Asked by lisa
What is the time to double when the initial investment is $1000 and the annual rate is 3.5%
Answers
Answered by
Reiny
solve for n ...
2000 = 1000(1.035)^n
YOu will have to use logs
2000 = 1000(1.035)^n
YOu will have to use logs
Answered by
Bosnian
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
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
Answered by
Bosnian
1.035^20.148791684000665883041124279516
=2
2*1000=2000
=2
2*1000=2000
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