Solve the equation: $\;$ $\log 2 + \log \left(4^{x-2} + 9\right) = 1 + \log \left(2^{x-2} + 1\right)$
Given equation: $\;\;$ $\log 2 + \log \left(4^{x-2} + 9\right) = 1 + \log \left(2^{x-2} + 1\right)$
i.e. $\;$ $\log \left(4^{x-2} + 9\right) - \log \left(2^{x-2} + 1\right) = 1 - \log 2$
i.e. $\;$ $\log \left[\dfrac{4^{x-2} + 9}{2^{x-2} + 1}\right] = \log 10 - \log2$
i.e. $\;$ $\log \left[\dfrac{\left(2^2\right)^{x-2} + 9}{2^{x-2} + 1}\right] = \log \left(\dfrac{10}{2}\right)$
i.e. $\;$ $\log \left[\dfrac{2^{2x-4} + 9}{2^{x-2} + 1}\right] = \log 5$ $\;\;\; \cdots \; (1)$
Taking antilog to base $10$ on both sides of equation $(1)$ gives
$\dfrac{2^{2x-4} + 9}{2^{x-2} + 1} = 5$
i.e. $\;$ $2^{2x-4} + 9 = 5 \times 2^{x-2} + 5$
i.e. $\;$ $\dfrac{2^{2x}}{2^4} - 5 \times \dfrac{2^x}{2^2} + 4 = 0$
i.e. $\;$ $2^{2x} - 5 \times 2^x \times 2^2 + 4 \times 2^4 = 0$
i.e. $\;$ $\left(2^x\right)^2 - 20 \times 20^x + 64 = 0$
i.e. $\;$ $\left(2^x\right)^2 - 16 \times 2^x - 4 \times 2^x + 64 = 0$
i.e. $\;$ $2^x \left(2^x - 16\right) - 4 \left(2^x - 16\right) = 0$
i.e. $\;$ $\left(2^x - 4\right) \left(2^x - 16\right) = 0$
i.e. $\;$ $2^x - 4 = 0$ $\;$ or $\;$ $2^x - 16 = 0$
i.e. $\;$ $2^x = 4 = 2^2$ $\;$ or $\;$ $2^x = 16 = 2^4$
$\implies$ $x = 2$ $\;$ or $\;$ $x = 4$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2, \; 4 \right\}$