User asks to solve part e of the problem.
We are asked to find the doubling time for the interest rate in the case of continuous compounding.
Recall that the formula for the continuously compounding interest rate is:
[tex]A=P\cdot e^{(r\cdot t\}}[/tex]
Then, we need to solve when the value "A" (Accrued value) doubles the principal P, that is:
[tex]2P=P\cdot e^{(r\cdot t\}}[/tex]
Dividing both sides by P and then applying natural logarithms (ln) on both sides, we get:
[tex]\begin{gathered} 2P=P\cdot e^{(r\cdot t\}} \\ 2=e^{(r\cdot t)} \\ \ln (2)=r\cdot t \\ \ln (2)=0.0395\cdot t \\ t=\frac{\ln l(2)}{0.0395} \\ t=17.548 \end{gathered}[/tex]
which gives us approximately 17.548 years which we round to two decimals as: 17.55 years
