# How do you solve 7^x=5^(x-4)?

Aug 22, 2015

Take $\log$ of both sides and use properties of $\log$ of exponents to find:

$x = \frac{- 4 \log \left(5\right)}{\log \left(7\right) - \log \left(5\right)} \approx - 19.133$

#### Explanation:

Take $\log$ of both sides to get:

$x \log \left(7\right) = \log \left({7}^{x}\right) = \log \left({5}^{x - 4}\right) = \left(x - 4\right) \log \left(5\right)$

$= x \log \left(5\right) - 4 \log \left(5\right)$

Subtract $x \log \left(5\right)$ from both sides to get:

$x \left(\log \left(7\right) - \log \left(5\right)\right) = - 4 \log \left(5\right)$

Divide both sides by $\left(\log \left(7\right) - \log \left(5\right)\right)$ to get:

$x = \frac{- 4 \log \left(5\right)}{\log \left(7\right) - \log \left(5\right)} \approx - 19.133$