# How do you solve for x in 18^(x-2) = 13^(-2x)?

Sep 28, 2015

$x = \frac{2 \ln \left(18\right)}{\ln \left(18\right) + 2 \ln \left(13\right)}$

#### Explanation:

Since 13 is a prime and 18 is a compound number without 13 as a prime factor you'll never be able to equate bases, therefore, we'll need to apply logs. I'm using the natural base due to preference you can use whatever base you want - that being said, I'd recommend using natural or decimal.

$\ln \left({18}^{x - 2}\right) = \ln \left({13}^{- 2 x}\right)$
$\left(x - 2\right) \ln \left(18\right) = \left(- 2 x\right) \ln \left(13\right)$
$x \ln \left(18\right) - 2 \ln \left(18\right) = - 2 x \ln \left(13\right)$

Isolating x

$x \ln \left(18\right) + 2 x \ln \left(13\right) = 2 \ln \left(18\right)$
$x \cdot \left(\ln \left(18\right) + 2 \ln \left(13\right)\right) = 2 \ln \left(18\right)$
$x = \frac{2 \ln \left(18\right)}{\ln \left(18\right) + 2 \ln \left(13\right)}$

You could break down some of these logs further but since the problem gave 18 and 13 as numbers, I'd leave the answer in terms of their logs.