algebra help needed

Question

A person invests in an account with intrest compounded continuously according to the formula A=F(T)=1090oe^0.028t where A is the amount in the account and T is in years F(0)=$10,900 , to the nearest cent find the amount in the account after 9 years $___ and to the nearest tenth of the year determine how long it will take for the investment to grow $13200 ____ years

Answer 1

Hello Angeles;

a) To find the amount compounded in 9 years, we plug in t = 9 there in the equation.

A = 1090 x e^(0.028t)

Calculator needed;

A = 1090 x e^{(0.028 x 9)}

A = $1404.06 (nearest cent)

b) to find the time, we will rearrange the equation;

A = 1090 x e^(0.028t)

13200 = 1090 x e^(0.028t) solve for t.

Divide both sides by 1090 and take ln of both sides.

0.028t = ln(13200/1090) (use calculator)

t ∼ 7.018 years.