If the third term in the binomial expansion of
${(1 + 2^{\log_{2} x})}^{5}$ is equal to 2560, then a possible value of $x$ is

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$4 \sqrt{2}$

$\frac{1}{8}$

$2 \sqrt{2}$

$\frac{1}{4}$

answer is D.

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Detailed Solution

The third term $T_{3}$ in the binomial expansion of

${(1 + x^{\log_{2} x})}^{5}$ is given by

$T_{3} =^{5} C_{2} {(x^{\log_{2} x})}^{2} = 10 {(\log_{2} x)}^{2}$

It is given that $T_{3} = 2560$

$\begin{array}{l} ∴ 10 {(x^{\log_{2} x})}^{2} = 2560 \Rightarrow {(x^{\log_{2} x})}^{2} = 256 \Rightarrow x^{\log_{2} x} = 16 \\ \Rightarrow \log_{2} x = \log_{x} 16 \Rightarrow \log_{2} x = \frac{4}{\log_{2} x} \\ \Rightarrow {(\log_{2} x)}^{2} = 4 \Rightarrow \log_{2} x = \pm 2 \Rightarrow x = 4, \frac{1}{4} \end{array}$