# How do you solve log_b(x-1)+log_b3=log_bx?

Jul 14, 2016

I'm assuming you want to solve for x...

#### Explanation:

${\log}_{b} \left(3\right) = {\log}_{b} \left(x\right) - {\log}_{b} \left(x - 1\right)$

${\log}_{b} \left(3\right) = {\log}_{b} \left(\frac{x}{x - 1}\right)$

Since we're in equal bases we can eliminate:

$3 = \frac{x}{x - 1}$

$3 \left(x - 1\right) = x$

$3 x - 3 = x$

$2 x = 3$

$x = \frac{3}{2}$

Hopefully this helps!