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
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