Larry has $2600 to invest and needs $3000 in 11 years. What annual rate of return will he need to get in order to accomplish his goal, if interest is compounded continuously?

Answer 1

let
2600(e^(11i) = 3000
e^(11i) = 1.153846
11i = ln(1.153846)
i = .013

a mere 1.3%