Question

If the third term in the binomial expansion of $( 1 + x^{\log_2 x} )^5$ equals $2560$, then a possible value of $x$ is :

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

Answer 1

Answer: (D)

$\left(1+x^{\log _{2} x}\right)^{5}$
$T_{3}={ }^{5} C_{2} \cdot\left(x^{\log _{2} x}\right)^{2}=2560$
$\Rightarrow 10 \cdot x^{2 \log _{2} x}=2560$
$\Rightarrow x^{2 \log _{2} x}=256$
$\Rightarrow 2\left(\log _{2} x\right)^{2}=\log _{2} 256$
$\Rightarrow 2\left(\log _{2} x\right)^{2}=8$
$\Rightarrow \left(\log _{2} x\right)^{2}=4 \quad \Rightarrow \quad \log _{2} x=2$ or $-2$
$x=4$ or $\frac{1}{4}$