# How do you solve log(x-1) - log2 = 12 + log3x ?

Aug 3, 2018

$x = \frac{1}{1 - 3 c}$ where $c = {e}^{12 + \log \left(2\right)}$

#### Explanation:

We will using that $\log \left(a\right) - \log \left(b\right) = \log \left(\frac{a}{b}\right)$ writing

$\log \left(x - 1\right) - \log \left(3 x\right) = 12 + \log \left(2\right)$ so

$\log \left(\frac{x - 1}{3 x}\right) = 12 + \log \left(2\right)$ and

$\frac{x - 1}{3 x} = {e}^{12 + \log \left(2\right)}$

let

$c = {e}^{12 + \log \left(2\right)}$

then we get

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

Multiplying by $3 x$

$x - 1 = 3 x c$ so $x - 3 x c = 1$

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

but the is negative, so no solutions!