$800 is deposited in a bank account which is compounded continuously at 8.5% annual interest rate. The future balance of the accourby the function: A = 800e0.085t. How long will it take for the initial deposit to double? Round off to the nearest tenth of a year.

Respuesta :

Given:

Function :

[tex]A=800e^{0.085t}[/tex]

Initial deposit =$800

Annual interest rate =8.5%

[tex]A=A_0e^{rt}[/tex]

Where,

[tex]\begin{gathered} A=\text{Amount after t time} \\ A_0=\text{Initial amount} \\ r=\text{interest rate} \\ t=\text{time} \end{gathered}[/tex][tex]\begin{gathered} r=\frac{8.5}{100} \\ r=0.085 \end{gathered}[/tex]

When deposit is double of initial deposit .

[tex]\begin{gathered} 2\times800=800e^{0.085t} \\ \frac{2\times800}{800}=e^{0.085t} \\ 2=e^{0.085t} \\ \ln 2=\ln e^{0.085t} \\ 0.085t=0.69314 \\ t=\frac{0.69314}{0.085} \\ t=8.15 \end{gathered}[/tex]

So after 8.15 year initial amount will be double.