[tex]\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad
\begin{cases}
A=\textit{accumulated amount}\to &\$14000\\
P=\textit{original amount deposited}\to &\$5000\\
r=rate\to 7\%\to \frac{7}{100}\to &0.07\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{monthly, thus twelve}
\end{array}\to &12\\
t=years
\end{cases}
\\\\\\
14000=5000\left(1+\frac{0.07}{12}\right)^{12t}\implies \cfrac{14000}{5000}=\left(1+\frac{0.07}{12}\right)^{12t}
\\\\\\
[/tex]
[tex]\bf \cfrac{14}{5}=\left( 1+\frac{7}{1200} \right)^{12t}\implies \cfrac{14}{5}=\left(\cfrac{1207}{1200} \right)^{12t}
\\\\\\
log\left(\frac{14}{5} \right)=log\left[ \left(\frac{1207}{1200} \right)^{12t} \right]\implies log\left(\frac{14}{5} \right)=12t\cdot log\left[ \left(\frac{1207}{1200} \right) \right]
\\\\\\
\cfrac{log\left(\frac{14}{5} \right)}{12\cdot log\left[ \left(\frac{1207}{1200} \right) \right]}=t\implies 14.75\approx t[/tex]
so, about 14 years and 9 months.
