Suppose that P is invested in a savings account in which int

Suppose that P is invested in a savings account in which interest, k, is compounded at 3% per year. The balance P(t) after time t, in years, is P(t)=P e ^kt.

a) What is the exponential growth function in terms of P and 0.03?

b)If $6000 is invested what is the balance after 9 years?

c)When will an investment of $6000 double itself?

Solution

P(t)=P e^ kt where k is the compound interest and t time in years

a) k = 3% = 0.03 So, exponential function : P(t) = Pe^0.03t

b) P = $6000 ; t = 9 yrs

P(9) = 6000e^(0.03*9) = 6000*1.31 = $ 7860

c) $ 6000 nvestment doubles.

12000 = 6000e^0.03t . Calculate t

2 = e^(0.03t)

Taking natural log on both sides:

ln2 = 0.03t ---> t = 23.1 years