Solve the equation: $\;$ $\dfrac{\log \sqrt{x + 7} - \log 2}{\log 8 - \log \left(x - 5\right)} = -1$
Given equation: $\;\;$ $\dfrac{\log \sqrt{x + 7} - \log 2}{\log 8 - \log \left(x - 5\right)} = -1$
i.e. $\;$ $\log \sqrt{x+7} - \log 2 = \log \left(x-5\right) - \log 8$
i.e. $\;$ $\log \sqrt{x+7} - \log \left(x - 5\right) - \log 2 + 3 \log 2 = 0$
i.e. $\;$ $\log \sqrt{x+7} - \log \left(x-5\right) + 2 \log 2 = 0$
i.e. $\;$ $\log \sqrt{x + 7} - \log \left(x - 5\right) + \log 4 = 0$
i.e. $\;$ $\log \left(\dfrac{4 \sqrt{x + 7}}{x - 5}\right) = 0$
i.e. $\;$ $\dfrac{4 \sqrt{x + 7}}{x - 5} = 10^0 = 1$
i.e. $\;$ $4 \sqrt{x + 7} = x - 5$
i.e. $\;$ $16 \left(x + 7\right) = x^2 - 10 x + 25$
i.e. $\;$ $x^2 - 26 x - 87 = 0$
i.e. $\;$ $\left(x - 29\right) \left(x + 3\right) = 0$
i.e. $\;$ $x = 29$ $\;$ or $\;$ $x = -3$
But, when $\;$ $x = -3$, $\;$ the term $\;$ $\log \left(x - 5\right)$ $\;$ in the given problem becomes
$\log \left(-3 - 5\right) = \log \left(-8\right)$
$\because \;$ logarithm of a negative number is not defined,
$\therefore \;$ $x = -3$ $\;$ is not a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{29 \right\}$