If {log _2}x + {log _x}2 = frac{{10}}{3} = {l | vedclass

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(d) We have ${\log _2}x + \frac{1}{{{{\log }_2}x}} = 3 + \frac{1}{3} = {\log _2}y + \frac{1}{{{{\log }_2}y}}$

$	herefore$ ${\log _2}x = 3,{\l

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None of these

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(d) We have ${\log _2}x + \frac{1}{{{{\log }_2}x}} = 3 + \frac{1}{3} = {\log _2}y + \frac{1}{{{{\log }_2}y}}$

$	herefore$ ${\log _2}x = 3,{\log _2}y = \frac{1}{3}$ $(x 
e y)$

==> $x = {2^3}$and$y = {2^{1/3}}$

$\Rightarrow x + y = 8 + {2^{1/3}}$.

If ${\log _2}x + {\log _x}2 = \frac{{10}}{3} = {\log _2}y + {\log _y}2$ and $x \ne y,$ then $x + y = $

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