# For the logarithms: 2^(x+1) = 3^x, how do you solve for x?

Oct 4, 2015

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

#### Explanation:

Take the log of the two sides

$\ln \left({2}^{x + 1}\right) = \ln \left({3}^{x}\right)$

Pass the exponents to the front of the log

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

Expand the left side

$x \ln \left(2\right) + \ln \left(2\right) = x \ln \left(3\right)$

Isolate x

$\ln \left(2\right) = x \ln \left(3\right) - x \ln \left(2\right)$

Put x in evidence

$\ln \left(2\right) = x \left(\ln \left(3\right) - \ln \left(2\right)\right)$

Pass that dividing

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

Or, if you want, you can use log properties to change that into other logs, like

$x = \ln \frac{2}{\ln \left(\frac{3}{2}\right)}$ or $x = {\log}_{\frac{3}{2}} \left(2\right)$