Solve the equation: $\;$ $3^{4x + 8} - 4 \times 3^{2x + 5} + 28 = 2 \log_2 \sqrt{2}$
Given equation: $\;\;$ $3^{4x + 8} - 4 \times 3^{2x + 5} + 28 = 2 \log_2 \sqrt{2}$
i.e. $\;$ $3^{4x} \times 3^8 - 4 \times 3^{2x} \times 3^5 + 28 = 2 \log_2 2^{\frac{1}{2}}$
i.e. $\;$ $\left(3^{2x}\right)^2 \times 6561 - 972 \times 3^{2x} + 28 = 2 \times \dfrac{1}{2} \log_2 2$
i.e. $\;$ $6561 \times \left(3^{2x}\right)^2 - 972 \times 3^{2x} + 28 = 1$
i.e. $\;$ $6561 \times \left(3^{2x}\right)^2 - 972 \times 3^{2x} + 27 = 0$
i.e. $\;$ $243 \times \left(3^{2x}\right)^2 - 36 \times 3^{2x} + 1 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $3^{2x} = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$243 p^2 - 36 p + 1 = 0$
i.e. $\;$ $\left(27 p - 1\right) \left(9p - 1\right) = 0$
i.e. $\;$ $p = \dfrac{1}{27}$, $\;$ or $\;$ $p = \dfrac{1}{9}$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = \dfrac{1}{27}$, $\;$ $3^{2x} = \dfrac{1}{27} = 3^{-3}$
$\implies$ $2x = -3$ $\implies$ $x = \dfrac{-3}{2}$
when $\;$ $p = \dfrac{1}{9}$, $\;$ $3^{2x} = \dfrac{1}{9} = 3^{-2}$
$\implies$ $2x = -2$ $\implies$ $x = -1$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{-1, \; \dfrac{-3}{2} \right\}$