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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

$$x1= 6$$

$$x2=-1$$

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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).

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  • $\begingroup$ I used delta, therfore: x1= 6, x2=-1 $\endgroup$ – alaina Oct 30 '15 at 19:01
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    $\begingroup$ 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 $x=6$. $\endgroup$ – Emilio Novati Oct 30 '15 at 19:08
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HINT: $$\log_a X = \log_a Y$$ implies $$X=Y$$

So in your case, you need to solve an equation.

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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 $x_2=-1$?

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