Solve the following system of equations: $\begin{cases} \log_4 x - \log_2 y = 0 \\ x^2 - 5y^2 + 4 = 0 \end{cases}$
Given system of equations:
$\log_4 x - \log_2 y = 0$ $\;\;\; \cdots \; (1)$
$x^2 - 5y^2 + 4 = 0$ $\;\;\; \cdots \; (2)$
Equation $(1)$ can be written as $\;\;$ $\dfrac{\log_2 x}{\log_2 4} = \log_2 y$
i.e. $\;$ $\dfrac{\log_2 x}{\log_2 2^2} = \log_2 y$
i.e. $\;$ $\dfrac{\log_2 x}{2 \log_2 2} = \log_2 y$
i.e. $\;$ $\dfrac{\log_2 x}{2} = \log_2 y$
i.e. $\;$ $\log_2 x = 2 \log_2 y$
i.e. $\;$ $\log_2 x = \log_2 y^2$ $\;\;\; \cdots \; (1a)$
Taking antilog to the base $2$ on both sides of equation $(1a)$ gives
$x = y^2$ $\;\;\; \cdots \; (3)$
In view of equation $(3)$, equation $(2)$ becomes
$x^2 - 5x^2 + 4 = 0$
i.e. $\;$ $\left(x - 4\right) \left(x - 1\right) = 0$
i.e. $\;$ $x = 4$ $\;$ or $\;$ $x = 1$
Substituting the values of $x$ in equation $(3)$ give
when $\;$ $x = 4$ $\implies$ $y^2 = 4$ $\implies$ $y = \pm 2$
when $\;$ $x = 1$ $\implies$ $y^2 = 1$ $\implies$ $y = \pm 1$
When $\;$ $y = -2$ $\;$ or $\;$ $y = -1$, $\;$ the term $\;$ $\log_2 y$ $\;$ in equation $(1)$ becomes $\;$ $\log_2 \left(-2\right)$ $\;$ and $\;$ $\log_2 \left(-1\right)$ $\;$ respectively.
But logarithm of a negative number is not defined.
$\implies$ $y = -2$ $\;$ and $\;$ $y = -1$ $\;$ are not valid solutions to the given system of equations.
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(4, 2\right), \; \left(1, 1\right) \right\}$