Solve the equation: $\;$ $\log_2 \left(4 \times 3^x - 6\right) - \log_2 \left(9^x - 6\right) = 1$
Given equation: $\;\;$ $\log_2 \left(4 \times 3^x - 6\right) - \log_2 \left(9^x - 6\right) = 1$
i.e. $\;$ $\log_2 \left(\dfrac{4 \times 3^x - 6}{9^x - 6}\right) = 1$
i.e. $\;$ $\dfrac{4 \times 3^x - 6}{9^x - 6} = 2^1 = 2$
i.e. $\;$ $4 \times 3^x - 6 = 2 \times 9^x - 12$
i.e. $\;$ $2 \times 3^{2x} - 4 \times 3^x - 6 = 0$
i.e. $\;$ $\left(3^x\right)^2 - 2 \times 3^x - 3 = 0$
i.e. $\;$ $\left(3^x\right)^2 - 3 \times 3^x + 3^x - 3 = 0$
i.e. $\;$ $3^x \left(3^x - 3\right) + 1 \left(3^x - 3\right) = 0$
i.e. $\;$ $\left(3^x + 1\right) \left(3^x - 3\right) = 0$
i.e. $\;$ $3^x = -1$ $\;$ or $\;$ $3^x = 3$
i.e. $\;$ $x = \log_3 \left(-1\right)$ $\;$ or $\;$ $x = \log_3 3 = 1$
$\because \;$ logarithim of a negative number is not defined,
$\therefore \;$ $x = \log_3 \left(-1\right)$ $\;$ is not a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{1 \right\}$