Solve the following system of equations: $\begin{cases} x^{\log y} = 2 \\ xy = 20 \end{cases}$
Given system of equations:
$x^{\log y} = 2$ $\;\;\; \cdots \; (1)$
$xy = 20$ $\;\;\; \cdots \; (2)$
Taking log on both sides of equation $(1)$ gives
$\log \left(x^{\log y}\right) = \log 2$
i.e. $\;$ $\log y \times \log x = \log 2$ $\;\;\; \cdots \; (3)$
Equation $(2)$ $\implies$ $x = \dfrac{20}{y}$ $\;\;\; \cdots \; (4)$
In view of equation $(4)$, equation $(3)$ becomes
$\log y \times \log \left(\dfrac{20}{y}\right) = \log 2$
i.e. $\;$ $\log y \left[\log 20 - \log y\right] = \log 2$
i.e. $\;$ $\left(\log y\right)^2 - \log 20 \; \log y + \log 2 = 0$
i.e. $\;$ $\left(\log y\right)^2 - \left[\log \left(10 \times 2\right)\right] \log y + \log 2 = 0$
i.e. $\;$ $\left(\log y\right)^2 - \left[\log 10 + \log 2\right] \log y + \log 2 = 0$
i.e. $\;$ $\left(\log y\right)^2 - \left[1 + \log 2\right] \log y + \log 2 = 0$
i.e. $\;$ $\left(\log y\right)^2 - \log y - \log 2 \; \log y + \log 2 = 0$
i.e. $\;$ $\log y \left(\log y - 1\right) - \log 2 \left(\log y - 1\right) = 0$
i.e. $\;$ $\left(\log y -1\right) \left(\log y - \log 2\right) = 0$
i.e. $\;$ $\log y = 1$ $\;$ or $\;$ $\log y = \log 2$
i.e. $\;$ $y = 10^1 = 10$ $\;$ or $\;$ $y = 2$
When $\;$ $y = 10$, $\;$ we have from equation $(4)$, $\;$ $x = \dfrac{20}{10} = 2$
When $\;$ $y = 2$, $\;$ we have from equation $(4)$, $\;$ $x = \dfrac{20}{2} = 10$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(2, \; 10\right), \; \left(10, \; 2\right) \right\}$