The question is:

Solve for $x$ in the following


Now, solving for $x$, we get;


$\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?

  • 1
    $\begingroup$ So your equation is : $-\log(x^2) - 12x + 36 = 2$ ? The notation is not really clear. $\endgroup$ – 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.

  • $\begingroup$ I got it now. My bad. Thanks for the help. $\endgroup$ – Abhishekstudent Mar 10 '17 at 14:32

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