continuous exponential growth model
[tex]A=Pe^{r\cdot t}[/tex]
since the size of the sample doubled:
A/P = 2
r = 8.5% = 0.085
e: eurler's number = 2.71828...
therefore,
[tex]\begin{gathered} \frac{A}{P}=e^{r\cdot t} \\ 2=e^{0.085\cdot t} \end{gathered}[/tex]
apply the law of exponents
[tex]\ln \mleft(2\mright)=0.085t[/tex][tex]t=\frac{\ln\left(2\right)}{0.085}[/tex]
So, t = 8.15467...
rounding to the nearest hundred
t = 8.15
