Solve the equation: $\;$ $\log \left(3^x - 2^{4-x}\right) = 2 + \dfrac{1}{4} \log 16 - \dfrac{x \log4}{2}$
Given equation: $\;\;$ $\log \left(3^x - 2^{4-x}\right) = 2 + \dfrac{1}{4} \log 16 - \dfrac{x \log4}{2}$
i.e. $\;$ $\log \left(3^x - 2^{4 - x}\right) - \dfrac{1}{4} \log 16 + x \times \dfrac{1}{2} \log 4 = 2$
i.e. $\;$ $\log \left(3^x - \dfrac{2^4}{2^x}\right) - \log \left(2^4\right)^{\frac{1}{4}} + x \times \log \left(2^2\right)^{\frac{1}{2}} = 2$
i.e. $\;$ $\log \left(\dfrac{6^x - 2^4}{2^x}\right) - \log 2 + \log 2^x = 2$
i.e. $\;$ $\log \left[\left(\dfrac{6^x - 2^4}{2^x}\right) \times \dfrac{1}{2} \times 2^x\right] = 2$
i.e. $\;$ $\log \left[\dfrac{6^x - 2^4}{2}\right] = 2$
i.e. $\;$ $\dfrac{6^x - 2^4}{2} = 10^2 = 100$
i.e. $\;$ $6^x - 16 = 200$
i.e. $\;$ $6^x = 216$
i.e. $\;$ $x = \log_6 216$
i.e. $\;$ $x = \log_6 6^3$
i.e. $\;$ $x = 3 \log_6 6$ $\implies$ $x = 3$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{3 \right\}$