Question

If ${\log }_{y}x-{\log }_{x}y=\frac{8}{3}$ and $xy=16$, then the values of $x$ and $y$ are

• A. $x=8,y=2$
• B. $x=\frac{1}{4},y=64$
• C. $x=4,y=4$
• D. $x=64,y=\frac{1}{4}$

Answer 1

Answer: (A), (B)

$x=8,y=2$
$x=\frac{1}{4},y=64$

Given that, $xy=16$ ....(1)
${\log }_{y}x-{\log }_{x}y=\frac{8}{3}$
$\Rightarrow {\log }_{y}x-\frac{1}{{\log }_{y}x}=\frac{8}{3}[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$
On substituting $t={\log }_{y}x$, we get
$\Rightarrow 3t^2-8t-3=0$
$\Rightarrow t=3,-\frac{1}{3}$
$\Rightarrow {\log }_{y}x=3$
$\Rightarrow x=y^3$ ....(2)
and ${\log }_{y}x=-\frac{1}{3}$
$\Rightarrow x^{3}y=1$ ....(3)

On solving (1) & (2), we get
$x=8,y=2$
On solving (1) & (3), we get
$x=\frac{1}{4},y=64$

Ans: A,B