# How do you solve  3^(2x) = 4^(x+1)?

Sep 18, 2015

$x = \frac{\log \left(2\right)}{\log \left(3\right) - \log \left(2\right)}$

#### Explanation:

Take the log of both sides
$\log \left({3}^{2 x}\right) = \log \left({4}^{x + 1}\right)$

Remember the property that $\log \left({m}^{n}\right) = n \log \left(m\right)$

$2 x \log \left(3\right) = \left(x + 1\right) \log \left(4\right)$

Expand

$2 x \log \left(3\right) = x \log \left(4\right) + \log \left(4\right)$

Put everything with an x to the same side

$2 x \log \left(3\right) - x \log \left(4\right) = \log \left(4\right)$

Put x on evidence, and put that 2 back in the log

$x \left(2 \log \left(3\right) - \log \left(4\right)\right) = \log \left(4\right)$

Isolate x

$x = \log \frac{4}{2 \log \left(3\right) - \log \left(4\right)}$

Remember that $4 = {2}^{2}$, so #log(4) = 2log(2)

$x = \frac{2 \log \left(2\right)}{2 \log \left(3\right) - 2 \log \left(2\right)}$

Put 2 in evidence and cancel it

$x = \frac{\log \left(2\right)}{\log \left(3\right) - \log \left(2\right)}$