how long does it take an investment of $8000 to double if it is invested at 7% interest compounded continuously. round to nearest year.
3 answers
e^(1.07t) = 2
Future value = Principle × e^(interest rate × time) =
16,000 = 8,000e^(0.07t)
2 = e^(0.07t)
ln (2) = 0.07t
0.693 = 0.07t
t = 0.693/0.07 = 9.90 years which rounds to 10 years.
QED
16,000 = 8,000e^(0.07t)
2 = e^(0.07t)
ln (2) = 0.07t
0.693 = 0.07t
t = 0.693/0.07 = 9.90 years which rounds to 10 years.
QED
Sorry about my typo. e^.07t is correct
2 answers