Question

$7100$ dollars is placed in an account with an annual interest rate of $7.75\%$. How much will be in the account after $30$ years, to the nearest cent?

Answer 1

Step-1: Calculating exponential growth:

Exponential function is $y=a{b}^x$, where $y$ is amount , $a$ is starting value, $b$is exponential growth and $x$ number of years, $r$ is rate of interest.
Exponential growth $\left(b\right)=1+r$
$$\begin{gathered} =1+\frac{7.75}{100}\left[\because 7.75\%=\frac{7.75}{100}\right] \\ =1+0.0775 \\ =1.0775 \end{gathered}$$

Step-2: Calculating the amount after $30$ years:

$$\begin{gathered} \Rightarrow y=a{b}^x \\ \Rightarrow y=7100{\left(1.0775\right)}^{30} \\ \Rightarrow y=66646.402 \\ \approx \$66646.4 \end{gathered}$$

Hence, the amount after $30$ years in the account is $\$66646.4$.