Question

If $x,y,z\ge 1$ such that ${\log }_{\frac{3}{4}}(xy)=\left({\log }_{\frac{4}{3}}x\right)\cdot \left({\log }_{4}y\right)$ and $2\left({\log }_{4}x{\log }_{2}z\right)={\log }_8(xz)-\frac{2}{3}$, then find the value of $\frac{2x+3y+4z}{3}$.

Answer 1

Answer: 7

${\log }_{\frac{3}{4}}(xy)={\log }_{\frac{4}{3}}x{\log }_{4}y\text{ LHS }\le 0\text{ and RHS }\ge 0$
$\Rightarrow {\log }_{\frac{3}{4}}(xy)=0={\log }_{\frac{4}{3}}x\cdot {\log }_{4}y\Rightarrow x=y=1$
$\therefore \text{ From }2{\log }_{4}x\cdot {\log }_{2}z={\log }_8(xz)-\frac{1}{3},\text{ put }x=1$
$0={\log }_8(xz)-\frac{2}{3}\Rightarrow {\log }_{2}z=2\Rightarrow z=4$
$\therefore \frac{2x+3y+4z}{3}=7$