Solve the equation: $\;$ $\left(\dfrac{5}{3}\right)^{x + 1} \times \left(\dfrac{9}{25}\right)^{x^2 + 2x - 11} = \left(\dfrac{5}{3}\right)^9$
Given equation: $\;\;$ $\left(\dfrac{5}{3}\right)^{x + 1} \times \left(\dfrac{9}{25}\right)^{x^2 + 2x - 11} = \left(\dfrac{5}{3}\right)^9$
i.e. $\;$ $\left(\dfrac{5}{3}\right)^{x + 1} \times \left[\left(\dfrac{3}{5}\right)^2\right]^{x^2 + 2x - 11} = \left(\dfrac{5}{3}\right)^9$
i.e. $\;$ $\dfrac{5^{x + 1}}{3^{x + 1}} \times \dfrac{3^{2x^2 + 4x - 22}}{5^{2x^2 + 4x - 22}} = \left(\dfrac{5}{3}\right)^9$
i.e. $\;$ $\dfrac{5^{x + 1 - 2x^2 - 4x + 22}}{3^{x + 1 - 2x^2 - 4x + 22}} = \left(\dfrac{5}{3}\right)^9$
i.e. $\;$ $\left(\dfrac{5}{3}\right)^{-2x^2 - 3x + 23} = \left(\dfrac{5}{3}\right)^9$
i.e. $\;$ $-2x^2 - 3x + 23 = 9$
i.e. $\;$ $2x^2 + 3x - 14 = 0$
i.e. $\;$ $\left(x - 2\right) \left(2x + 7\right) = 0$
i.e. $\;$ $x = 2$ $\;\;$ or $\;\;$ $x = \dfrac{-7}{2}$