Solve the equation: $\;$ $\log x - \dfrac{1}{2} \log \left(x - \dfrac{1}{2}\right) = \log \left(x + \dfrac{1}{2}\right) - \dfrac{1}{2} \left(x + \dfrac{1}{8}\right)$
Given equation: $\;\;$ $\log x - \dfrac{1}{2} \log \left(x - \dfrac{1}{2}\right) = \log \left(x + \dfrac{1}{2}\right) - \dfrac{1}{2} \left(x + \dfrac{1}{8}\right)$
i.e. $\;$ $\log x - \log \left(x + \dfrac{1}{2}\right) = \dfrac{1}{2} \log \left(x - \dfrac{1}{2}\right) - \dfrac{1}{2} \log \left(x + \dfrac{1}{8}\right)$
i.e. $\;$ $\log \left(\dfrac{x}{x + \frac{1}{2}}\right) = \dfrac{1}{2} \log \left(\dfrac{x - \frac{1}{2}}{x + \frac{1}{8}}\right)$
i.e. $\;$ $2 \log \left(\dfrac{2x}{2x + 1}\right) = \log \left(\dfrac{\frac{2x - 1}{2}}{\frac{8x + 1}{8}}\right)$
i.e. $\;$ $2 \log \left(\dfrac{2x}{2xx + 1}\right) = \log \left[\dfrac{4 \left(2x - 1\right)}{8x + 1}\right]$
i.e. $\;$ $\log \left(\dfrac{2x}{2x + 1}\right)^2 = \log \left(\dfrac{8x - 4}{8x + 1}\right)$
i.e. $\;$ $\dfrac{4x^2}{4x^2 + 4x + 1} = \dfrac{8x - 4}{8x + 1}$
i.e. $\;$ $32 x^3 + 4x^2 = 32 x^3 + 32 x^2 + 8x - 16 x^2 - 16x - 4$
i.e. $\;$ $12 x^2 - 8x -4 = 0$
i.e. $\;$ $\left(3x + 1\right) \left(x - 1\right) = 0$
i.e. $\;$ $x = \dfrac{-1}{3}$ $\;$ or $\;$ $x = 1$
When $\;$ $x = \dfrac{-1}{3}$, $\;$ there are terms like $\;$ $\log \left(\dfrac{-1}{3}\right)$ $\;$ in the given problem.
But logarithim of a negative number is not defined.
$\therefore \;$ $x = \dfrac{-1}{3}$ $\;$ is not a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{1 \right\}$