precal question

Question

Find the time required for an investment of $4000 to grow to $7000 at an interest rate of 7.5% per year, compounded quarterly. (Round your answer to two decimal places.)

 yr

Answer 1

Hi Vivianna

7.53 years

You can use the Formula for Compounded Interest

A = P(1 + r/n)^nt

A = Amount (Goal) = 7000

P = Original Investment = 4000

r = rate write is as a decimal = .075

n = number of times interest is compounded per unit t = 4 (quarterly)

t = time in years

7000 = 4000(1 + .075/4)^4t

Divide both sides by 4000

7000/4000 = (1 + .075/4)^4t

1.75 = (1 + .0.01875)^4t

1.75 =(1.01875)^4t

To solve for the exponent you will to use logs, and you can look up the rules for logs

log(1.75)/log(1.01875) = 4t

30.12511695 = 4t

Divide both sides by 4

30.12511695/4 = 7.53 years

You can check this by plugging into the Compound Interest Formula to calculate A

A = P(1 + r/n)^nt

A = 4000(1 + .075/4)^4(7.5313)

A = 4000(1.01875)^30.125

A = 6999.985