# How do you solve log_2 (x - 1) – log_2 (x – 4) = log_2 4?

Dec 1, 2015

$x = 5$

#### Explanation:

We have that $\log \left(a\right) - \log \left(b\right) = \log \left(\frac{a}{b}\right)$, so

${\log}_{2} \left(x - 1\right) - {\log}_{2} \left(x - 4\right) = {\log}_{2} \left(\frac{x - 1}{x - 4}\right)$

The equation has thus become

${\log}_{2} \left(\frac{x - 1}{x - 4}\right) = {\log}_{2} \left(4\right)$

This means that

$\frac{x - 1}{x - 4} = 4$

If $x \setminus \ne 4$, multiply both sides by $x - 4$, obtaining

$x - 1 = 4 x - 16$, and solve as usual

$- 3 x = - 15 \setminus \to x = 5$

If $x = 5$, the expression becomes
${\log}_{2} \left(5 - 1\right) - {\log}_{2} \left(5 - 4\right) = {\log}_{2} \left(4\right)$
${\log}_{2} \left(4\right) - {\log}_{2} \left(1\right) = {\log}_{2} \left(4\right)$
which is true, since ${\log}_{a} \left(1\right) = 0$ for every $a$.