If the third term in the binomial expansion o | vedclass

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Question

In the expansion of $\left(1+x^{\log _{2} x}\right)^{5}$

third term say $\mathrm{T}_{3}=^{5} \mathrm{C}_{2}\left(\mathrm{x}^{\log _{2} \mathrm{x}}\

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

In the expansion of $\left(1+x^{\log _{2} x}\right)^{5}$

third term say $\mathrm{T}_{3}=^{5} \mathrm{C}_{2}\left(\mathrm{x}^{\log _{2} \mathrm{x}}\right)^{2}=2560$

$\Rightarrow\left(x^{\log x}\right)^{2}=256$

taking lograthium to the base $2$ on both sides

$\Rightarrow 2\left(\log _{2} x\right)^{2}=8 \Rightarrow\left(\log _{2} x\right)=\pm 2$

$\Rightarrow x=4, \frac{1}{4}$

Here $x=\frac{1}{4}$

If the third term in the binomial expansion of ${\left( {1 + {x^{{{\log }_2}\,x}}} \right)^5}$ equals $2560$, then a possible value of $x$ is

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