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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♀️

Maisha Aziz
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    shouldn't you include your working so that we can locate where you are wrong? – Siong Thye Goh May 29 '18 at 17:22
  • I can’t upload a picture because new user. I’ll write it down I guess. – Maisha Aziz May 29 '18 at 17:25
  • Welcome to Math.SE Maisha. Please use mathjax when formatting questions, and include ANY and ALL working you have done. – Rhys Hughes May 29 '18 at 17:27
  • Your equation is correct, given that particular question. However, the two solutions to that equation are complex. Since it's very unlikely that you're expected to find complex solutions, I'd say the question is wrong. Can you check that you haven't misplaced a sign somewhere? – Rhys Hughes May 29 '18 at 17:35
  • Yeah the question is typed correctly. The problem is for our course there is no markscheme so either way I don’t know if I am correct for any of my answers. ☹️ – Maisha Aziz May 29 '18 at 17:42
  • You shouldn't write $\log(x^2-4x-3x+12)/x-5$ if you mean $\log(x^2-4x-3x+12)/(x-5).$ Those are two different things. – Michael Hardy May 29 '18 at 17:59

3 Answers3

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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.

Michael Hardy
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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}

José Carlos Santos
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    Sounds like the issue is that the resulting equation has only complex roots, so maybe the original question has a typo. – Alex R. May 29 '18 at 17:27
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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 $0$.

BallBoy
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