Question

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?

Answer 1

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