we know that
The compound interest formula is equal to
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]
[tex]A=P(1+\frac{r}{n})^{nt}[/tex]where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
n is the number of times interest is compounded per year
in this problem we have
P=9,136.00
A=14,942.00
r=11%=11/100=0.11
n=4
substitute in the formula above
[tex]\begin{gathered} 14,942=9,136(1+\frac{0.11}{4})^{4t} \\ \frac{14,942}{9,136}=((\frac{4.11}{4})^4)^t \\ \\ \text{Apply log both sides} \\ \log (\frac{14,942}{9,136})=t\cdot\log (\frac{4.11}{4})^4 \\ \text{solve for t} \\ t=\log (\frac{14,942}{9,136})\text{ : }\cdot\log (\frac{4.11}{4})^4 \\ t=4.53\text{ years} \end{gathered}[/tex]therefore
the answer is
