Question

The number of (ordered pairs) solutions of the simultaneous equation
$4{\log }_{y}x-5{\log }_{x}y=19$
${\log }_{2}x+{\log }_{2}y=6$, is

Answer 1

${\log }_{2}x+{\log }_{2}y=6\Rightarrow {\log }_{2}xy=6\Rightarrow xy=2^6=64\cdots \left(1\right)$

$4{\log }_{y}x-5{\log }_{x}y=19\Rightarrow 4{\log }_{y}x-\frac{5}{{\log }_{y}x}=19\Rightarrow 4({\log }_{y}x{)}^2-19{\log }_{y}x-5=0\Rightarrow ({\log }_{y}x-5\left)\right(4{\log }_{y}x+1)=0\Rightarrow {\log }_{y}x=5\text{or}{\log }_{y}x=-\frac{1}{4}$

$\text{When}{\log }_{y}x=5\Rightarrow x=y^5\text{From eqn(1),}{y}^5.y=2^6\Rightarrow y=\pm 2\Rightarrow y=2(\because y\text{can't be negative})\Rightarrow x=2^5=32$

$\text{When}{\log }_{y}x=-\frac{1}{4}\Rightarrow x=y^{-\frac{1}{4}}\Rightarrow x^4=\frac{1}{y}\text{From eqn(1),}x\times \frac{1}{x^4}=2^6\Rightarrow x=\frac{1}{4}\Rightarrow y=2^6\times 4=256$
$\therefore$ There are two solutions : $(32,2)\&(\frac{1}{4},256)$