Question

If $x, y, z \geq 1$ such that $\log _{\frac{3}{4}}(x y)=\left(\log _{\frac{4}{3}} x\right) \cdot\left(\log _4 y\right)$ and $2\left(\log _4 x \log _2 z\right)=\log _8(x z)-\frac{2}{3}$, then find the value of $\frac{2 x+3 y+4 z}{3}$.

Answer 1

$ \log _{\frac{3}{4}}(x y)=\log _{\frac{4}{3}} x \log _4 y \text { LHS } \leq 0 \text { and RHS } \geq 0 $
$\Rightarrow \log _{\frac{3}{4}}(x y)=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(x z)-\frac{1}{3}, \text { put } x=1 $
$0=\log _8(x z)-\frac{2}{3} \Rightarrow \log _2 z=2 \Rightarrow z=4$
$\therefore \frac{2 x+3 y+4 z}{3}=7 $