Question

If $4^{{\log }_{9}x}-6x^{{\log }_{9}2}+2^{{\log }_{3}27}=0,$ then the possible value(s) of $x$ is/are

• A. $2$
• B. $2^9$
• C. $9$
• D. $81$

Answer 1

Answer: (C), (D)

The correct options are
C $9$
D $81$

$4^{{\log }_{9}x}-6x^{{\log }_{9}2}+2^{{\log }_{3}27}=0\Rightarrow 4^{{\log }_{9}x}-6\cdot 2^{{\log }_{9}x}+2^{{\log }_{3}27}=0(\because a^{{\log }_{b}c}=c^{{\log }_{b}a})$
Let $y=2^{{\log }_{9}x}$
Then $y^2-6y+8=0$
$\Rightarrow (y-2)(y-4)=0$
$\Rightarrow y=2,4$

So, $2^{{\log }_{9}x}=4$
$\Rightarrow {\log }_{9}x=2$
$\Rightarrow x=9^2=81$

When $2^{{\log }_{9}x}=2$
$\Rightarrow {\log }_{9}x=1$
$\Rightarrow x=9$