Question

If $\log _3\left(\log _2 x\right)+\log _{\frac{1}{3}}\left(\log _{\frac{1}{2}} y\right)=1$ and $x y^2=9$, then

• A. xy = 81
• B. xy = 9
• C. number of possible ordered pairs (x, y) is 2.
• D. number of possible ordered pairs (x, y) is 1.

Answer 1

Answer: (A), (D)

$ \log _3\left(\log _2 x\right)-\log _3\left(\log _2 \frac{1}{y}\right)=1$
$\log _3\left(\frac{\log _2 x}{\log _2 \frac{1}{y}}\right)=1 \Rightarrow \log _{\frac{1}{y}} x=3 \Rightarrow x y^3=1$
Also, $ x y^2=9$
$\therefore y=\frac{1}{9} \text { and } x=729 $
$\therefore x y=81 $
$(x, y)=\left(729, \frac{1}{4}\right)$