Solve the equation: $\;$ $2^{\left(2 \log 4x\right) - 1} - 7^{\log 4x} = 7^{\left(\log 4x\right) - 1} - 3 \times 4^{\log 4x}$
Given equation: $\;\;$ $2^{\left(2 \log 4x\right) - 1} - 7^{\log 4x} = 7^{\left(\log 4x\right) - 1} - 3 \times 4^{\log 4x}$
i.e. $\;$ $\left(2^2\right)^{\log 4x} \times 2^{-1} - 7^{\log 4x} = 7^{\log 4x} \times 7^{-1} - 3 \times 4^{\log 4x}$
i.e. $\;$ $4^{\log 4x} + 3 \times 4^{\log 4x} = \dfrac{7^{\log 4x}}{7} + 7^{\log 4x}$
i.e. $\;$ $4^{\log 4x} \left(\dfrac{1}{2} + 3\right) = 7^{\log 4x} \left(\dfrac{1}{7} + 1\right)$
i.e. $\;$ $4^{\log 4x} \times \dfrac{7}{2} = 7^{\log 4x} \times \dfrac{8}{7}$
i.e. $\;$ $\left(\dfrac{4}{7}\right)^{\log 4x} = \dfrac{8 \times 2}{7 \times 7} = \left(\dfrac{4}{7}\right)^2$
$\implies$ $\log_{10} 4x = 2$
i.e. $\;$ $4x = 10^2 = 100$ $\implies$ $x = 25$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{25 \right\}$