Solve the equation: $\;$ $\log_2 \left(4^x + 1\right) = x + \log_2 \left(2^{x+3} - 6\right)$
Given equation: $\;\;$ $\log_2 \left(4^x + 1\right) = x + \log_2 \left(2^{x+3} - 6\right)$
i.e. $\;$ $\log_2 \left(4^x + 1\right) - \log_2 \left(2^{x+3} - 6\right) = x$
i.e. $\;$ $\log_2 \left[\dfrac{4^x + 1}{2^{x+3} - 6}\right] = x$
i.e. $\;$ $\dfrac{4^x + 1}{2^{x+3} - 6} = 2^x$
i.e. $\;$ $4^x + 1 = 2^x \left(2^{x+3} - 6\right)$
i.e. $\;$ $4^x + 1 = 2^{2x+3} - 6 \times 2^x$
i.e. $\;$ $\left(2^2\right)^x + 1 = \left(2^2\right)^x \times 2^3 - 6 \times 2^x$
i.e. $\;$ $8 \times \left(2^x\right)^2 - \left(2^x\right)^2 - 6 \times 2^x - 1 = 0$
i.e. $\;$ $7 \times \left(2^x\right)^2 - 6 \times 2^x - 1 = 0$
i.e. $\;$ $7 \times \left(2^x\right)^2 - 7 \times 2^x + 2^x - 1 = 0$
i.e. $\;$ $7 \times 2^x \left(2^x - 1\right) + 1 \left(2^x - 1\right) = 0$
i.e. $\;$ $\left(7 \times 2^x + 1\right) \left(2^x - 1\right) = 0$
i.e. $\;$ $2^x = \dfrac{-1}{7}$ $\;$ or $\;$ $2^x = 1$
i.e. $\;$ $x = \log_2 \left(\dfrac{-1}{7}\right)$ $\;$ or $\;$ $x = \log_2 1$
Logarithim of a negative number is not defined.
$\therefore \;$ $x = \log_2 \left(\dfrac{-1}{7}\right)$ $\;$ is not a valid solution.
When $\;$ $x = \log_2 1$ $\implies$ $x = 0$
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{0 \right\}$