Did I solve this logarithmic equation problem corectly and if so why doesn't commpleting the square work?

Question

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

Answer 1

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.

Answer 2

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.

Answer 3

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