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

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(A) Given ${2^x} = {4^y} = {8^z}$.Writing in terms of base $2$,we get ${2^x} = {2^{2y}} = {2^{3z}}$.Equating the exponents,$x = 2y = 3z = k$ (let).The

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$11/96$

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(A) Given ${2^x} = {4^y} = {8^z}$.Writing in terms of base $2$,we get ${2^x} = {2^{2y}} = {2^{3z}}$.Equating the exponents,$x = 2y = 3z = k$ (let).Then $x = k$,$y = k/2$,and $z = k/3$.Given $xyz = 288$,so $k 	imes (k/2) 	imes (k/3) = 288$.$k^3 / 6 = 288 \implies k^3 = 1728 \implies k = 12$.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 find the value of $\frac{1}{{2x}} + \frac{1}{{4y}} + \frac{1}{{8z}}$.

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