Solve the equation: $\;$ $1 - \dfrac{1}{2} \log \left(2x - 1\right) = \dfrac{1}{2} \log \left(x - 9\right)$
Given equation: $\;\;$ $1 - \dfrac{1}{2} \log \left(2x - 1\right) = \dfrac{1}{2} \log \left(x - 9\right)$
i.e. $\;$ $\dfrac{1}{2} \left[\log \left(x - 9\right) + \log \left(2x - 1\right)\right] = 1$
i.e. $\;$ $\log \left[\left(x - 9\right) \left(2x - 1\right)\right] = 2$ $\;\;\; \cdots \; (1)$
i.e. $\;$ $\left(x - 9\right) \left(2x - 1\right) = 10^2 = 100$
i.e. $\;$ $2 x^2 - 19 x + 9 = 100$
i.e. $\;$ $2 x^2 - 19x - 91 = 0$
i.e. $\;$ $\left(2x + 7\right) \left(x - 13\right) = 0$
i.e. $\;$ $x = \dfrac{-7}{2}$ $\;$ or $\;$ $x = 13$
When $\;$ $x = \dfrac{-7}{2}$, $\;$ the term $\;$ $\log \left(2x - 1\right)$ $\;$ in the given problem becomes
$\log \left[2 \times \left(\dfrac{-7}{2}\right) - 1\right] = \log \left(-8\right)$
But logarithm of a negative number is not defined.
$\therefore \;$ $x = \dfrac{-7}{2}$ $\;$ is not a valid solution.
When $\;$ $x = 13$, $\;$ the given problem [equation (1)] becomes
$\log \left[\left(13-9\right) \left(26-1\right) \right] = 2$
i.e. $\;$ $\log \left[4 \times 25\right] = 2$
i.e. $\log 100 = 2$ $\;$ which is true.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{13 \right\}$