Solve the equation: $\;$ $7^{x+2} - \dfrac{1}{7} \times 7^{x+1} - 14 \times 7^{x-1} + 2 \times 7^x = 48$
Given equation: $\;\;$ $7^{x+2} - \dfrac{1}{7} \times 7^{x+1} - 14 \times 7^{x-1} + 2 \times 7^x = 48$
i.e. $\;$ $7^x \times 7^2 - 7^{-1} \times 7^x \times 7^1 - 2 \times 7 \times 7^x \times 7^{-1} + 2 \times 7^x = 48$
i.e. $\;$ $49 \times 7^x - 7^x - 2 \times 7^x + 2 \times 7^x = 48$
i.e. $\;$ $48 \times 7^x = 48$
i.e. $\;$ $7^x = 1 = 7^0$
$\implies$ $x = 0$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = 0$