Solve the equation: $\;$ $\log_2 \left(4^x + 4\right) = \log_2 2^x + \log_2 \left(2^{x+1} - 3\right)$
Given equation: $\;\;$ $\log_2 \left(4^x + 4\right) = \log_2 2^x + \log_2 \left(2^{x+1} - 3\right)$
i.e. $\;$ $\log_2 \left(4^x + 4\right) - \log_2 2^x - \log_2 \left(2^{x+1} - 3\right) = 0$
i.e. $\;$ $\log_2 \left[\dfrac{4^x + 4}{2^x \left(2^{x+1} - 3\right)}\right] = 0$
i.e. $\;$ $\dfrac{4^x + 4}{2^{2x + 1} - 3 \times 2^x} = 2^0 = 1$
i.e $\;$ $\left(2^2\right)^x + 4 = 2^{2x + 1} - 3 \times 2^x$
i.e. $\;$ $\left(2^x\right)^2 + 3 \times 2^x - \left(2^2\right)^x \times 2 + 4 = 0$
i.e. $\;$ $\left(2^x\right)^2 - 2 \times \left(2^x\right)^2 + 3 \times 2^x + 4 = 0$
i.e. $\;$ $\left(2^x\right)^2 - 3 \times 2^x - 4 = 0$ $\;\;\; \cdots \; (1)$
Let $\;$ $2^x = p$ $\;\;\; \cdots \; (2)$
Then equation $(1)$ becomes
$p^2 - 3p -4 = 0$
i.e. $\;$ $\left(p + 1\right) \left(p - 4\right) = 0$
i.e. $\;$ $p = -1$ $\;$ or $\;$ $p = 4$
When $\;$ $p = -1$, $\;$ we have from equation $(2)$,
$2^x = -1$ $\implies$ $x = \log_2 \left(-1\right)$
But, logarithim of a negative number is not defined.
$\therefore \;$ $p = -1$ $\;$ does not give a valid solution.
When $\;$ $p = 4$, $\;$ we have from equation $(2)$,
$2^x = 4 = 2^2$ $\implies$ $x = 2$
Check
Substituting $\;$ $x = 2$ $\;$ in the given problem, we get,
$\log_2 \left(4^2 + 4\right) = \log_2 2^2 + \log_2 \left(2^{2+1} - 3\right)$
i.e. $\;$ $\log_2 20 = \log_2 4 + \log_2 5$
i.e. $\;$ $\log_2 20 = \log_2 \left(4 \times 5\right)$
i.e. $\;$ $\log_2 20 = \log_2 20$ $\;\;\;$ which is true.
$\therefore \;$ $x = 2$ $\;$ is a valid solution.
$\therefore \;$ The solution to the given equation is $\;\;$ $x = \left\{2 \right\}$