so, we know the earned interest is 120, and we know the original deposited amount is 900 bucks, so, the accumulated amount will then be 900 + 120 or 1020 bucks.
[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 &\$1020\\
P=\textit{original amount deposited}\to &\$900\\
r=rate\to 6\%\to \frac{6}{100}\to &0.06\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years
\end{cases}[/tex]
[tex]\bf 1020=900\left(1+\frac{0.06}{1}\right)^{1\cdot t}\implies \cfrac{1020}{900}=(1.06)^t\implies \cfrac{102}{90}=(1.06)^t
\\\\\\
log\left(\frac{102}{90} \right)=log(1.06^t)\implies log\left(\frac{102}{90} \right)=t\cdot log(1.06)
\\\\\\
\cfrac{log\left(\frac{102}{90} \right)}{log(1.06)}=t\implies 20.92 \approx t[/tex]
