# Logarithmic Equation: How to solve for x

Equation: $$\log_a (x) + \log_a (x-4) = \log_a (x+6)$$

Progress

$$\log_a (x^2-4x) = \log_a (x+6)$$ $$x^2-5x-6=0$$

Delta

$$x1= 6$$

$$x2=-1$$

Your first step is correct. Now, if you search only real solutions, you have $$x^2-4x=x+6$$ can you solve? (be care to the the acceptability of the solutions).
• well. Now you have to verify that the $\log$ are defined . i.e. $x>0$, $x-4>0$ and $x+6>0$. So the solution is $x=6$. – Emilio Novati Oct 30 '15 at 19:08
HINT: $$\log_a X = \log_a Y$$ implies $$X=Y$$
Now that you've found two candidates, you need to check them in the original equation. We need to be able to take the logarithm of $x, x-4$, and $x+6$. For $x_1=6$, this means taking the log of $6, 2$, and $12$, which is fine. What about $x_2=-1$?