Aimathic – Logarithmic scales – Free Math Worksheets

Logarithmic scales

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Difficulty: 1 – 5

52614312

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

- 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.

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. 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 \(\log(x) \in [5, 6)\).

Answer

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

52760512

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.

52763512

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 \(10\).

52843312

In acoustics, sound intensity \(I\) and sound level \(L\) 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> 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 function that gives \(I\) in terms of \(L\). What intensity corresponds to \(50\,\text{dB}\)?

Hints

- 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 \(10\).

Solution

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 \(L=50\), \(I=10^{\frac{50-120}{10}}=10^{-7}\,\text{W}/\text{m}^2\).

Answer

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 \(50\,\text{dB}\), \(I=10^{-7}\,\text{W}/\text{m}^2\).

52843412

A digital light sensor measures illuminance \(x\), in lux (\(\text{lx}\)), and produces an output voltage \(U\), in millivolts (\(\text{mV}\)). 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 \(100\,\text{mV}\). Find the corresponding illuminance.

Hints

- 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.

Solution

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 \(x=10^{4.75}\approx56{,}234.13\,\text{lx}\).

Answer

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

52851912

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 \(I=5\times10^6 I_0\). Find its sound level. b) A sound barrier reduces traffic noise from \(84\,\text{dB}\) to \(66\,\text{dB}\). By what factor is the sound intensity reduced? 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 formula and simplify the ratio. - For part b, compare the two decibel levels and use the definition of a common logarithm. - In part c, combine the difference of two logarithms into one logarithm. - Here, \(\log\) means the logarithm to base \(10\).

Solution

1. a) \(L=10\log(5\times10^6)=10(\log(5)+6)\approx66.99\,\text{dB}\). 2. b) 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 \(63\). 3. c) 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) \(L\approx66.99\,\text{dB}\), or about \(67\,\text{dB}\) b) The intensity is reduced by a factor of \(10^{1.8}\approx63.10\). c) A \(10\,\text{dB}\) decrease gives \(I_2=\frac{1}{10}I_1\).

52852712

In chemistry, the hydronium-ion concentration \(c(H_3O^+)\), measured in \(\text{mol}/\text{L}\), is often described using pH: \(\text{pH}=-\log(c(H_3O^+))\) a) Find the pH of each liquid. (1) Stomach acid with \(c(H_3O^+)=0.032\,\text{mol}/\text{L}\) (2) A soap solution with \(c(H_3O^+)=10^{-10}\,\text{mol}/\text{L}\) b) Soil for rhododendrons should have a pH of about \(5.5\). Find the corresponding hydronium-ion concentration. c) Lemon juice has pH \(2.4\), while vinegar has pH \(2.9\). By what factor do their hydronium-ion concentrations differ?

Hints

- Account for the negative sign in front of the logarithm. - To solve for the concentration, rewrite the logarithmic equation in exponential form. - If \(\log(x)=y\), then \(x=10^y\). - Use the quotient of powers property when comparing concentrations.

Solution

1. a)(1) \(\text{pH}=-\log(0.032)\approx1.49\). 2. a)(2) \(\text{pH}=-\log(10^{-10})=10\). 3. b) 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. c) The concentrations are \(c_1=10^{-2.4}\) and \(c_2=10^{-2.9}\). Their ratio is \(\frac{c_1}{c_2}=10^{-2.4-(-2.9)}=10^{0.5}\approx3.16\). 5. The hydronium-ion concentration in lemon juice is about \(3.16\) times that in vinegar.

Answer

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.

52852812

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 normal conversation has intensity about \(10^{-7}\,\text{W}/\text{m}^2\). 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?

Hints

- Simplify the ratio inside the logarithm before evaluating it. - Isolate the logarithmic expression before rewriting the equation in exponential form. - Use the quotient property to simplify the difference of two logarithms. - In part c, compare the level for \(I\) with the level for \(2I\).

Solution

1. a) \(L=10\log\left(\frac{10^{-7}}{10^{-12}}\right)=10\log(10^5)=50\,\text{dB}\). 2. b) \(120=10\log\left(\frac{I}{10^{-12}}\right)\), 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. c) The 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) \(50\,\text{dB}\) b) \(1\,\text{W}/\text{m}^2\) c) About \(3.01\,\text{dB}\)

52760612

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. 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 the factor \(2\) or \(n\). - In part b, round up because the number of machines must be a whole number. - What increase results from multiplying intensity by \(10\)?

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 \(\frac{10}{\ln(10)}\ln(n)\geq15\). 4. This gives \(\ln(n)\geq1.5\ln(10)\), so \(n\geq e^{1.5\ln(10)}=10^{1.5}\approx31.62\). 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

52852012

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 \(75\,\text{dB}\). c) How many identical speakers must operate together to produce a total sound level of \(85\,\text{dB}\)?

Hints

- In part a, solve the equation for the ratio \(\frac{I}{I_0}\). - 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 only on the factor by which intensity changes.

Solution

1. a) \(75=10\log\left(\frac{I}{I_0}\right)\), so \(\log\left(\frac{I}{I_0}\right)=7.5\). Therefore, \(\frac{I}{I_0}=10^{7.5}\approx31{,}622{,}776.60\). 2. b) \(L_{\text{new}}=10\log\left(\frac{2I}{I_0}\right)=10\log(2)+10\log\left(\frac{I}{I_0}\right)\). 3. Thus, \(L_{\text{new}}=10\log(2)+75\approx78.01\,\text{dB}\). 4. c) 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) \(\frac{I}{I_0}=10^{7.5}\approx31{,}622{,}776.60\) b) \(L_{\text{new}}\approx78.01\,\text{dB}\) c) \(10\) speakers

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