What is the time to double when the initial investment is $1000 and the annual rate is 3.5%

Answer 1

solve for n ...
2000 = 1000(1.035)^n

YOu will have to use logs

Answer 2

1000*1.035^n=2000 Divide with 1000

1.035^n=2

n*log(1.035)=log(2)

Divide with log (1.035)

n=log(2)/log(1.35)
=20.148791684000665883041124279516

OR

1000*1.035^n=2000 Divide with 1000

1.035^n=2

n*ln(1.035)=ln(2)

Divide with ln (1.035)

n=ln(2)/ln(1.35)=20.148791684000665883041124279516

Answer 3

1.035^20.148791684000665883041124279516
=2

2*1000=2000