Question

Oliver invested $970 in an account paying an interest rate of 7 1/2 % compounded continuously. Carson invested $970 in an account paying an interest rate of 7 3/8% compounded annually. To the nearest of a hundredth of a year, how much longer would it take for Carson's money to double than for Oliver's money to double?

Answer 1

Answer:

0.50 or about half a year longer.

Step-by-step explanation:

We can write an equation to model bot investments.

Oliver invested $970 in an account paying an interest rate of 7.5% compounded continuously.

Recall that continuous compound is given by the equation:

$A = Pe^{rt}$

Where A is the amount afterwards, P is the principal amount, r is the rate, and t is the time in years.

Since the initial investment is $970 at a rate of 7.5%:

$A = 970e^{0.075t}$

Carson invested $970 in an account paying an interest rate of 7.375% compounded annually.

Recall that compound interest is given by the equation:

$\displaystyle A = P\left(1+\frac{r}{n}\right)^{nt}$

Where A is the amount afterwards, P is the principal amount, r is the rate, n is the number of times compounded per year, and t is the time in years.

Since the initial investment is $970 at a rate of 7.375% compounded annually:

$\displaystyle A = 970\left(1+\frac{0.07375}{1}\right)^{(1)t}=970(1.07375)^t$

When Oliver's money doubles, he will have $1,940 afterwards. Hence:

$1940= 970e^{0.075t}$

Solve for t:

$\displaystyle 2 = e^{0.075t}$

Take the natural log of both sides:

$\ln\left (2\right) = \ln\left(e^{0.075t}\right)$

Simplify:

$\ln(2) = 0.075t\Rightarrow \displaystyle t = \frac{\ln(2)}{0.075}\text{ years}$

When Carson's money doubles, he will have $1,940 afterwards. Hence:

$\displaystyle 1940=970(1.07375)^t$

Solve for t:

$2=(1.07375)^t$

Take the natural log of both sides:

$\ln(2)=\ln\left((1.07375)^t\right)$

Simplify:

$\ln(2)=t\ln\left((1.07375)\right)$

Hence:

$\displaystyle t = \frac{\ln(2)}{\ln(1.07375)}$

Then it will take Carson's money:

$\displaystyle \Delta t = \frac{\ln(2)}{\ln(1.07375)}-\frac{\ln(2)}{0.075}=0.4991\approx 0.50$

About 0.50 or half a year longer to double than Oliver's money.