Question

7) A savings account earns interest that is compounded continuously at 3%. If you deposit $1000,how long does it take for your principal investment to triple? Round to the nearest hundredth.

Answer 1

The equation for the amount of money A in an account that earns interest compounded continuously, is:

`A=A_0e^(rt)`

Where A_0 represents the initial deposit, r is the interest rate and t represents time.

If the annual interest is 3%, then the interest rate r is:

`r=(3)/(100)=0.03`

If the amount of money triples after a time t, then the savings account will be worth $3000 at that time. Substitute A=3000, A_0=1000 and r=0.03:

`3000=1000\cdot e^(0.03t)`

Isolate t to find the amount of time needed for the investment to triple:

`\begin{gathered} \Rightarrow(3000)/(1000)=e^(0.03t) \\ \Rightarrow3=e^(0.03t) \end{gathered}`

Take the natural logarithm to both sides of the equation:

`\ln (3)=\ln (e^(0.03t))`

The natural logarithm of e^0.03t is equal to 0.03t. Then:

`\begin{gathered} 0.03t=\ln (3) \\ \Rightarrow t=(\ln (3))/(0.03) \end{gathered}`

Use a calculator to evaluate the expression:

`\begin{gathered} \Rightarrow t=(1.0986\ldots)/(0.03) \\ \Rightarrow t=36.62\ldots \end{gathered}`

Therefore, to the nearest hundredth, the amount of time needed for the investment to triple, is:

`36.62\text{ years}`