Solve the equation: $\;$ $\left(\dfrac{5}{12}\right)^x \times \left(\dfrac{6}{5}\right)^{x - 1} = \left(0.3\right)^{-1}$
Given equation: $\;\;$ $\left(\dfrac{5}{12}\right)^x \times \left(\dfrac{6}{5}\right)^{x - 1} = \left(0.3\right)^{-1}$
i.e. $\;$ $\left(\dfrac{5}{3 \times 2^2}\right)^x \times \left(\dfrac{3 \times 2}{5}\right)^{x - 1} = \left(\dfrac{3}{10}\right)^{-1}$
i.e. $\;$ $\dfrac{5^x}{3^x \times 2^{2x}} \times \dfrac{3^{x-1} \times 2^{x-1}}{5^{x-1}} = \left(\dfrac{3}{5 \times 2}\right)^{-1}$
i.e. $\;$ $5^{x - x + 1} \times 3^{x - 1 - x} \times 2^{x - 1 - 2x} = \dfrac{3^{-1}}{5^{-1} \times 2^{-1}}$
i.e. $\;$ $\dfrac{5^1 \times 3^{-1} \times 2^{-x - 1} \times 5^{-1} \times 2^{-1}}{3^{-1}} = 1$
i.e. $\;$ $5^{1 - 1} \times 3^{-1 + 1} \times 2^{-x - 1 - 1} = 1$
i.e. $\;$ $5^0 \times 3^0 \times 2^{-x - 2} = 1$
i.e. $\;$ $1 \times 1 \times 2^{-x - 2} = 2^0$
i.e. $\;$ $-x - 2 = 0$
i.e. $\;$ $x = -2$