Solve the equation: $\;$ $5^{x-1} = 10^x \times 2^{-x} \times 5^{x + 1}$
Given equation: $\;\;$ $5^{x-1} = 10^x \times 2^{-x} \times 5^{x + 1}$
i.e. $\;$ $5^{x - 1} = \left(5 \times 2\right)^x \times 2^{-x} \times 5^{x + 1}$
i.e. $\;$ $5^{x - 1} = 5^x \times 2^x \times 2^{-x} \times 5^{x + 1}$
i.e. $\;$ $5^{x - 1} = 5^{x + x + 1} \times 2^{x - x}$
i.e. $\;$ $5^{x - 1} = 5^{2x + 1} \times 2^0$
i.e. $\;$ $5^{x - 1} = 5^{2x + 1} \times 1$
$\implies$ $x - 1 = 2x + 1$
i.e. $\;$ $x = -2$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = -2$