## Suppose that $20,000 is invested at 9% interest, compounded annually. After time t, in years, it grows to the amount A given by the f Question Suppose that$20,000 is invested at 9% interest, compounded annually. After time t, in years, it grows to the amount A given by the function A(t) = $20,000(1.09) a) Find the amount of time after which there will be$200,000 in the account.
b) Find the doubling time.
After what amount of time will there be $200,000 in the account? in progress 0 4 days 2021-07-19T06:26:11+00:00 1 Answers 0 views 0 ## Answers ( ) 1. Answer: Years = {log(total) -log(Principal)} ÷ log(1 + rate) Years = [log (200,000) -log (20,000)] / log (1.09) Years = [5.3010299957 -4.3010299957] / 0.037426497941 Years = 1 / 0.037426497941 Years = 26.7190374471 ************************************************* Finding Doubling Time : (We’ll calculate$100 becoming \$200)

Years = {log(total) -log(Principal)} ÷ log(1 + rate)

Years = log (200) -log (100) / log(1.09)

Years =  (2.3010299957 -2) / 0.037426497941

Years = .3010299957 / 0.037426497941

Years 8.0432317278

It takes 8.0432317278 years for money to double at 9% per year.

Source:

http://www.1728.org/compint2.htm

Step-by-step explanation: