If log 3( log 2 x)+ log (1/3)( log (1/2) y)=1 and x y2=9, then

Question

If log 3( log 2 x)+ log (1/3)( log (1/2) y)=1 and x y2=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.

Tardigrade Admin · Accepted Answer

log 3( log 2 x)- log 3( log 2 (1/y))=1 log 3(( log 2 x/ log 2 (1)y))=1 ⇒ log (1/y) x=3 ⇒ x y3=1 Also, x y2=9 ∴ y=(1/9) text and x=729 ∴ x y=81 (x, y)=(729, (1/4))

Q. If $lo g_{3} (lo g_{2} x) + lo g_{\frac{1}{3}} (lo g_{\frac{1}{2}} y) = 1$ and $x y^{2} = 9$ , then

xy = 81

xy = 9

number of possible ordered pairs (x, y) is 2.

number of possible ordered pairs (x, y) is 1.

$lo g_{3} (lo g_{2} x) - lo g_{3} (lo g_{2} \frac{1}{y}) = 1$
$lo g_{3} (\frac{l o g _{2} x}{l o g _{2} \frac{1}{y}}) = 1 \Rightarrow lo g_{\frac{1}{y}} x = 3 \Rightarrow x y^{3} = 1$
Also, $x y^{2} = 9$
$∴ y = \frac{1}{9} and x = 729$
$∴ x y = 81$
$(x, y) = (729, \frac{1}{4})$

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