# How do you solve log_b9+log_bx^2=log_bx?

Jul 25, 2016

$x = \frac{1}{9}$

#### Explanation:

Since $\log a + \log b = \log \left(a b\right)$, you can write

${\log}_{b} 9 {x}^{2} = {\log}_{b} x$

that's equivalent to:

$9 {x}^{2} = x \mathmr{and} x > 0$

and the solution is:

$x = \frac{1}{9}$