# Thread: [SOLVED] Checking my x solutions?

1. ## [SOLVED] Checking my x solutions?

Hi again all,

I've got:

log(base3) (x) + log(base3) (x + 2) = log(base3) (8)

I think I've solved it by turning into

x^2 + 2x - 8 = 0

so I get x = 2 and x = -4

The x = -4 doesn't work in the original equation but it does work in subsequent steps in the equation (for example in log(base3) (x(x + 2)) = log(base3) (8))

Does this matter or is this still not a solution?

2. Originally Posted by DannyMath
Hi again all,

I've got:

log(base3) (x) + log(base3) (x + 2) = log(base3) (8)

I think I've solved it by turning into

x^2 + 2x - 8 = 0

so I get x = 2 and x = -4

The x = -4 doesn't work in the original equation but it does work in subsequent steps in the equation (for example in log(base3) (x(x + 2)) = log(base3) (8))

Does this matter or is this still not a solution?
The domain is $\displaystyle x > 0$ because a negative log gives a complex number. Hence -4 is not in the domain so it cannot be a real solution