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♀️
On
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 $0$.
On
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.
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}