# How do you solve 7^(x - 2) = 5^x?

Feb 24, 2016

Convert to logarithmic form.

#### Explanation:

$\log {7}^{x - 2} = \log {5}^{x}$

$\left(x - 2\right) \log 7 = x \log 5$

$x \log 7 - 2 \log 7. = x \log 5$

$x \log 7 - x \log 5 = 2 \log 7$

$x \left(\log 7 - \log 5\right) = 2 \log 7$

$x = \frac{2 \log 7}{\log 7 - \log 5}$

You can simplify the answer further by using the rules $a \log n = \log {n}^{a}$ and ${\log}_{a} n - {\log}_{a} m = {\log}_{a} \left(\frac{n}{m}\right)$

$x = \log {7}^{2} / \log \left(\frac{7}{5}\right)$

$x = \log \frac{49}{\log} \left(\frac{7}{5}\right)$

Hopefully this helps!