The quadratic equation is giving me error can you please help me locate where I am wrong Question. Solve $$\log(x-3) + \log (x-4) - \log(x-5)=0.$$ 
Attempt. I got $$x^2-8x+17=0.$$
$$\log(x-3)(x-4)/(x-5)=0$$
$$\log(x^2-4x-3x+12)/x-5=0$$
$$x^2-7x+12= 10^0 (x-5)$$
$$x^2-7x-x+12+5=0$$
$$x^2-8x+17=0$$
Hi guys update: apparently the answer was equation is undefined‍♀️
 A: You have $x^2-8x+17=0.$ Completing the square, you get
\begin{align}
& (x^2 - 8x + 16) + 1=0 \\
& (x-4)^2+1=0 \\
& (x-4)^2 = -1 \\
& x-4 = \pm i \\
& x = 4\pm i.
\end{align}
If you substitute that into the original equation, you're taking the logarithm of a complex number. How to do that is moderately problematic, and only if you've examined that question does it make sense for you to be assigned this problem. So there is a possibility that something is wrong with the statement of the problem.
A: What's the problem? We have\begin{align}\log(x-3)+\log(x-4)-\log(x-5)=0&\iff\log\bigl((x-3)(x-4)\bigr)=\log(x-5)\\&\iff\log(x^2-7x+12)=\log(x-5)\\&\iff x^2-7x+12=x-5\\&\iff x^2-8x+17=0.\end{align}
A: Assuming you're looking for real solutions, this equation has no solutions, which is why you're bothered. Here's a way to show it has no solutions:
Since $\log$ is an increasing function, 
$$ \log(x-4) > \log(x-5)$$
$$\log(x-4) - \log(x-5) > 0$$
Now if $\log(x-5)$ is real, we must have $x-5 > 0$, so $x-3 > 2 > 1$; thus $\log(x-3) > 0$. Adding this in gives
$$ \log(x-3) + \log(x-4) - \log(x-5) > 0$$
So it cannot equal $0$.
