Question

If $4{\log }_{\frac{x}{2}}\left(\sqrt{x}\right)+2{\log }_{4x}\left(x^2\right)=3{\log }_{2x}\left(x^3\right)$, then the value/s of x is/are

• A. 1
• B. 2
• C. $\frac{1}{2}$
• D. 4
• E. $\frac{1}{8}$

Answer 1

Answer: (A), (D), (E)

$\frac{1}{8}$

Given that,$4{\log }_{\frac{x}{2}}\left(\sqrt{x}\right)+2{\log }_{4x}\left(x^2\right)=3{\log }_{2x}\left(x^3\right)\Rightarrow \frac{4{\log }_2\left(\sqrt{x}\right)}{{\log }_2(x/2)}+\frac{2{\log }_2\left(x^2\right)}{{\log }_2\left(4x\right)}=\frac{3{\log }_2\left(x^3\right)}{{\log }_2\left(2x\right)}\Rightarrow \frac{4\times \frac{1}{2}{\log }_{2}x}{{\log }_2\left(x\right)-1}+\frac{4{\log }_2\left(x\right)}{2+{\log }_2\left(x\right)}=\frac{9{\log }_2\left(x\right)}{1+{\log }_2\left(x\right)}\mathrm{Let}{\log }_{2}x=t.\mathrm{The}\mathrm{given}\mathrm{equation}\mathrm{can}\mathrm{be}\mathrm{expressed}\mathrm{as}\frac{2t}{t-1}+\frac{4t}{t+2}=\frac{9t}{t+1}\mathrm{or},t\left(\frac{2}{t-1}+\frac{4}{t+2}-\frac{9}{t+1}\right)=0\mathrm{Either}t=0\mathrm{or}\frac{2}{t-1}+\frac{4}{t+2}-\frac{9}{t+1}=0\mathrm{If}\frac{2}{t-1}+\frac{4}{t+2}-\frac{9}{t+1}=0\Rightarrow \frac{2t+4+4t-4}{\left(t-1\right)\left(t+2\right)}=\frac{9}{t+1}\Rightarrow \frac{6t}{t^2+t-2}=\frac{9}{t+1}\Rightarrow 6t^2+6t=9t^2+9t-18\Rightarrow t^2+t-6=0\Rightarrow \left(t+3\right)\left(t-2\right)=0\Rightarrow t=2\mathrm{or}-3\Rightarrow {\log }_{2}x=2\mathrm{or}-3\Rightarrow x=2^2\mathrm{or}{2}^{-3}=4\mathrm{or}\frac{1}{8}\mathrm{If}t=0\Rightarrow {\log }_{2}x=0\Rightarrow x=1$