Answer
a. $\$ 7328.73$
b. about $3\displaystyle \frac{1}{2}$ years
Work Step by Step
If a principal $P$ is invested in an account paying an annual interest rate $r$,
and the interest is compounded continuously, then the amount after t years is
$A(t)=Pe^{rt}$
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a.
$A(2)=6500e^{0.06(2)}\approx\$ 7328.73$
b. Solve for t after inserting given values
$8000=6500e^{0.06t} \qquad.../\div 6500$
$\displaystyle \frac{16}{13}=e^{0.06t} \qquad$ ... apply ln() to both sides
$\displaystyle \ln(\frac{16}{13})=0.06t \qquad$ ... $/\div 0.06$
$t=\displaystyle \frac{\ln(\frac{16}{13})}{0.06}\approx 3.46$.
The investment grows to $\$ 8000$ in about $3\displaystyle \frac{1}{2}$ years.