Answer
6 years and 4 months
Work Step by Step
If a principal $P$ is invested in an account paying an annual interest rate $r$, compounded $n$ times a year, then after $t$ years the amount $A(t)$ in the account is
$A(t)=P(1+\displaystyle \frac{r}{n})^{nt}$
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Solve for t after inserting given values
$8000=5000(1+\displaystyle \frac{0.075}{4})^{4t}$
$8000=5000(1.01875^{4t}) \qquad$ ... $/\div 5000$
$1.6=1.01875^{4t} \qquad$ ... apply log() to both sides
$\log 1.6=4t\log 1.01875 \qquad$ ... $/\div(4\log 1.01875)$
$t=\displaystyle \frac{\log 1.6}{4\log 1.01875}\approx 6.33$ years.
The investment will increase to $\$ 8000$ in approximately $6.33$ years,
(1/3 of a year = 4 months)
or 6 years and 4 months
