Solve the equation: $\;$ $7^{\log x} - 5^{\left(\log x\right) + 1} = 3 \times 5^{\left(\log x\right) - 1} - 13 \times 7^{\left(\log x\right) - 1}$
Given equation: $\;\;$ $7^{\log x} - 5^{\left(\log x\right) + 1} = 3 \times 5^{\left(\log x\right) - 1} - 13 \times 7^{\left(\log x\right) - 1}$
i.e. $\;$ $7^{\log x} - 5^{\log x} \times 5^1 = 3 \times 5^{\log x} \times 5^{-1} - 13 \times 7^{\log x} \times 7^{-1}$
i.e. $\;$ $7^{\log x} \left(1 + \dfrac{13}{7}\right) = 5^{\log x} \left(\dfrac{3}{5} + 5\right)$
i.e. $\;$ $7^{\log x} \times \dfrac{20}{7} = 5^{\log x} \times \dfrac{28}{5}$
i.e. $\;$ $\left(\dfrac{7}{5}\right)^{\log x} = \dfrac{28 \times 7}{20 \times 5} = \dfrac{49}{25} = \left(\dfrac{7}{5}\right)^2$
$\implies$ $\log_{10} x = 2$
i.e. $\;$ $x = 10^2 = 100$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{100 \right\}$