Algebra 2 - Logirithims- What did I do wrong - Easy Best ANSWER?

Question

How do you solve- what is the answer to:

log2(x) + log2(x + 2) = 3



I did this:

log2(x) + log2(x + 2) = 3
log2((x)(x+2)) = 3
log2(x^2 +2x) = 3
2^3 = x^2 +2x
8 = x^2 +2x
0 = x^2+2x - 8
o = (x+4)(x-2)
x = 4, -2

and I am wrong - please Help!!! Ist correct answer gets best answer

Do both -4 and 2 work in the original equation?

Answer 1

Let log be log to base 2 in the following:-
log x + log(x + 2) = 3
log [ (x) (x + 2) ] = 3
(x) (x + 2) = 2³
(x) (x + 2) = 8
x² + 2x - 8 = 0
(x + 4) (x - 2) = 0
x = - 4 , x = 2

Accept + ve value of x = 2

Answer 2

Look at your next to last step where you have factored the polynomial

0 = (x + 4) * (x -2)

This means either x + 4 = 0 or x - 2 = 0.

So, if x + 4 = 0, then x = -4, and
if x - 2 = 0, the x = 2.

You just got the signs turned around.

Answer 3

0 = (x+4)(x-2)
Doesn't that make x = -4 and 2