- #1
venik
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Homework Statement
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.
Homework Equations
F=Pe^rt
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.