Question

If the minimum value of the expression $\frac{2^{{\log }_{\sqrt{2}}{x}^2}-3^{{\log }_{27}(x^2+1{)}^3}-2x}{7^{2{\log }_{49}{x}^2}-x-1}$ is $k,$ then $4k$ equals to

Answer 1

$y=\frac{2^{{\log }_{\sqrt{2}}{x}^2}-3^{{\log }_{27}(x^2+1{)}^3}-2x}{7^{2{\log }_{49}{x}^2}-x-1}$
$=\frac{2^{{\log }_2{x}^4}-3^{{\log }_3(x^2+1)}-2x}{7^{{\log }_7{x}^2}-x-1}$
$=\frac{x^4-(x^2+2x+1)}{x^2-x-1}(\because a^{{\log }_{a}b}=b)$
$=\frac{x^4-(x+1{)}^2}{x^2-x-1}=\frac{(x^2-x-1)(x^2+x+1)}{x^2-x-1}$
$=x^2+x+1$

$\Rightarrow y={(x+\frac{1}{2})}^2+\frac{3}{4}$
$\therefore y_{min}=\frac{3}{4}=k$
$\Rightarrow 4k=3$