# How do you solve #log _ 2 (4x-8)=1#?

##### 1 Answer

Jan 31, 2016

# x = 5/2 #

#### Explanation:

using :

# log_b a = n hArr a = b^n # then

# log_2 (4x - 8 ) = 1 → 4x - 8 = 2^1 = 2 # so 4x - 8 = 2

# → 4x = 10 → x = 10/4 = 5/2 #