Question

Find the value of $x$, if $4^{{\log }_{9}x}-6x^{{\log }_{9}2}+2^{{\log }_{9}27}=0$

Answer 1

$4^{{\log }_{9}x}-6x^{{\log }_{9}2}+2^{{\log }_{9}27}=0$
$\Rightarrow {\left(2^{{\log }_{9}x}\right)}^2-62^{{\log }_{9}x}+8=0$
Substitute $2^{{\log }_{9}x}=y$, we get

$\Rightarrow y^2-6y+8=0$
$\Rightarrow$ $y=4$ or $2$
When $y=4$

$2^{{\log }_{9}x}=2^2$

$\Rightarrow {\log }_{9}x=2$

$\Rightarrow x=81$

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

$\Rightarrow {\log }_{9}x=1$

$\Rightarrow x=9$