Solve the following system of equations: $\begin{cases} \log_4 x - \log_x y = \dfrac{7}{6} \\ xy = 16 \end{cases}$
Given system of equations:
$\log_4 x - \log_x y = \dfrac{7}{6}$ $\;\;\; \cdots \; (1)$
$xy = 16$ $\;\;\; \cdots \; (2)$
Equation $(1)$ $\implies$ $6 \log_4 x - 6 \log_x y = 7$ $\;\;\; \cdots \; (3)$
Equation $(2)$ $\implies$ $y = \dfrac{16}{x}$ $\;\;\; \cdots \; (4)$
In view of equation $(4)$, equation $(3)$ becomes
$6 \log_4 x - 6 \log_x \left(\dfrac{16}{x}\right) = 7$
i.e. $\;$ $6 \log_4 x - 6 \log_x 16 + 6 \log_x x = 7$
i.e. $\;$ $6 \log_4 x - 6 \left(\dfrac{\log_4 16}{\log_4 x}\right) + 6 \times 1 = 7$
i.e. $\;$ $6 \log_4 x - 6 \left(\dfrac{2}{\log_4 x}\right) = 1$
i.e. $\;$ $6 \left(\log_4 x\right)^2 - \log_4 x - 12 = 0$
i.e. $\;$ $6 \left(\log_4 x\right)^2 - 9 \log_4 x + 8 \log_4 x - 12 = 0$
i.e. $\;$ $3 \log_4 x \left(2 \log_4 x - 3\right) + 4 \left(2 \log_4 x - 3\right) = 0$
i.e. $\;$ $\left(3 \log_4 x + 4\right) \left(2 \log_4 x - 3\right) = 0$
i.e. $\;$ $\log_4 x = \dfrac{-4}{3}$ $\;$ or $\;$ $\log_4 x = \dfrac{3}{2}$
i.e. $\;$ $x = 4^{\frac{-4}{3}} = \left(4^{-4}\right)^{\frac{1}{3}} = \dfrac{1}{\sqrt[3]{256}}$ $\;$ is an irrational number
or $\;$ $x = 4^{\frac{3}{2}} = \left(4^3\right)^{\frac{1}{2}} = \sqrt{64} = 8$
When $\;$ $x = \dfrac{1}{\sqrt[3]{256}}$, $\;$ the base of the term $\;$ $\log_x y$ $\;$ in equation $(1)$ becomes irrational.
But, base of a logarithm cannot be irrational.
$\therefore \;$ $x = \dfrac{1}{\sqrt[3]{256}}$ $\;$ is not a valid solution.
When $\;$ $x = 8$, $\;$ we have from equation $(4)$, $\;$ $y = \dfrac{16}{8} = 2$
$\therefore \;$ The solution to the given system of equations is $\;\;$ $x = \left\{\left(8, \; 2\right) \right\}$