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

0.50 or about half a year longer.

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+(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+(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 = (\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 = (\ln(2))/(\ln(1.07375))`

Then it will take Carson's money:

`\displaystyle \Delta t = (\ln(2))/(\ln(1.07375))-(\ln(2))/(0.075)=0.4991\approx 0.50`

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