Solve the equation: $\;$ $\log_2 \left(4^{x+1} + 4\right) \times \log_2 \left(4^x + 1\right) = \log_{\frac{1}{\sqrt{2}}} \sqrt{\dfrac{1}{8}}$
Given equation: $\;\;$ $\log_2 \left(4^{x+1} + 4\right) \times \log_2 \left(4^x + 1\right) = \log_{\frac{1}{\sqrt{2}}} \sqrt{\dfrac{1}{8}}$
i.e. $\;$ $\log_2 \left(4^x \times 4 + 4\right) \times \log_2 \left(4^x + 1\right) = \dfrac{\log_2 \sqrt{\dfrac{1}{8}}}{\log_2 \dfrac{1}{\sqrt{2}}}$
i.e. $\;$ $\log_2 \left[4 \left(4^x + 1\right)\right] \times \log_2 \left(4^x + 1\right) = \dfrac{\log_2 2^{\frac{-3}{2}}}{\log_2 2^{\frac{-1}{2}}}$
i.e. $\;$ $\left[\log_2 4 + \log_2 \left(4^x + 1\right)\right] \times \log_2 \left(4^x + 1\right) = \dfrac{\dfrac{-3}{2} \log_2 2}{\dfrac{-1}{2} \log_2 2}$
i.e. $\;$ $\log_2 2^2 \times \log_2 \left(4^x + 1\right) + \left[\log_2 \left(4^x + 1\right)\right]^2 = 3$
i.e. $\;$ $\left[\log_2 \left(4^x + 1\right)\right]^2 + 2 \log_2 \left(4^x + 1\right) - 3 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $\log_2 \left(4^x + 1\right) = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$p^2 + 2p - 3 = 0$
i.e. $\;$ $\left(p + 3\right) \left(p - 1\right) = 0$
i.e. $\;$ $p = -3$ $\;$ or $\;$ $p = 1$
Substituting the value of $p$ in equation $(2)$ gives
when $\;$ $p = -3$,
$\log_2 \left(4^x + 1\right) = -3$
i.e. $\;$ $4^x + 1 = 2^{-3} = \dfrac{1}{8}$
i.e. $\;$ $4^x = \dfrac{-7}{8}$ $\implies$ $x = \log_4 \left(\dfrac{-7}{8}\right)$
But, logarithim of a negative number is not defined.
$\therefore \;$ $p = -3$ $\;$ is not a valid solution.
when $\;$ $p = 1$,
$\log_2 \left(4^x + 1\right) = 1$
i.e. $\;$ $4^x + 1 = 2^1 = 2$
i.e. $\;$ $4^x = 1$
i.e. $\;$ $x = \log_4 1$ $\implies$ $x = 0$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{0 \right\}$