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  1. $\log(x) + \log(x-1) = \log(4x)$
  2. $\log((x)(x-1))=\log(4x)$ use the law of logs
  3. $\log(x^2 -x)=\log(4x)$ distribute the x
  4. $x^2-x = 4x$ use the one to one nature of logs
  5. $x^2 = 5x$ add x
  6. $x=5$ Divide by x
amWhy
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HiShin
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    You might indicate at the last step how you know $x\neq0$, but otherwise, it's okay. Formatting tip: if you use $\log$ instead of $log$ your posts will look better. Same goes for other standard functions like $\sin$, $\max$ etc. – saulspatz May 19 '20 at 17:38
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    Steps 1,2,3,4,5 are okay. Then go on with $x(x-5)=0$ so that $x=5$ or $x=0$. Substituting in the original expressions shows that $x=5$ is indeed a solution and that $x=0$ falls off because $\log$ is only defined on positive numbers. Division by an unkown $x$ is tricky and should be avoided. – drhab May 19 '20 at 17:40
  • $x=5$ is correct. What do you mean by completing the square? – Karl May 19 '20 at 17:40
  • You can divide by $x$ if you know than it can't be zero. $x=5$ is the solution. What is your question? – Vasili May 19 '20 at 17:41
  • You could also have started by writing $\log(4x)=\log(4)+\log(x)$ and subtracting $\log(x)$ from both sides. – Karl May 19 '20 at 17:43
  • @imranfat when I do that I get x = 2.5 after completing the square – HiShin May 19 '20 at 17:51
  • @amWhy. Realize it now, deleted. – imranfat May 19 '20 at 17:54
  • @HiShin. No need to complete the square, just factor. – imranfat May 19 '20 at 17:54
  • HiShin No, you get $x(x-5) = 0 \implies (x=0 \text{ or } x=5)$. Since $x = 0$ is not in the domain of the original function (because $\log(0)$ is undefined, only $x=5$ remains as a valid solution, just as drhab wrote in the second comment below your post. – amWhy May 19 '20 at 17:55
  • Can you indicate which step you wish to "complete the square"... – Saket Gurjar May 19 '20 at 18:00
  • Cool I get that stuff, however why doesn't completing the square work? even if its unnecessary, shouldn't it still work? – HiShin May 19 '20 at 18:02
  • @HiShin Do you mean while solving $x^2=5x$? – Saket Gurjar May 19 '20 at 18:04
  • @Saket Gurjar when I get to step 4. I can either add $\ x $ or subtract $\ 4x $ , if I subtract $\ 4x $ then I get $\ x^2 - 5x = 0$. why doesn't completing the square work to tell me what $\ x $ is – HiShin May 19 '20 at 18:14
  • THANK YOU. I forgot to balance the equation.... – HiShin May 19 '20 at 18:18
  • @Hishin I've expanded my comment into an answer. – Deepak May 19 '20 at 18:21

3 Answers3

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Your solution is correct except with the omission of the justification of why division by $x$ is permissible in the second last step. Division is only allowed when $x\neq 0$. Here you must state that the original equation disallows $x=0$ as then $\log x$ would be undefined, so division by (nonzero) $x$ is justified. Then your solution would be completely correct.

As to your question about completing the square, it certainly "works":

$x^2 = 5x \\ \implies x^2 - 5x = 0 \\ \implies x^2 - 5x + (\frac 52)^2 - \frac{25}4 = 0 \\ \implies (x-\frac 52)^2 = \frac {25}4 \\ \implies x-\frac 52 = \pm \frac 52 \\ \implies x=0 \ \mathrm{or} \ x=5$

(and you reject $0$ as $\log 0$ is undefined).

But this is completely and utterly unnecessary in solving the problem.

Deepak
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Completing the square method does work.

$$x^2=5x$$

$$x^2-5x+\frac{25}{4}=\frac{25}{4}$$

$$\left(x-\frac{5}{2}\right)^2=\left(\frac{5}{2}\right)^2$$

$$(x)(x-5)=0$$

Here remember to discard the $x=0$ solution.

Saket Gurjar
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Firstly, you should indicate that $\ x$ $\neq$ 0 if that is true then division by X is allowed

Secondly, completing the square does work make sure that you are balancing the equation.

Both Saket Gurjar and Deepak answer the question correctly

HiShin
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