Question

Integral value of $x$ which satisfies the equation ${\log }_{6}54+{\log }_{x}16={\log }_{\sqrt{2}}x-{\log }_{36}\left(4/9\right)$ is

Answer 1

${\log }_{6}54+{\log }_{x}16={\log }_{\sqrt{2}}x-{\log }_{36}\left(4/9\right)$
$\Rightarrow {\log }_{6}54+\frac{4}{{\log }_{2}x}=2{\log }_{2}x-\frac{1}{2}{\log }_6\left(\frac{4}{9}\right)$
put $t={\log }_{2}x$ to obtain ${\log }_{6}54+\frac{4}{t}=2t-\frac{1}{2}{\log }_6\left(\frac{4}{9}\right)$
$\Rightarrow {\log }_6\left(\frac{54\times 2}{3}\right)=2t-\frac{4}{t}$
$\Rightarrow t^2-t-2=0$
$\Rightarrow t=-1,2$
$\Rightarrow x=\frac{1}{2},4$
Ans: 4