Exponential and Logarithmic Word Problems with Step-by-Step Solutions

Question: The population \( P \) of a city increases according to the formula:

\[ P = 5000 e^{a t} \]

where \( t \) is in years and \( t = 0 \) corresponds to 1980. In 1990, the population was 10000. Find the value of the constant \( a \) and approximate your answer to 3 decimal places.

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Solution:

If \( t = 0 \) corresponds to 1980, then 1990 corresponds to \( t = 10 \). We know that when \( t = 10 \), the population \( P = 10000 \). Substituting these values into the formula yields:

\[ 5000 e^{10a} = 10000 \]

Divide both sides of the equation by 5000 and simplify:

\[ e^{10a} = \frac{10000}{5000} \] \[ e^{10a} = 2 \]

Using the natural logarithm \( \ln \), rewrite the exponential form as:

\[ 10a = \ln 2 \]

Solve for \( a \) and approximate to 3 decimal places:

\[ a = \frac{\ln 2}{10} \approx 0.069 \]

Question: The populations \( P_1 \) and \( P_2 \) of two cities are given by the formulas:

\[ P_1 = 10000 e^{k t} \] \[ P_2 = 20000 e^{0.01 t} \]

where \( k \) is a constant and \( t \) is the time in years with \( t = 0 \) corresponding to the year 2000. Find the constant \( k \) so that the two populations are equal in the year 2020. Approximate your answer to 3 decimal places.

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Solution:

If \( t = 0 \) corresponds to 2000, then 2020 corresponds to \( t = 20 \). We need \( P_1 = P_2 \) when \( t = 20 \).

\[ 10000 e^{20k} = 20000 e^{20 \times 0.01} \] \[ 10000 e^{20k} = 20000 e^{0.2} \]

Divide both sides by \( 20000 e^{20k} \) and simplify:

\[ \frac{10000}{20000} = \frac{e^{0.2}}{e^{20k}} \] \[ \frac{1}{2} = e^{0.2 - 20k} \]

Convert the exponential form into logarithmic form:

\[ 0.2 - 20k = \ln\left(\frac{1}{2}\right) \]

Solve for \( k \):

\[ 20k = 0.2 - \ln(0.5) \] \[ k = \frac{0.2 - \ln(0.5)}{20} \] \[ k \approx 0.045 \]

Question: The level of sound \( D \) in decibels is defined as:

\[ D = 10 \log\left(\frac{I}{10^{-16}}\right) \]

where \( I \) is the sound intensity in watts per square centimeter. Determine the level in decibels of a sound with intensity \( I = 10^{-8}\ \text{watts/cm}^2 \).

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Solution:

Substitute \( I = 10^{-8} \ \text{watts/cm}^2 \) into the formula:

\[ D = 10 \log\left(\frac{10^{-8}}{10^{-16}}\right) \]

Simplify the fraction using exponent rules \( \dfrac{x^a}{x^b} = x^{a-b} \):

\[ D = 10 \log\left(10^{-8 - (-16)}\right) = 10 \log\left(10^{8}\right) \]

Since \( \log_{10}(10^8) = 8 \):

\[ D = 10 \times 8 = 80 \ \text{decibels} \]

Question: Two sounds of intensities \( I_1 \) and \( I_2 \) have decibel levels of 60 and 80 respectively. Use the decibel formula from Problem 3 to determine the ratio of the intensities \( \dfrac{I_2}{I_1} \).

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Solution:

Write the equations for both sounds:

\[ 60 = 10 \log\left(\frac{I_1}{10^{-16}}\right) \] \[ 80 = 10 \log\left(\frac{I_2}{10^{-16}}\right) \]

Subtract the first equation from the second:

\[ 80 - 60 = 10 \log\left(\frac{I_2}{10^{-16}}\right) - 10 \log\left(\frac{I_1}{10^{-16}}\right) \] \[ 20 = 10 \left[ \log\left(\frac{I_2}{10^{-16}}\right) - \log\left(\frac{I_1}{10^{-16}}\right) \right] \]

Divide by 10 and apply the quotient rule \( \log a - \log b = \log\left(\dfrac{a}{b}\right) \):

\[ 2 = \log\left( \frac{\frac{I_2}{10^{-16}}}{\frac{I_1}{10^{-16}}} \right) \] \[ 2 = \log\left(\frac{I_2}{I_1}\right) \]

Convert from logarithmic to exponential form:

\[ \frac{I_2}{I_1} = 10^2 = 100 \]

Question: The spread of a virus through a city is modeled by:

\[ N = \frac{15000}{1 + 100 e^{-0.5t}} \]

where \( N \) is the number of people infected after \( t \) days. How many days will it take for 2000 people to be infected? Approximate your answer to 3 decimal places.

