Answer
a. $\$ 6,435.09$
b. about 8.24 years
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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(a)
$A(3)=5000(1+\displaystyle \frac{0.085}{4})^{4(3)}\\=5000(1.02125^{12})\\=6435.09$.
The amount after 3 years is $\$ 6,435.09.$
(b) Solve for t after inserting given values
$10,000=5000(1+\displaystyle \frac{0.085}{4})^{4t}$
$10,000=5000(1.02125^{4t}) \qquad.../\div 5000$
$ 2=1.02125^{4t}\qquad$ ... apply log() to both sides
$\log 2=4t\log 1.02125 \qquad/\div(4\log 1.02125)$
$t=\displaystyle \frac{\log 2}{4\log \mathrm{l}.02125}\approx 8.24$ years.
The investment will double in about 8.24 years.