# Shouldn't the logarithm have no solution?

The question is:

Solve for $x$ in the following

$\log_{10}{(x^2-12x+36)}=2$

Now, solving for $x$, we get;

$(x-6)^2=100$

$\implies x=-4$or $x=16$

But, these are the roots of that equation. So, if I put the value of $x$ in the original equation to check, I get;

$\log{0}=100$ which can not be true.

SO, why are we counting both the roots as solutions?

• So your equation is : $-\log(x^2) - 12x + 36 = 2$ ? The notation is not really clear. – Zubzub Mar 10 '17 at 14:28

Assuming you're trying to solve $$\log (x^2-12x+36)=2$$ And the $\log$ is base $10$, I'm not sure how you get $\log 0=100$. If you put both $-4, 16$ back in, we get $$\log 100=2$$ As expected.