[tex]\qquad \textit{Amount for Exponential Decay} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{current amount}\dotfill &\$2300\\ P=\textit{initial amount}\dotfill &\$15100\\ r=rate\to 12.25\%\to \frac{12.25}{100}\dotfill &0.1225\\ t=years \end{cases} \\\\\\ 2300=15100(1 - 0.1225)^{t}\implies \cfrac{2300}{15100}=(1 - 0.1225)^{t}\implies \cfrac{23}{151}=0.8775^t[/tex]
[tex]\log\left( \cfrac{23}{151} \right)=\log(0.8775^t)\implies \log\left( \cfrac{23}{151} \right)=t\log(0.8775) \\\\\\ \cfrac{~~\log\left( \frac{23}{151} \right)~~}{\log(0.8775)}=t\implies 14.4\approx t\qquad \textit{about 14 years and 146 days}[/tex]
