If log_2 x + log_x 2 = frac{10}{3} = log_2 y | vedclass

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(D) Let $t = \log_2 x$. Then the equation becomes $t + \frac{1}{t} = \frac{10}{3}$.Multiplying by $3t$,we get $3t^2 - 10t + 3 = 0$.Factoring the quadr

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(D) Let $t = \log_2 x$. Then the equation becomes $t + \frac{1}{t} = \frac{10}{3}$.Multiplying by $3t$,we get $3t^2 - 10t + 3 = 0$.Factoring the quadratic equation: $3t^2 - 9t - t + 3 = 0 \Rightarrow 3t(t - 3) - 1(t - 3) = 0$.This gives $(3t - 1)(t - 3) = 0$,so $t = 3$ or $t = 1/3$.Since $x 
eq y$,we assign $\log_2 x = 3$ and $\log_2 y = 1/3$ (or vice versa).Thus,$x = 2^3 = 8$ and $y = 2^{1/3} = \sqrt[3]{2}$.Therefore,$x + y = 8 + \sqrt[3]{2}$.

If $\log_2 x + \log_x 2 = \frac{10}{3} = \log_2 y + \log_y 2$ and $x \neq y$,then $x + y = $

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