Question

If the third term in the binomial expansion of ${(1+x^{{\log }_{2}x})}^5$ equals $2560$, the a possible value of $x$ is :

• A. $\frac{1}{8}$
• B. $\frac{1}{4}$
• C. $2\sqrt{2}$
• D. $4\sqrt{2}$

Answer 1

The correct option is B $\frac{1}{4}$

$T_3={}^5{C}_2{\left(x^{{\log }_{2}x}\right)}^2=2560$
$\Rightarrow \frac{5!}{3!2!}{x}^{2{\log }_{2}x}=2560$
$\Rightarrow (x^{{\log }_{2}x}{)}^2=256$
$\Rightarrow x^{{\log }_{2}x}=16$
Taking ${\log }_2$ both sides
$\Rightarrow {\log }_{2}x\cdot {\log }_{2}x=4{\log }_{2}2$
$\Rightarrow {\left({\log }_{2}x\right)}^2=4$
$\Rightarrow {\log }_{2}x=\pm 2$
$\Rightarrow {\log }_{2}x=2,{\log }_{2}x=-2$
$x=4,x=\frac{1}{4}$