[tex]\bf \qquad \textit{Amount for Exponential Decay, \boxed{\textit{3 years}}} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&15000\\ P=\textit{initial amount}\dotfill &P\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &3\\ \end{cases} \\\\\\ 15000=P(1-r)^3 \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf \qquad \textit{Amount for Exponential Decay, \boxed{\textit{7 years}}} \\\\ A=P(1 - r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill&8000\\ P=\textit{initial amount}\dotfill &P\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\dotfill &7\\ \end{cases} \\\\\\ 8000=P(1-r)^7 \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]\bf 15000=P(1-r)^3\implies \cfrac{15000}{(1-r)^3}=P\qquad \qquad \stackrel{\textit{doing some substitution}}{8000=\left( \cfrac{15000}{(1-r)^3} \right)(1-r)^7} \\\\\\ \cfrac{8000}{15000}=\cfrac{(1-r)^7}{(1-r)^3}\implies \cfrac{8}{15}=(1-r)^{7-3}\implies \cfrac{8}{15}=(1-r)^{4}[/tex]
[tex]\bf \sqrt[4]{\cfrac{8}{15}}=1-r\implies r=1-\sqrt[4]{\cfrac{8}{15}}\implies r\approx 0.15 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ 15000=P(1-r)^3\implies 15000=P(1-0.15)^3\implies 15000=P(0.85)^3 \\\\\\ \cfrac{15000}{0.85^3}=P\implies 24424.995\approx P[/tex]
