# Ask Uncle Colin: How do I find the power?

Dear Uncle Colin,

The equation $67.5 = 10(1.0915)^{10-n} + 30(1.0915)^{10-2n}$ cropped up in a question. Excel can solve that numerically, but I can’t solve it on paper! Any ideas?

Problems Occur When Exponentials Recur

Hi, POWER, and thanks for your message!

That’s an ugly one. First thing: beware of the booby-trap; you might be tempted to take logs of everything straight away, but it doesn’t work. The right-hand side is the sum of two terms, and the logarithm of all of that isn’t any simpler than the original expression. (You didn’t think it was log(first term) + log(second term), did you?)

Instead, this is an ungainly quadratic - the numbers are horrible, so I’m going to throw in some letters to make it simpler. Let’s let $k = 1.0915^{10}$ and $x = 1.0915^{-n}$. This makes the equation $67.5 = 10kx + 30 kx^2$, which is a quadratic in $x$ that can be (pretty) easily solved.

If we have $30kx^2 + 10kx - 67.5 = 0$, I’d probably reach straight for the formula with $a = 30k$, $b = 10k$ and $c = -67.5$, to give $x = \frac{-10k \pm \sqrt{100k^2 + 8100k}}{60k}$. That simplifies to $x = \frac{-1 \pm \sqrt{1 +\frac{81}{k}}}{6}$.

$x$ needs to be positive, so only the positive root makes sense, and we’ve got $x = \frac{1}{6}\left(-1 + \sqrt{1 + 81(1.0915)^{-10}}\right)$. That turns out to be 0.815790 and change (to the disgust of the Mathematical Ninja, I used a calculator.)

We’ve now got $0.815790 = 1.0915^{-n}$, which is a fairly straightforward logs problem with the answer $n = -\frac{\log(0.815790)}{\log(1.0915)} = 2.32543$.

Phew! Hope that helps,

Uncle Colin

* Edited 2017-01-18 to fix LaTeX.