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Basic of Logarithms
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જો ${2^x} = {4^y} = {8^z}$ અને $xyz = 288,$ તો ${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
