જો ${2^x} = {4^y} = {8^z}$ અને $xyz = 288,$ તો ${1 \over {2x}} + {1 \over {4y}} + {1 \over {8z}} = $
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d
(d) \({2^x} = {2^{2y}} = {2^{3z}}\,i.e.,\,x = 2y = 3z = k\) (say).
(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, \({1 \over {2x}} + {1 \over {4y}} + {1 \over {8z}} = {{11} \over {96}}\)
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