1. The problem statement, all variables and given/known data This is a question I *might* have already got the answer to, I'd just like for someone very good with calculus and algebra to verify/answer the question themselves. To be more exact with the problem, we are putting 1 dollar (a variable principle) into a continuously compounded interest account at rate r. At what moment does one year's interest (e^r-1), equal our principle of 1 dollar per year. Or at what time can we stop putting the dollar in completely. I solved it as the most complicated substitution problem I, personally, have ever done. Feel free to do it any way you please, but I'm looking for both answers and/or possible mistakes in my math. 2. Relevant equations F=Pe^rt 3. The attempt at a solution Let F = Final, P = principle, r = interest rate, y = years. Given that: F = Pe^(ry) And P1 = ($)1 x y (P1 because I'm going to have to use another P later) Then replacing 1y for P we get F = ye^(ry) This gives us F for any time y, and rate r. But we want a specific F, to get that I first defined what P2 is required for the next year's interest to be equal to the 1 dollar we are putting in every year. P2(e^r-1) = 1 P2=1/(e^r-1) In order to substitute this into our original equation we must substitute P2 into a separate F = Pe^(rt2) We know that t2 = 1 because in the question we asked when does (e^r-1) of the *last* year equal $1. We get F = e^r/(e^r-1) Then substitute this final into our F = ye^ry we get e^r/(e^r-1) = ye^ry 0 = ye^ry - e^r/(e^r-1) at 15% interest I get y = 3.96 years. Graphing on my calculator because as far as I know that is unsolvable algebraically at 8% interest I get y = 7.27 years. Sounds too good to be true. Put $20k in a savings account for 4-7 years and you will raise your wage $20k/year for the rest of your life? Start investing! It will be free soon. lol. This is only my most recent approach to this problem. Other answers which I have debunked are (for 8% interest) 13.8 years, 20 years, and 7.5 years. I'd like confirmation, or a correct answer please.