Solve the following system of equations: $\begin{cases} 5 \log_2 x = \log_2 y^3 - \log_{\sqrt{2}} 2 \\ \log_2 y = 8 - \log_{\sqrt{2}} x \end{cases}$
Given system of equations:
$5 \log_2 x = \log_2 y^3 - \log_{\sqrt{2}} 2$ $\;\;\; \cdots \; (1)$
$\log_2 y = 8 - \log_{\sqrt{2}} x$ $\;\;\; \cdots \; (2)$
We have from equation $(1)$, $\;$ $5 \log_2 x = 3 \log_2 y - \log_{\sqrt{2}} \left(\sqrt{2}\right)^2$
i.e. $\;$ $5 \log_2 x - 3 \log_2 y = - 2$ $\;\;\; \cdots \; (3)$
We have from equation $(2)$, $\;$ $\log_2 y = 8 - \dfrac{\log_2 x}{\log_2 \sqrt{2}}$
i.e. $\;$ $\log_2 y = 8 - \dfrac{\log_2 x}{\log_2 2^{\frac{1}{2}}}$
i.e. $\;$ $\log_2 y = 8 - \dfrac{\log_2 x}{\dfrac{1}{2}}$
i.e. $\;$ $2 \log_2 x + \log_2 y = 8$ $\;\;\; \cdots \; (4)$
Multiplying equation $(4)$ by $3$ gives
$6 \log_2 x + 3 \log_2 y = 24$ $\;\;\; \cdots \; (5)$
Adding equations $(3)$ and $(5)$ gives
$11 \log_2 x = 22$
i.e. $\;$ $\log_2 x = 2$
i.e. $\;$ $x = 2^2 = 4$
Substituting the value of $x$ in equation $(4)$ gives
$2 \log_2 4 + \log_2 y = 8$
i.e. $\;$ $\log_2 y = 8 - 2 \log_2 2^2$
i.e. $\;$ $\log_2 y = 8 - 2 \times 2 = 4$
i.e. $\;$ $y = 2^4 = 16$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(4, \; 16\right) \right\}$