Question

You deposit $8000 in an account earning 5% interest compounded continuously. The amount of money in the account after t years is given by Aſt) 8000e0.05 How much will you have in the account in 4 years? $ decimal places. How long will it be until you have $14900 in the account? How long does it take for the money in the account to double?

Answer 1

`\begin{gathered} A(t)=8000e^{0.05\text{ t}} \\ \text{when t= 4 years, one has} \\ A(4)=8000e^(0.05\cdot4) \\ A(4)=8000e^(0.2) \\ A(4)=8000(1.22) \\ A(4)=9760 \\ In\text{ 4 years you will have 9760 dollars} \end{gathered}`
`\begin{gathered} \text{Now, we must find the time when you have 14900. In this case:} \\ 14900=8000e^{0.05\text{ t}} \\ \text{then,} \\ e^{0.05\text{ t}}=(14900)/(8000) \\ e^{0.05\text{ t}}=1.8625 \\ 0.05t=Ln\text{ (1.8625) }\longleftarrow\text{ natural logarithm} \\ 0.05t=0.6219 \\ t=(0.6219)/(0.05) \\ t=12.43 \\ \text{Hence in 12.43 years you will have 14900 dollars} \end{gathered}`
`\begin{gathered} \text{Now, we must find when you have 8000}\cdot2=16000\text{ in your account, Hence,} \\ 16000=8000e^{0.05\text{ t}} \\ e^{0.05\text{ t}}=(16000)/(8000) \\ e^{0.05\text{ t}}=2 \\ 0.05t=Ln(2)\text{ }\longleftarrow n\text{atural logarithm} \\ 0.05t=0.6931 \\ t=(0.6931)/(0.05) \\ t=13.86 \\ \text{THEREFORE, in 13.86 years you will have 16000 dollars} \end{gathered}`