Solve the equation: $\;$ $2 \log_2 \left(\dfrac{x - 7}{x - 1}\right) + \log_2 \left(\dfrac{x - 1}{x + 1}\right) = 1$
Given equation: $\;\;$ $2 \log_2 \left(\dfrac{x - 7}{x - 1}\right) + \log_2 \left(\dfrac{x - 1}{x + 1}\right) = 1$
i.e. $\;$ $\log_2 \left(\dfrac{x - 7}{x - 1}\right)^2 + \log_2 \left(\dfrac{x - 1}{x + 1}\right) = 1$
i.e. $\;$ $\log_2 \left[\left(\dfrac{x - 7}{x - 1}\right)^2 \times \left(\dfrac{x - 1}{x + 1}\right)\right] = 1$
i.e. $\;$ $\dfrac{\left(x - 7\right)^2}{\left(x + 1\right) \left(x - 1\right)} = 2^1 = 2$
i.e. $\;$ $x^2 - 14x + 49 = 2x^2 - 2$
i.e. $\;$ $x^2 + 14x - 51 = 0$
i.e. $\;$ $\left(x + 17\right) \left(x - 3\right) = 0$
i.e. $\;$ $x + 17 = 0$ $\;$ or $\;$ $x - 3 = 0$
i.e. $\;$ $x = -17$ $\;$ or $\;$ $x = 3$
When $\;$ $x = 3$, $\;$ the term $\;$ $\log_2 \left(\dfrac{x - 7}{x - 1}\right)$ $\;$ in the given problem becomes
$\log_2 \left(\dfrac{3 - 7}{3 - 1}\right) = \log \left(-2\right)$
But, logarithm of a negative number is not defined.
When $\;$ $x = -17$, $\;$ the given problem becomes
$2 \log_2 \left(\dfrac{-17 - 7}{-17 - 1}\right) + \log_2 \left(\dfrac{-17 - 1}{-17 + 1}\right) = 1$
i.e. $\;$ $\log_2 \left(\dfrac{-24}{-18}\right)^2 + \log_2 \left(\dfrac{-18}{-16}\right) = 1$
i.e. $\;$ $\log_2 \left(\dfrac{4}{3}\right)^2 + \log_2 \left(\dfrac{9}{8}\right) = 1$
i.e. $\;$ $\log_2 \left(\dfrac{16}{9} \times \dfrac{9}{8}\right) = 1$
i.e. $\;$ $\log_2 2 = 1$
i.e. $\;$ $1 = 1$ $\;\;$ which is true.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-17 \right\}$