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

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(d) ${2^x} = {2^{2y}} = {2^{3z}}\,i.e.,\,x = 2y = 3z = k$ (say).

Then $xyz = {{{k^3}} \over 6} = 288$, So $k = 12$

.$	herefore x = 12,y = 6,z =

Vedclass · Accepted Answer

$11/96$

Vedclass · Answer

(d) ${2^x} = {2^{2y}} = {2^{3z}}\,i.e.,\,x = 2y = 3z = k$ (say).

Then $xyz = {{{k^3}} \over 6} = 288$, So $k = 12$

.$	herefore x = 12,y = 6,z = 4$ Therefore, ${1 \over {2x}} + {1 \over {4y}} + {1 \over {8z}} = {{11} \over {96}}$

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

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