Question

How do you solve `log_2 (x-2)+log_2 (x+4)=4`?

Answer 1

If `log_2(x−2)+log_2(x+4)=4`
then `x = 4`
as can be seen by replacing `x` with `4` in the sum from the original equation.
`log_2(4-2) + log_2(4+4)`
= `log_2(2) + log_2(8)`

Since `2^1 = 2` and `2^3 = 8`
(and therefore `log_2(2) = 1` and `log_2(8) = 3`

when x = 4,
`log_2(x−2)+log_2(x+4)= 1+3 = 4`

How to get there :
Let `2^j = x-2` (i.e. `j = log_2(x-2)`)
and
let `x^k = x+4` (i.e. `k = log_2(x+4)`)

From the given equation
`j + k = 4`

`(2^j) * (2^k) = (x-2)(x+4)`
combining, we get:
`2^(j+k) = x^2 + 2x - 8`
but, since `j+k = 4`
`2^4 = 16 = x^2 + 2x - 8`
therefore
`x^2 + 2x -24 = 0`

Simple factoring then gives
`(x - 4) (x + 6) = 0`
so
`x = 4` or `x = -6`

If `x = -6`
then
`log_2(x−2)` and `log_2(x+4)=4`
would be `log_2`'s of negative values (impossible).

Therefore
`x = 4`