Exponential Model Word Problems Practice

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.

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.

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.

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.