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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 

User Chris Schwerdt
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1 Answer

11 votes
11 votes

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} \\ (6000)/(4000)=(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 \\ (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.