Question

The equation $4{\log }_{\frac{x}{2}}\left(\sqrt{x}\right)+2{\log }_{4x}\left(x^2\right)=3{\log }_{2x}\left(x^3\right)$ is having

• A. all rational roots
• B. two rational and one irrational roots
• C. one rational and two irrational roots
• D. all irrational roots

Answer 1

The correct option is A all rational roots

$4{\log }_{\frac{x}{2}}\left(\sqrt{x}\right)+2{\log }_{4x}\left(x^2\right)=3{\log }_{2x}\left(x^3\right)$
$\Rightarrow \frac{4{\log }_2\left(\sqrt{x}\right)}{{\log }_2\left(x/2\right)}+\frac{2{\log }_2\left(x^2\right)}{{\log }_2\left(4x\right)}=\frac{3{\log }_2\left(x^3\right)}{{\log }_2\left(2x\right)}$
$\Rightarrow \frac{4\times \frac{1}{2}{\log }_2\left(x\right)}{{\log }_2\left(x\right)-1}+\frac{4{\log }_2\left(x\right)}{2+{\log }_2\left(x\right)}=\frac{9{\log }_2\left(x\right)}{1+{\log }_2\left(x\right)}$

Put ${\log }_{2}x=t$
$\frac{2t}{t-1}+\frac{4t}{t+2}=\frac{9t}{t+1}$
$\Rightarrow t=0\text{or}\frac{2}{t-1}+\frac{4}{t+2}=\frac{9}{t+1}$
$\Rightarrow \frac{2t+4+4t-4}{(t-1)(t+2)}=\frac{9}{t+1}$
$\Rightarrow t^2+t-6=0$
$\Rightarrow (t-2)(t+3)=0$
$\Rightarrow t=0,2,-3$
$\Rightarrow x=1,4,\frac{1}{8}$