[tex]\qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &250\\ r=rate\to 16\%\to \frac{16}{100}\dotfill &0.16\\ t=\textit{elapsed time} \end{cases} \\\\\\ A=250(1 + 0.16)^{t}\implies A=250(1.16)^t \\\\[-0.35em] ~\dotfill[/tex]
[tex]A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\dotfill &1000\\ P=\textit{initial amount}\dotfill &250\\ r=rate\to 16\%\to \frac{16}{100}\dotfill &0.16\\ t=\textit{elapsed time} \end{cases} \\\\\\ 1000=250(1.16)^t\implies \cfrac{1000}{250}=1.16^t\implies 4=1.16^t \\\\\\ \log(4)=\log(1.16^t)\implies \log(4)=t\log(1.16) \\\\\\ \cfrac{\log(4)}{\log(1.16)}=t\implies \stackrel{\textit{about 9 years and 4 months}}{9.34\approx t}[/tex]
