Jessica deposited $785 into an account that pays 6.2% interest compounded continuously. How long will it take until Jessica has $1000?
785 e^(.062t) = 1000
e^(.02t) = 1.27388..
take ln of both sides
ln(e^(.062t)) = ln 1.27388...
.062t lne = ln 1.27388.. , but lne = 1
t = ln1.27388.. /.062 = appr 3.9 years