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

Jan 9, 2016

$x = 1$

#### Explanation:

Typically, it is difficult to solve equations that combine polynomials, logs, and exponentials.

To make progress with this equation, you must apply log rules.

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

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

Because you have two equal log statements, then the pieces inside must be equal as well...

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

simplifying...

$\frac{2 x}{{2}^{x}} = x$

doing algebra...

$\frac{2 x}{x} = {2}^{x}$

$2 = {2}^{x}$

$x = 1$