1

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.

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    $log , y=0$ iff $y=1$. Complete the argument by solving $x^{2}(8x+1)=(2x-1) (4x^{2}+4x+1)$. This gives $x=1$ or $x=-\frac 13$. But negative values are not allowed because the given equation involves $log (x-\frac 1 2)$. – Kavi Rama Murthy May 03 '22 at 09:41
  • There's a better way to simply an equation with log functions. For example, $\log x - \log y = \log \frac{x}{y}$ and $2\log z = \log z^2$. – Eddy Piedad May 03 '22 at 10:00

2 Answers2

1

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.

Eddy Piedad
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0

$$2\log\left(x\right)\ -\ \log\left(2x-1\right)\ -\ 2\log\left(2x+1\right)\ +\ \log\left(8x+1\right)\ =\ 0$$ $$\log\left(x^2\right)\ -\ \log\left(2x-1\right)\ -\ \log\left(4x^2 + 4x +1\right)\ +\ \log\left(8x+1\right)\ =\ 0$$

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, $$n\log |x|\equiv \log |x|^n.$$

ryang
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