Solving $\log(x) -\frac{1}{2}\log(x-\frac{1}{2}) = \log(x+\frac{1}{2}) - \frac{1}{2}\log(x+\frac{1}{8})$ 
Find $x$ in the equation $\log(x) -\frac{1}{2}\log(x-\frac{1}{2}) = \log(x+\frac{1}{2}) - \frac{1}{2}\log(x+\frac{1}{8})$.

My attempt:
$$\log\left(x\right)\ -\ \frac{1}{2}\ \log\left(x-\frac{1}{2}\right)\ =\ \log\left(x+\frac{1}{2}\right)\ -\ \frac{1}{2}\log\left(x\ +\ \frac{1}{8}\right)$$
$$\log\left(x\right)\ -\ \frac{1}{2}\log\left(\frac{2x-1}{2}\right)\ =\ \log\left(\frac{2x+1}{2}\right)\ -\ \frac{1}{2}\log\left(\frac{8x+1}{8}\right)$$
$$\log\left(x\right)\ -\ \frac{1}{2}\log\left(2x-1\right)\ +\ \frac{1}{2}\log\left(2\right)\ =\ \log\left(2x+1\right)\ -\ \log\left(2\right)\ -\ \frac{1}{2}\log\left(8x+1\right)\ +\ \frac{1}{2}\log\left(8\right)$$
$$\log\left(x\right)\ -\ \frac{1}{2}\log\left(2x-1\right)\ =\ \log\left(2x+1\right)\ -\ \frac{1}{2}\log\left(8x+1\right)$$
$$\log\left(x\right)\ -\ \frac{1}{2}\log\left(2x-1\right)\ -\ \log\left(2x+1\right)\ +\ \frac{1}{2}\log\left(8x+1\right)=0$$
$$2\log\left(x\right)\ -\ \log\left(2x-1\right)\ -\ 2\log\left(2x+1\right)\ +\ \log\left(8x+1\right)\ =\ 0$$
$$\log\left(x^2\right)\ -\ \log\left(2x-1\right)\ -\ \log\left(4x^2 + 4x +1\right)\ +\ \log\left(8x+1\right)\ =\ 0$$
$$\log\left(x^2(8x+1)\right)\ -\ \log\left(2x-1\right)\ -\ \log\left(4x^2 + 4x +1\right) = 0$$
$$\log\left(\frac{x^{2}\left(8x+1\right)}{(2x-1)(4x^2 + 4x + 1)}\right)=\ 0\ \tag{1}$$

I'm not getting any idea what to do with this further. Also, the original equation has only $x = 1$ as a solution but the equation $(1.)$ has $2$ solutions $x = 1; x= -1/3$.
I think there might be a simpler solution to this equation rather than my cumbersome one.
Not sure what's the mistake.
 A: Simplifying $$\log(x) -\frac{1}{2}\log(x-\frac{1}{2}) = \log(x+\frac{1}{2}) - \frac{1}{2}\log(x+\frac{1}{8})$$
into
$$\log\left(\frac{x}{(x-\frac{1}{2})^\frac{1}{2}}\right) = \log\left(\frac{x+\frac{1}{2}}{(x+\frac{1}{8})^\frac{1}{2}}\right)$$
And taking the exponential,
$$\frac{x}{(x-\frac{1}{2})^\frac{1}{2}} = \frac{x+\frac{1}{2}}{(x+\frac{1}{8})^\frac{1}{2}}$$
Taking the square of both sides,
$$\frac{x^2}{x-\frac{1}{2}} = \frac{x^2+x+\frac{1}{4}}{x+\frac{1}{8}}$$
Simplifying the equation, we have
$$-3x^2 + 2x +1 = 0$$
Use quadratic equation to find two solutions $$x=1, x=-1/3$$
that you were looking for. I hope this helps.
A: 
$$2\log\left(x\right)\ -\ \log\left(2x-1\right)\ -\ 2\log\left(2x+1\right)\ +\ \log\left(8x+1\right)\ =\ 0$$
$$\log\left(x^2\right)\ -\ \log\left(2x-1\right)\ -\ \log\left(4x^2 + 4x +1\right)\ +\ \log\left(8x+1\right)\ =\ 0$$

Note that $$\log\left(x^2\right)=t\kern.6em\not\kern-.6em\implies2\log(x)=t,$$ but that $$2\log(x)=t\iff\log\left(x^2\right)=t \;\text{ and }\;x>0.$$ Neglecting the implicit condition $x>0$ was how, in the above step, you introduced the extraneous solution $x= -\frac13.$
In general, for $n\in\mathbb Z,$ $$n\log x\not\equiv \log x^n;$$ instead, $$n\log |x|\equiv \log |x|^n.$$
