Question

Dennis invest $4000 into an account that pays at a 3.5% interest rate compounded continuously. How many years will it take until Dennis has $6000 in his account? Round your answer to the nearest year 

Answer 1

Answer:

12 years

Explanation:

For an investment whose interest is compounded continuously, the amount in the account after t years is determined using the formula:

$A(t)=P_oe^{rt}\text{ where }\begin{cases}{P_o=\text{ The amount invested}} \\ r=Interest\text{ }{Rate} \\ {t}=Time\end{cases}$

In our given problem:

• A(t) = $6,000

,

• Po = $4000

,

• r = 3.5% = 0.035

We want to find the value of t.

Substitute the given values into the formula:

$6000=4000e^{0.035t}$

Then solve for t:

$\begin{gathered} \text{ Divide both sides by 4000} \\ \frac{6000}{4000}=\frac{4000e^{0.035t}}{4000} \\ 1.5=e^{0.035t} \\ \text{ Take the ln of both sides:} \\ \ln(1.5)=\ln(e^{0.035t}) \\ 0.035t=\ln(1.5) \\ \text{ Divide both sides by }0.035 \\ \frac{0.035t}{0.035}=\frac{\operatorname{\ln}(1.5)}{0.035} \\ t=11.58 \\ t\approx12\text{ years} \end{gathered}$

It will take Dennis 12 years (rounded to the nearest year) before he has $6,000 in his account.