# How do you solve  log_2 (x - 2) + 5 = 8 - log_2 4?

Jul 18, 2015

x=4

#### Explanation:

$\log \left(a b\right) = \log a + \log b$____(1)

${\log}_{a} \left(b\right) = x \implies {a}^{x} = b$_______(2)

${\log}_{2} \left(x - 2\right) + 5 = 8 - {\log}_{2} \left(4\right)$

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

$\implies {\log}_{2} \left(\left(x - 2\right) 4\right) = 3$ [since from (1)]

$\implies {2}^{3} = 4 \left(x - 2\right)$[since from (2)]

$\implies 8 = 4 x - 8$

$\implies 4 x = 16$

$\implies x = 4$