Exponential Model Word Problems Practice

1. Noah borrows $\$2000$ from his father and agrees to repay the loan and any interest determined by his father as soon as he has the money.

   The relationship between the amount of money, $A$, in dollars that Noah owes his father (including interest), and the elapsed time, $t$, in years, is modeled by the following equation.

   $A=2000e^{0.1t}$

   How long did it take Noah to pay off his loan if the amount he paid to his father was equal to $\$2450$?

   Give an exact answer expressed as a natural logarithm.

2. Katya is a ranger at a nature reserve in Siberia, Russia, where she studies the changes in the reserve's bear population over time.

   The relationship between the elapsed time $t$, in years, since the beginning of the study and the bear population $B(t)$, on the reserve is modeled by the following function.

   $B(t)=5000 \cdot 2^{-0.05t}$

   In how many years will the reserve's bear population be $2000$?

   Round your answer, if necessary, to the nearest hundredth.

3. Harper uploaded a funny video of her dog onto a website.

   The relationship between the elapsed time, $d$, in days, since the video was first uploaded, and the total number of views, $V(d)$, that the video received is modeled by the following function.

   $V(d)=4^{{1.25d}}$

   How many views will the video receive after $6$ days?

   Round your answer, if necessary, to the nearest hundredth.

4. A huge ice glacier in the Himalayas initially covered an area of $45$ square kilometers. Because of changing weather patterns, this glacier begins to melt, and the area it covers begins to decrease exponentially.

   The relationship between $A$, the area of the glacier in square kilometers, and $t$, the number of years the glacier has been melting, is modeled by the following equation.

   $A=45e^{-0.05t}$

   How many years will it take for the area of the glacier to decrease to $15$ square kilometers?

   Give an exact answer expressed as a natural logarithm.
   

Source: Khan Academy, https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:logs/x2ec2f6f830c9fb89:exp-models/e/exponential-models-word-problems
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1. Thinking about the problem

   We want to know how many years, $t$, it took for Noah to repay his debt, $A$, of $\$2450$.

   So we need to find the value of $t$ for which $A=2450$.

   Substituting $2450$ in for $A$ in the model gives us the following equation.

   $2450=2000e^{0.1t}$

   Solving the equation

   We can solve the equation as shown below

   $\begin{aligned}2000e^{0.1t}&=2450\\\\e^{0.1t}&=1.225\\\\0.1t&=\ln\left(1.225\right)\\\\t&=10\cdot{{\,\ln\left(1.225\right)}}\end{aligned}$

   It will take $10\cdot {{\,\ln\left(1.225\right)}}$ years for Noah to repay his loan.

   The expression above represents an exact solution to the problem. We can use a calculator to approximate the value of the expression, but this will be a rounded inexact answer.

   The answer

   The answer is $10\cdot{{\ln\left(1.225\right)}}$ years.

   ---

2. Thinking about the problem

   We want to know how many years, $t$, it will take for the bear population, $B(t)$, to reach $2000$.

   So we need to find the value of $t$ for which $B(t)=2000$.

   Substituting $2000$ in for $B(t)$ in the function gives us the following equation.

   $2000=5000 \cdot 2^{-0.05t}$

   Solving the equation

   We can solve the equation as shown below.

   $\begin{aligned}5000\cdot2^{-0.05t}&=2000\\\\2^{-0.05t}&=0.4\\\\-0.05t&=\log_2(0.4)\\\\t&=\dfrac{\log_2(0.4)}{-0.05}\end{aligned}$

   Changing the base to approximate the solution

   Since most calculators only calculate logarithms in base
   $10$ and base $e$, let's change the base. 

   $\begin{aligned}t&=\dfrac{\log_2(0.4)}{-0.05}\\\\&=\dfrac{1}{-0.05}\cdot \dfrac{\log(0.4)}{\log(2)}\\\\&\approx 26.44\end{aligned}$

   The bear population in the reserve will be at $2000$ bears after $26.44$ years.

   ---

3. Thinking about the problem

   We want to find the number of video views received after $6$ days.

   In other words, we are given a $d$ value of $6$ days and want to find the number of video views associated with that input, or $V(6)$.

   To do this, we can substitute ${6}$ in for $d$ and evaluate.

   $V({6})=4 ^{1.25({6})}$

   Evaluating the expression

   We can evaluate the expression as shown below.

   $\begin{aligned}V(6)&=4 ^{1.25(6)}\\\\&=4^{{7.5}}\\\\&=32{,}768\end{aligned}$

   After $6$ days, the video will receive $32{,}768$ views.

   ---

4. Thinking about the problem

   We want to know how many years, $t$, it will take for the area of the glacier, $A$, to decrease to $15$ square kilometers.

   So we need to find the value of $t$ for which $A=15$.

   Substituting $15$ in for $A$ in the model gives us the following equation.

   $15=45e^{-0.05t}$

   Solving the equation

   We can solve the equation as shown below.

   $\begin{aligned}45e^{-0.05t}&=15\\\\e^{-0.05t}&=\dfrac{1}{3}\\\\-0.05t&=\ln\left(\dfrac{1}{3}\right)\\\\t&=-20\cdot {\ln\left(\dfrac{1}3\right)}\end{aligned}$

   It will take $-20\cdot {\ln\left(\dfrac{1}3\right)}$ years for the area of the glacier to decrease to $15$ square kilometers.

   The expression above represents an exact solution to the equation. We can use a calculator to approximate the value of the expression, but this will be a rounded inexact answer.

   The answer

   The answer is $-20\cdot {\ln\left(\dfrac{1}3\right)}$ years.