Problem-Solving with Exponential Functions

We have to deal with problem-solving in many real-world situations. Therefore, it is important to know the steps you must take when problem-solving depending on the type of problem. Let's use exponential functions to solve the following problems:

Suppose [tex]$\$[/tex]4000[tex]$ is invested at a $[/tex]6\%[tex]$ interest rate compounded annually. How much money will there be in the bank at the end of five years? At the end of 20 years?

\ \textless \ strong\ \textgreater \ Read the problem and summarize the information:\ \textless \ /strong\ \textgreater \
$[/tex]\[tex]$4000$[/tex] is invested at a [tex]$6\%$[/tex] interest rate compounded annually. We want to know how much money we will have after 5 years and after 20 years.

Assign variables:
- [tex]$x=$[/tex] time in years
- [tex]$y=$[/tex] amount of money in the investment account

We start with [tex]$\$[/tex]4000[tex]$ and each year we apply a $[/tex]6\%[tex]$ interest rate on the amount in the bank. The pattern is that each year we multiply the previous amount by a factor of $[/tex]100\% + 6\% = 106\% = 1.06[tex]$.

\ \textless \ strong\ \textgreater \ Complete a table of values by continuing to multiply each year's amount by 1.06:\ \textless \ /strong\ \textgreater \

\begin{tabular}{|l|l|}
\hline Time (Years) & Investment Amount (\$[/tex]) \\
\hline 0 & 4000 \\
\hline 1 & 4240 \\
\hline 2 & 4494.40 \\
\hline 3 & 4764.06 \\
\hline 4 & 5049.91 \\
\hline 5 & 5352.90 \\
\hline
\end{tabular}

Using the table, we see that at the end of five years we have [tex]$\$[/tex]5352.90[tex]$ in the investment account.

In the case of five years, we don't need an equation to solve the problem, we can just multiply several times. However, if we want the amount at the end of 20 years, it becomes too difficult to constantly multiply. We can use a formula instead.

Since we take the original investment and keep multiplying by the same factor of 1.06, this means we can use exponential notation:
$[/tex][tex]$
y=4000(1.06)^x
$[/tex][tex]$

To find the amount after five years we use $[/tex]x=5[tex]$ in the equation:
$[/tex][tex]$
y=4000(1.06)^5=\$[/tex]
[tex]$
$\square$ (Notice, this matches what we found in the table for 5 years)

To find the amount after 20 years we use $x=20$ in the equation:
$[/tex]
y=4000(1.06)^{20}=\[tex]$
$[/tex][tex]$
$[/tex]\square[tex]$ (Round to two decimal places when referring to money)

To check our answers we can plug in some low values of $[/tex]x[tex]$ to see if they match the values in the table. The answers make sense because, after the first year, the amount goes up by $[/tex]\[tex]$240$[/tex] ([tex]$6\%$[/tex] of [tex]$\$[/tex]4000[tex]$). The amount of increase gets larger each year and that makes sense because the interest is $[/tex]6\%$ of an amount that is larger and larger every year.