Answer
$r\approx 0.0408\approx 4.08\%$
Work Step by Step
Recall:
For continuous compounding, the formula that gives the future value of an investment is $$A=Pe^{rt}$$
where
$P$=principal amount invested
$r$=annual interest rate
$t$=time in years
$A$=future value of the investment
In the given problem, we have:
$P=\$350$
$A=\$429.20$
$t=6$ years
Substitute these values into the formula above to obtain:
\begin{align*}
A&=Pe^{rt}\\
429.20&=350\left(e^{r\cdot 5}\right)\\\\
429.20&=350\left(e^{5r}\right)\\\\
\frac{429.20}{350}&=\frac{350e^{5r}}{350}\\\\
\frac{429.20}{350}&=e^{5r}\\\\
\end{align*}
Take the natural logarithm of both sides to obtain:
\begin{align*}
\ln{\left(\frac{429.20}{350}\right)}&=\ln{e^{5r}}\\\\
\ln{\left(\frac{429.20}{350}\right)}&=5r\ln{e}\\\\
\ln{\left(\frac{429.20}{350}\right)}&=5r(1)\\\\
\ln{\left(\frac{429.20}{350}\right)}&=5r\\\
\frac{\ln{\left(\frac{429.20}{350}\right)}}{5}&=\frac{5r}{5}\\\
\frac{\ln{\left(\frac{429.20}{350}\right)}}{5}&=r\\\
\end{align*}
Use a calculator to evaluate and obtain:
$r\approx 0.0408\approx 4.08\%$
