If 2^x = 4^y = 8^z and xyz = 288,then frac{1} | vedclass

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(D) Given $2^x = 4^y = 8^z$. Expressing all in base $2$: $2^x = 2^{2y} = 2^{3z}$. This implies $x = 2y = 3z = k$ (let $k$ be a constant). Then $x = k$

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$11/96$

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(D) Given $2^x = 4^y = 8^z$. Expressing all in base $2$: $2^x = 2^{2y} = 2^{3z}$. This implies $x = 2y = 3z = k$ (let $k$ be a constant). Then $x = k$,$y = k/2$,and $z = k/3$. Given $xyz = 288$,so $k 	imes (k/2) 	imes (k/3) = 288$. $\frac{k^3}{6} = 288 \implies k^3 = 1728 \implies k = 12$. Thus,$x = 12$,$y = 6$,and $z = 4$. Now,calculate $\frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = \frac{1}{2(12)} + \frac{1}{4(6)} + \frac{1}{8(4)} = \frac{1}{24} + \frac{1}{24} + \frac{1}{32}$. $= \frac{2}{24} + \frac{1}{32} = \frac{1}{12} + \frac{1}{32} = \frac{8 + 3}{96} = \frac{11}{96}$.

If $2^x = 4^y = 8^z$ and $xyz = 288$,then $\frac{1}{2x} + \frac{1}{4y} + \frac{1}{8z} = $

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