Question

Find the time required for an investment of 5000 dollars to grow to 6600 dollars at an interest rate of 7.5percent per year, compounded quarterly.Your answer is t =years.

Answer 1

STEP - BY - STEP EXPLANATION

What to find?

Time required for an investment of 5000 dollars to grow to 6600 dollars at an interest rate of 7.5%

Given:

Principal (P) = 5000

Amount (I) = 6600

Rate = 7.5% =0.075

t=?

n=4

Step 1

State the formula to use.

$A=p(1+\frac{r}{n})^{nt}$

Step 2

Substitute the values into the formula.

$6600=5000(1+\frac{0.075}{4})^{4t}$

Step 3

Divide both-side of the equation by 5000.

$\frac{6600}{5000}=\frac{5000}{5000}(1+\frac{0.075}{4})^{4t}$$1.32=(1+0.01875)^{4t}$$1.32=(1.01875)^{4t}$

Step 4

Take the natural log of both-side.

$In(1.32)=4tIn(1.01875)$

Step 5

Divide both-side by In(1.01875).

$4t=\frac{In(1.32)}{In(1.01875)}$$4t=14.9454$

Step 6

Divide both-side of the equation by 4

$t\approx3.7$

ANSWER

t = 3.7 years