Question

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

Answer 1

`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`