Question

Solve

$\log 2\left(2x^2\right)\log 2\left(16x\right)=\frac{9}{2}{\log }^{2}2x$

Answer 1

${\log }_2\left(2x^2\right)\log 2\left(16x\right)=\frac{9}{2}{\log }^{2}2x$

$\Rightarrow ({\log }_{2}2+2{\log }_{2}x)(4{\log }_{2}2+{\log }_{2}x)=\frac{9}{2}{\left({\log }_{2}x\right)}^2$

$\Rightarrow (1+2{\log }_{2}x)(4+{\log }_{2}x)=\frac{9}{2}{\left({\log }_{2}x\right)}^2$

$\Rightarrow 2({\log }_{2}x{)}^2+9({\log }_{2}x)+4=\frac{9}{2}({\log }_{2}x{)}^2$

$\Rightarrow 4({\log }_{2}x{)}^2+18({\log }_{2}x)+8=9({\log }_{2}x{)}^2$

$\Rightarrow 5({\log }_{2}x{)}^2-18({\log }_{2}x)-8=0$

$\Rightarrow 5({\log }_{2}x{)}^2-20{\log }_{2}x+2{\log }_{2}x-8=0$

$\Rightarrow 5{\log }_{2}x({\log }_{2}x-4)+2({\log }_{2}x-4)=0$

$\Rightarrow ({\log }_{2}x-4)(5{\log }_{2}x+2)=0$

$\Rightarrow {\log }_{2}x=4$

$\Rightarrow x=2^4$

$\Rightarrow x=16$

And, $\Rightarrow 5{\log }_{2}x+2=0$

$\Rightarrow x=2^{-2/5}$

So, $\Rightarrow x=16,2^{-2/5}$