Algebra help needed

Question

A person invests in an account with interest compound continuously according to the formula A=F(T)=10900e^0.028t where A is the amount in the account and T is in years. Find F(0)=$__,

to the nearest cent find the amount in the account after 9 years $____,to the nearest tenth of a year determine how long it will take for the investment to grow to $13200. _____ years.

Answer 1

F(0) = $10,900e^.028(0)

= 10,900e^0

= 10,900(1)

= $10,900

= the initial amount at time t=0

e^0 = 1, any number to the zero power = 1

in 9 years t = 9

F(9) = 10900e^.028(9) = 10900e^.252

= about $14,023.90

it reaches 13200 when

13,200= 10900e^.028t, solve for t

132/109 = e^.028t

.028t = ln(132/109)

t = ln(132/109)/.028

= 6.837644298

= about 6.8 years

= about 6 years and 10 months

Answer 2