#### Basic math problem

You invest $X per year in an account that yields R in gains. How long does it take for the gain in a year to be greater than $X?

Another way of asking this is 'when does R times the future value of investing X each year exceed X?'

The future value of a regular yearly investment where N = number of years, X = yearly investment, and R = growth rate of the investment is:

\[FV = \frac{X*((1+R)^N - 1)}{R}\]

What we're looking for is the number of years it takes for R times that to exceed X. That is, we want to solve:

\[\frac{R*X*((1+R)^N - 1)}{R} > X\]

Noticing that the R's cancel in numerator and denominator and dividing both sides by X you get:

\[((1+R)^N - 1) > 1\]

Adding 1 to both sides:

\[((1+R)^N) > 2\]

Simplifying:

\[log((1+R)^N) > log(2)\]

\[N*log((1+R)) > log(2)\]

\[N > \frac{log(2)}{log(1+R)}\]

This is kind of cool. You might not recognize it right away, but that says 'N is greater than the doubling time of the investment'. It's really cool that it works out that way. Because of some pretty good approximations that work out, that means that the investment growth takes over the new money moved in after roughly '72 divided by annual interest rate' years.

For a quick concrete example of what this means...say you invest $10,000 per year into an account yielding 6%. The time it takes for the 6% yield each year to exceed $10,000 is log(2)/log(1.06) which is ~12 years.

#### Simple plot

Here's a simple interactive plot showing the breakdown between money invested and money from gains for an annual $10,000 investment using the interest rate that you enter below:

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