Solve the following system of equations: $\begin{cases} \log_2 x + 2 \log_2 y = 3 \\ x^2 + y^4 = 16 \end{cases}$
Given system of equations:
$\log_2 x + 2 \log_2 y = 3$ $\;\;\; \cdots \; (1)$
$x^2 + y^4 = 16$ $\;\;\; \cdots \; (2)$
Equation $(1)$ $\implies$ $\log_2 x + \log_2 y^2 = 3$
i.e. $\;$ $\log_2 \left(x y^2\right) = 3$
i.e. $\;$ $x y^2 = 2^3 = 8$
i.e. $\;$ $y^2 = \dfrac{8}{x}$ $\implies$ $y^4 = \dfrac{64}{x^2}$ $\;\;\; \cdots \; (3)$
Substituting the value of $y^4$ from equation $(3)$ in equation $(2)$ gives
$x^2 + \dfrac{64}{x^2} = 16$
i.e. $\;$ $x^4 - 16 x^2 + 64 = 0$
i.e. $\;$ $\left(x^2 - 8\right)^2 = 0$
i.e. $\;$ $x^2 - 8 = 0$
i.e. $\;$ $x^2 = 8$ $\implies$ $x = \pm 2 \sqrt{2}$
When $x = -2 \sqrt{2}$, the term $\log_2 x$ in equation $(1)$ becomes $\log_2 \left(-2\sqrt{2}\right)$.
But logarithm of a negative number is not defined.
$\implies$ $x - 2 \sqrt{2}$ is not a valid solution.
When $x = + 2 \sqrt{2}$, we have from from equation $(1)$
$\log_2 2 \sqrt{2} + 2 \log_2 y = 3$
i.e. $\;$ $\log_2 2 + \log_2 2^{\frac{1}{2}} + 2 \log_2 y = 3$
i.e. $\;$ $1 + \dfrac{1}{2} + 2 \log_2 y = 3$
i.e. $\;$ $2 \log_2 y = \dfrac{3}{2}$
i.e. $\;$ $\log_2 y = \dfrac{3}{4}$
$\implies$ $y = 2^{\frac{3}{4}} = \left(2^3\right)^{\frac{1}{4}} = \sqrt[4]{8}$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2 \sqrt{2}, \; \sqrt[4]{8}\right) \right\}$