same here, the interest earned is 210, the principal is 2200, so the accumulated amount is 2410.
[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 &\$2410\\
P=\textit{original amount deposited}\to &\$2200\\
r=rate\to 10\%\to \frac{10}{100}\to &0.10\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{monthly, thus twelve}
\end{array}\to &12\\
t=years
\end{cases}[/tex]
[tex]\bf 2410=2200\left(1+\frac{0.1}{12}\right)^{12t}\implies \cfrac{2410}{2200}=\left(1+\frac{1}{120} \right)^{12t}
\\\\\\
\cfrac{241}{220}=\left(\cfrac{121}{120} \right)^{12t}\implies log\left( \frac{241}{220} \right)=log\left[ \left(\frac{121}{120} \right)^{12t} \right]\\\\\\
log\left( \frac{241}{220} \right)=12t\cdot log\left(\frac{121}{120} \right)\implies
\cfrac{log\left( \frac{241}{220} \right)}{12 log\left(\frac{121}{120} \right)}=t\implies 0.915 \approx t[/tex]
so, about 10 months and 27 days.
