How do you solve log_b(x+2)-log_b4=log_b3x?

Sep 26, 2016

$x = \frac{2}{11}$

Explanation:

${\log}_{b} \left(x + 2\right) - {\log}_{b} 4 = {\log}_{b} 3 x$

$\Leftrightarrow {\log}_{b} \left(\frac{x + 2}{4}\right) = {\log}_{b} 3 x$

or $\frac{x + 2}{4} = 3 x$

or $x + 2 = 12 x$

or $11 x = 2$ i.e. $x = \frac{2}{11}$