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

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Basic of Logarithms

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

A

$11/48$

B

$11/24$

C

$11/8$

D

$11/96$

Solution

(d) ${2^x} = {2^{2y}} = {2^{3z}}\,i.e.,\,x = 2y = 3z = k$ (say).

Then $xyz = {{{k^3}} \over 6} = 288$, So $k = 12$

.$\therefore x = 12,y = 6,z = 4$ Therefore, ${1 \over {2x}} + {1 \over {4y}} + {1 \over {8z}} = {{11} \over {96}}$

Standard 11

Mathematics

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