Question

How do you solve `2 log_4 x =1+ log_4 (x + 8)`?

Answer 1

Solve using the following properties:
• `alogn = logn^a`
• `log_a(n) - log_a(m) = log_a(n/m)`

Explanation:

`log_4(x^2) - log_4(x + 8) = 1`

`log_4((x^2)/(x + 8)) = 1`

`(x^2)/(x + 8) = 4`

`x^2 = 4(x + 8)`

`x^2 = 4x + 32`

`x^2 - 4x - 32 = 0`

`(x - 8)(x + 4) = 0`

`x = 8 and x = -4`

Checking in the original equation, we find that only `x = 8` works, so the `x = -4` is extraneous and you therefore will not include it inside the solution set.

Hopefully this helps!