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

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Answer 1 · 2016-01-31T10:05:48

$x = 5/2$

using : $log_b a = n hArr a = b^n$

then $log_2 (4x - 8 ) = 1 → 4x - 8 = 2^1 = 2$

so 4x - 8 = 2 $→ 4x = 10 → x = 10/4 = 5/2$

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