is this? b) When two identical speakers operate together, the sound intensity doubles. Find the new sound level if one speaker produces \(75\,\text{dB}\). Round to the nearest hundredth of a decibel. c) How many identical speakers must operate together to produce a total sound level of \(85\,\text{dB}\)?

Hints

- 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.

Solution

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 \(n=10\).

Answer

a) \(\frac{I}{I_0}=10^{7.5}\) b) \(L_{\text{new}}\approx78.01\,\text{dB}\) c) \(10\) speakers
Logarithmic scales 21
In chemistry, \(\text{pH}=-\log(c)\), where \(c\) is the hydronium-ion concentration expressed as a number in \(\text{mol}/\text{L}\). Use the base-\(10\) logarithm. a) Find the pH of each liquid. (1) Stomach acid with a hydronium-ion concentration of \(0.032\,\text{mol}/\text{L}\). Round to the nearest hundredth. (2) A soap solution with a concentration of \(10^{-10}\,\text{mol}/\text{L}\). b) Soil for rhododendrons should have a pH of about \(5.5\). Find the corresponding hydronium-ion concentration in \(\text{mol}/\text{L}\). Express it in scientific notation with its coefficient rounded to the nearest hundredth. c) Lemon juice has pH \(2.4\), while vinegar has pH \(2.9\). By what factor do their hydronium-ion concentrations differ? Round to the nearest hundredth.

Hints

- What does the minus sign do to the logarithm when the numerical concentration is less than one? - How can you undo a base-ten logarithm when pH is known? - Keep the concentration in the stated unit when reporting your result. - How do two powers of ten compare when their exponents differ?

Solution

1. For 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 the numerical concentration: \(c=10^{-\text{pH}}\). At pH \(5.5\), \(c=10^{-5.5}\approx3.16\times10^{-6}\), so the concentration is approximately \(3.16\times10^{-6}\,\text{mol}/\text{L}\). 4. For lemon juice and vinegar, the numerical concentrations are \(c_1=10^{-2.4}\) and \(c_2=10^{-2.9}\). 5. Their ratio is \(\frac{c_1}{c_2}=10^{-2.4-(-2.9)}=10^{0.5}\approx3.16\). Lemon juice has the greater concentration.

Answer

a) (1) \(\text{pH}\approx1.49\); (2) \(\text{pH}=10\) b) About \(3.16\times10^{-6}\,\text{mol}/\text{L}\) c) Lemon juice has about \(3.16\) times the hydronium-ion concentration of vinegar.
Logarithmic scales 22
Sound intensity level \(L\), in decibels (\(\text{dB}\)), is calculated by \(L=10\log\left(\frac{I}{I_0}\right)\), where \(I\) is the sound intensity and \(I_0=10^{-12}\,\text{W}/\text{m}^2\) is the reference intensity at the threshold of hearing. a) A conversation-level sound has intensity about \(10^{-6}\,\text{W}/\text{m}^2\). Find its sound intensity level. b) A sound level of \(120\,\text{dB}\) is commonly associated with the threshold of pain. Find the corresponding intensity \(I\). c) If the intensity doubles, by how many decibels does the sound intensity level increase? Round to the nearest hundredth of a decibel.

Hints

- 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.

Solution

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 \(10\log\left(\frac{2I}{I_0}\right)-10\log\left(\frac{I}{I_0}\right)=10\log(2)\approx3.01\,\text{dB}\).

Answer

a) \(60\,\text{dB}\) b) \(1\,\text{W}/\text{m}^2\) c) About \(3.01\,\text{dB}\)
Logarithmic scales 23
The two panels show the same four marked measurements \(p,q,r,s\). One panel uses a linear horizontal scale and the other uses a base-10 logarithmic horizontal scale. Do not assume the panel order. a) Identify the logarithmic panel and justify your choice using the displayed numerical scale and marker spacing. b) Describe what happens to the spacing of \(p,q,r,s\) on the linear panel compared with the logarithmic panel. c) Which panel is more informative when the purpose is to compare repeated multiplicative changes across several orders of magnitude? Explain.
Figure for problem 555630

Hints

- 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.

Solution

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.

Answer

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.
Logarithmic scales 24
Sound level \(L\), in decibels, can be written using the natural logarithm as \(L=\frac{10}{\ln(10)}\ln\!\left(\frac{I}{I_0}\right)\), where \(I\) is sound intensity and \(I_0\) is a reference intensity. One jackhammer produces a sound level of \(85\,\text{dB}\). a) Find the total sound level when a second identical jackhammer operates beside the first. Assume the intensities add. Round to the nearest hundredth of a decibel. b) Find the minimum number of identical jackhammers that must operate simultaneously to produce a sound level of at least \(100\,\text{dB}\).

Hints

- 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.

Solution

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 \(32\).

Answer

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

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