Solve the equation: $\;$ $\log_7 \left(2^x - 1\right) + \log_7 \left(2^x - 7\right) = 1$
Given equation: $\;\;$ $\log_7 \left(2^x - 1\right) + \log_7 \left(2^x - 7\right) = 1$
i.e. $\;$ $\log_7 \left[\left(2^x - 1\right) \left(2^x - 7\right)\right] = 1$
$\implies$ By definition, $\;$ $2^{2x} - 8 \times 2^x + 7 = 7^1 = 7$
i.e. $\;$ $2^{2x} - 8 \times 2^x = 0$
i.e. $\;$ $2^x \left(2^x - 8\right) = 0$
i.e. $\;$ $2^x = 0$ $\;$ or $\;$ $2^x = 8$
i.e. $\;$ $x = \log_2 \left(0\right)$ $\;$ or $\;$ $2^x = 2^3$
But $\;$ $\log_2 \left(0\right)$ $\;$ is not defined and is therefore not a valid solution.
Now, $\;$ $2^x = 2^3$ $\implies$ $x = 3$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{3 \right\}$