Solve the equation: $\;$ $\log_3 \left(4 \times 3^x - 1\right) = 2x + 1$
Given equation: $\;\;$ $\log_3 \left(4 \times 3^x - 1\right) = 2x + 1$
i.e. $\;$ $4 \times 3^x - 1 = 3^{2x+1}$
i.e. $\;$ $4 \times 3^x - 1 = 3^{2x} \times 3^1$
i.e. $\;$ $3 \times \left(3^x\right)^2 - 4 \times 3^x + 1 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $3^x = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$3p^2 - 4p + 1 = 0$
i.e. $\;$ $\left(p - 1\right) \left(3p - 1\right) = 0$
i.e. $\;$ $p = 1$ $\;$ or $\;$ $p = \dfrac{1}{3}$
Substituting the values of $p$ in equation $(2)$ give
when $\;$ $p = 1$, $\;$ then $\;$ $3^x = 1 = 3^0$ $\implies$ $x = 0$
when $\;$ $p = \dfrac{1}{3}$, $\;$ then $\;$ $3^x = \dfrac{1}{3} = 3^{-1}$ $\implies$ $x = -1$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-1, \; 0 \right\}$