[tex]\bf \qquad \textit{Simple Interest Earned}\\\\
A = Pe^{rt}\qquad
\begin{cases}
A=\textit{accumulated amount}\to 1000000\\
P=\textit{original amount deposited}\to& \$1300\\
r=rate\to 17\%\to \frac{17}{100}\to &0.017\\
t=years
\end{cases}[/tex]
[tex]\bf 1000000=1300e^{0.17t}\implies \cfrac{1000000}{1300}=e^{0.17t}\implies \cfrac{10000}{13}=e^{0.17t}
\\\\\\
ln\left( \frac{1000000}{1300} \right)=ln(e^{0.17t})\implies ln\left( \frac{1000000}{1300} \right)=0.17t
\\\\\\
\cfrac{ln\left( \frac{1000000}{1300} \right)}{0.17}=t\impliedby years[/tex]
now, how many days? simply multiply t * 365.
