Solve the equation: $\;$ $\log_5 \left(\dfrac{2 + x}{10}\right) = \log_5 \left(\dfrac{2}{x + 1}\right)$
Given equation: $\;\;$ $\log_5 \left(\dfrac{2 + x}{10}\right) = \log_5 \left(\dfrac{2}{x + 1}\right)$
i.e. $\;$ $\log_5 \left(\dfrac{2 + x}{10}\right) - \log_5 \left(\dfrac{2}{x + 1}\right) = 0$
i.e. $\;$ $\log_5 \left[\dfrac{\dfrac{2 + x}{10}}{\dfrac{2}{x + 1}}\right] = 0$
i.e. $\;$ $\log_5 \left[\dfrac{\left(2 + x\right) \left(x + 1\right)}{20}\right] = 0$
i.e. $\;$ $\dfrac{\left(2 + x\right) \left(x + 1\right)}{20} = 5^0 = 1$
i.e. $\;$ $x^2 + 3x + 2 = 20$
i.e. $\;$ $x^2 + 3x - 18 = 0$
i.e. $\;$ $\left(x + 6\right) \left(x - 3\right) = 0$
i.e. $\;$ $x = -6$ $\;$ or $\;$ $x = 3$
When $\;$ $x = -6$, $\;$ the terms in the given equation become
$\log_5 \left(\dfrac{2 + x}{10}\right) = \log_5 \left(\dfrac{2 - 6}{10}\right) = \log_5 \left(\dfrac{-4}{10}\right)$
and $\;$ $\log_5 \left(\dfrac{2}{x + 1}\right) = \log_5 \left(\dfrac{2}{-6 + 1}\right) = \log_5 \left(\dfrac{-2}{5}\right)$
But, logarithim of a negative number is not defined.
$\therefore \;$ $x = -6$ is not a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{3 \right\}$