[tex]~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\dotfill &\$3000\\ P=\textit{original amount deposited}\dotfill &\$2000\\ r=rate\to 9.5\%\to \frac{9.5}{100}\dotfill &0.095\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{semiannually, thus twice} \end{array}\dotfill &2\\ t=years \end{cases}[/tex]
[tex]3000=2000\left(1+\frac{0.095}{2}\right)^{2\cdot t} \implies \cfrac{3000}{2000}=1.0475^{2t}\implies \cfrac{3}{2}=1.0475^{2t} \\\\\\ \log\left( \cfrac{3}{2} \right)=\log(1.0475^{2t})\implies \log\left( \cfrac{3}{2} \right)=t\log(1.0475^{2}) \\\\\\ \cfrac{\log\left( \frac{3}{2} \right)}{\log(1.0475^{2})}=t\implies 4.37\approx t\qquad \textit{about 4 years and 4 months and a half}[/tex]
