Solve the following system of equations: $\begin{cases} 4^{\frac{x}{y}} - 3 \times 4^{\frac{5y-x}{y}} = 16 \\ \sqrt{x} - \sqrt{2y} = \sqrt{12} - \sqrt{8} \end{cases}$
Given system of equations:
$4^{\frac{x}{y}} - 3 \times 4^{\frac{5y-x}{y}} = 16$ $\;\;\; \cdots \; (1)$
$\sqrt{x} - \sqrt{2y} = \sqrt{12} - \sqrt{8}$ $\;\;\; \cdots \; (2)$
We have from equation $(1)$
$4^{\frac{x}{y}} - 3 \times 4^5 \times 4^{\frac{-x}{y}} = 16$
i.e. $\;$ $4^{\frac{x}{y}} - \dfrac{3072}{4^{\frac{x}{y}}} = 16$
i.e. $\;$ $\left(4^{\frac{x}{y}}\right)^2 - 16 \times 4^{\frac{x}{y}} - 3072 = 0$
i.e. $\;$ $\left(4^{\frac{x}{y}}\right)^2 - 64 \times 4^{\frac{x}{y}} + 48 \times 4^{\frac{x}{y}} - 3072 = 0$
i.e. $\;$ $4^{\frac{x}{y}} \left(4^{\frac{x}{y}} - 64\right) + 48 \left(4^{\frac{x}{y}} - 64\right) = 0$
i.e. $\;$ $\left(4^{\frac{x}{y}} - 64\right) \left(4^{\frac{x}{y}} + 48\right) = 0$
i.e. $\;$ $4^{\frac{x}{y}} = 64$ $\;$ or $\;$ $4^{\frac{x}{y}} = -48$
i.e. $\;$ $\dfrac{x}{y} = \log_4 64$ $\;$ or $\;$ $\dfrac{x}{y} = \log_4 \left(-48\right)$
But logarithm of a negative number is not defined.
$\therefore \;$ $\dfrac{x}{y} = \log_4 \left(-48\right)$ $\;$ is not a valid solution.
Now, $\;$ $\dfrac{x}{y} = \log_4 64$ $\implies$ $\dfrac{x}{y} = \log_4 4^3$
i.e. $\;$ $\dfrac{x}{y} = 3 \log_4 4 = 3$
$\implies$ $x = 3y$ $\;\;\; \cdots \; (3)$
Substituting the value of $x$ in equation $(2)$ gives
$\sqrt{3y} - \sqrt{2y} = \sqrt{12} - \sqrt{8}$
i.e. $\;$ $\sqrt{y} \left(\sqrt{3} - \sqrt{2}\right) = 2 \sqrt{3} - 2 \sqrt{2}$
i.e. $\;$ $\sqrt{y} \left(\sqrt{3} - \sqrt{2}\right) = 2 \left(\sqrt{3} - \sqrt{2}\right)$
i.e. $\;$ $\sqrt{y} = 2$ $\implies$ $y = 4$
Substituting the value of $y$ in equation $(3)$ gives $\;$ $x = 3 \times 4 = 12$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(12, \; 4\right) \right\}$