Asked by tailor
How long it will take for an investment of 1000 dollars to double in value if the interest rate is 9.5 percent per year, compounded continuously?
Answers
Answered by
Reiny
1000 e^(.095t) = 2000
e^.095t = 2
ln both sides
ln (e^(.095t)) = ln 2
.095t lne = ln2, but remember lne = 1
.095t = ln2
t = ln2/.095 = appr 7.3 years
e^.095t = 2
ln both sides
ln (e^(.095t)) = ln 2
.095t lne = ln2, but remember lne = 1
.095t = ln2
t = ln2/.095 = appr 7.3 years
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