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Solution:

Set \( N = 2000 \) and solve for \( t \):

\[ 2000 = \frac{15000}{1 + 100 e^{-0.5 t}} \]

Multiply both sides by the denominator and divide by 2000:

\[ 1 + 100 e^{-0.5 t} = \frac{15000}{2000} \] \[ 1 + 100 e^{-0.5 t} = 7.5 \]

Subtract 1 and divide by 100:

\[ 100 e^{-0.5 t} = 6.5 \] \[ e^{-0.5 t} = 0.065 \]

Use \( \ln \) to solve for the exponent:

\[ -0.5 t = \ln(0.065) \] \[ t = \frac{\ln(0.065)}{-0.5} \approx 5.467 \]

It takes approximately 5.467 days.

Question: The amount of a radioactive material decays according to:

\[ A(t) = A_0 e^{-k t} \]

where \( A_0 \) is the initial amount, \( k \) is a positive constant, and \( t \) is the time in days. Find a formula for the half-life of the material.

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Solution:

The half-life \( T \) is the time required for the material to decay to half its initial amount. At \( t = T \), the amount remaining is \( \dfrac{A_0}{2} \):

\[ \frac{A_0}{2} = A_0 e^{-kT} \]

Divide both sides by \( A_0 \):

\[ \frac{1}{2} = e^{-kT} \]

Use \( \ln \) to convert to logarithmic form:

\[ -kT = \ln\left(\frac{1}{2}\right) \]

Using \( \ln(1/2) = \ln(1) - \ln(2) = -\ln 2 \):

\[ -kT = -\ln 2 \]

Solve for \( T \):

\[ T = \frac{\ln 2}{k} \]

Question: A logistic equation has the form:

\[ y = \frac{a}{1 + b e^{-kt}} \]

where \( a \), \( b \), and \( k \) are constants. Solve the equation for \( t \).

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Solution:

Multiply both sides by \( 1 + b e^{-kt} \) and divide by \( y \):

\[ 1 + b e^{-kt} = \frac{a}{y} \]

Subtract 1 from both sides:

\[ b e^{-kt} = \frac{a}{y} - 1 \]

Divide both sides by \( b \):

\[ e^{-kt} = \frac{\frac{a}{y} - 1}{b} \]

Take the natural logarithm of both sides:

\[ -kt = \ln\left(\frac{\frac{a}{y} - 1}{b}\right) \]

Solve for \( t \):

\[ t = -\frac{1}{k} \ln\left(\frac{\frac{a}{y} - 1}{b}\right) \]

Question: The voltage \( V(t) \) across an electrical component is given by:

\[ V(t) = V_0 e^{- a t } \]

where \( a \) is a positive constant depending on the circuit and \( V_0 \) is the initial voltage. Find \( t \) for which \( V(t) \) is 50% of \( V_0 \).

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Solution:

Set \( V(t) = 0.5 V_0 \):

\[ 0.5 V_0 = V_0 e^{-at} \]

Divide both sides by \( V_0 \):

\[ e^{-at} = 0.5 = \frac{1}{2} \]

Use \( \ln \) to convert to logarithmic form:

\[ -at = \ln\left(\frac{1}{2}\right) = -\ln 2 \]

Solve for \( t \):

\[ t = \frac{\ln 2}{a} \]

Question: The signal ratio in decibels of an electronic system is given by:

\[ 10 \log\left(\frac{P_{o}}{P_{i}}\right) \]

where \( P_o \) is the output power and \( P_i \) is the input power. Find the input power \( P_i \) if the output power is 10 mW and the signal ratio is 10 decibels.

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Solution:

Substitute the signal ratio (10 dB) and \( P_o = 10 \):

\[ 10 = 10 \log\left(\frac{10}{P_i}\right) \]

Divide both sides by 10:

\[ 1 = \log\left(\frac{10}{P_i}\right) \]

Rewrite the base-10 logarithm in exponential form:

\[ 10^1 = \frac{10}{P_i} \] \[ 10 = \frac{10}{P_i} \]

Multiply both sides by \( P_i \) and divide by 10:

\[ P_i = 1 \text{ mW} \]

Question: The current \( i \) flowing in an RL circuit is given by:

\[ i = \frac{E}{R} \left[ 1 - e^{-\frac{Rt}{L}} \right] \]

where \( E \) is the applied voltage, \( R \) is the resistance, and \( L \) is the inductance. Express \( i \) in terms of \( E \) and \( R \) when \( t = \dfrac{L}{R} \).

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Solution:

Substitute \( t = \dfrac{L}{R} \) into the exponent:

\[ i = \frac{E}{R} \left( 1 - e^{-\frac{R}{L} \cdot \frac{L}{R}} \right) \]

The \( R \) and \( L \) variables in the exponent cancel:

\[ i = \frac{E}{R} \left( 1 - e^{-1} \right) \]

Since \( e^{-1} \approx 0.3678 \):

\[ i = \frac{E}{R} (1 - 0.3678) \] \[ i \approx 0.632 \frac{E}{R} \]