Solve the equation: $\;$ $9^{\log_3 \left(1 - 2x\right)} = 5x^2 - 5$
Given equation: $\;\;$ $9^{\log_3 \left(1 - 2x\right)} = 5x^2 - 5$
i.e. $\;$ $\left(3^2\right)^{\log_3 \left(1 - 2x\right)} = 5x^2 - 5$
i.e. $\;$ $\left(3\right)^{2 \log_3 \left(1 - 2x\right)} = 5x^2 - 5$
i.e. $\;$ $3^{\log_3 \left(1 - 2x\right)^2} = 5x^2 - 5$
i.e. $\;$ $\left(1 - 2x\right)^2 = 5x^2 - 5$
i.e. $\;$ $1 - 4x + 4x^2 = 5x^2 - 5$
i.e. $\;$ $x^2 + 4x - 6 = 0$
i.e. $\;$ $x = \dfrac{-4 \pm \sqrt{16 + 24}}{2}$
i.e. $\;$ $x = \dfrac{-4 \pm \sqrt{40}}{2}$
i.e. $\;$ $\dfrac{-4 \pm 2 \sqrt{10}}{2}$
i.e. $\;$ $x = -2 \pm \sqrt{10}$
Now, $\;$ $\sqrt{10} \approx 3.16, \;\;\; -2 + \sqrt{10} = 1.16$
and $\;$ $\log_3 \left(1 - 2x\right) = \log_3 \left(1 - 2.32\right) = \log_3 \left(-1.32\right)$
But, logarithim of a negative number is not defined.
$\therefore \;$ $x = -2 + \sqrt{10}$ $\;$ is not a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-2 - \sqrt{10} \right\}$