Solve the equation: $\;$ $\log \left(5^{x-2} + 1\right) = x + \log 13 - 2 \log 5 + \left(1-x\right) \log 2$
Given equation: $\;\;$ $\log \left(5^{x-2} + 1\right) = x + \log 13 - 2 \log 5 + \left(1-x\right) \log 2$
i.e. $\;$ $\log \left(5^{x-2} + 1\right) - \log 2^{1-x} - \log 13 + \log 5^2 = x$
i.e. $\;$ $\log \left[\dfrac{\left(5^{x-2} + 1\right) \times 5^2}{2^{1-x} \times 13}\right] = x$
i.e. $\;$ $\log \left[\dfrac{\left(5^x + 25\right) \times 2^x}{26}\right] = x$
i.e. $\;$ $\log \left[\dfrac{10^x + 2^x \times 25}{26}\right] = x$
i.e. $\;$ $\dfrac{10^x + 2^x \times 25}{26} = 10^x$
i.e. $\;$ $10^x + 2^x \times 25 = 26 \times 10^x$
i.e. $\;$ $25 \times 2^x = 25 \times 10^x$
i.e. $\;$ $2^x = 10^x$
i.e. $\;$ $2^x = 2^x \times 5^x$
i.e. $\;$ $2^x \left(5^x - 1\right) = 0$
i.e. $\;$ $2^x = 0$ $\;$ or $\;$ $5^x - 1 = 0$
When $\;$ $2^x = 0$,
$\implies$ $x = \log_2 0$
But $\;$ $\log_2 0$ $\;$ is not defined.
$\therefore \;$ $2^x = 0$ $\;$ does not give a valid solution.
When $\;$ $5^x - 1 = 0$,
$\implies$ $5^x = 1$
i.e. $\;$ $x = \log_5 1 = 0$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{0 \right\}$