Question

If the fourth term in the expansion of ${(\sqrt{\frac{1}{x^{\log x+1}}}+x^{1/12})}^6$ is equal to $200$ and $x>1$, then $x$ is equal to

• A. ${10}^{\sqrt{2}}$
• B. $10$
• C. ${10}^4$
• D. None of these

Answer 1

Answer: (B)

$10$

Given $T_4=200$

${\Rightarrow }^6{C}_3{\left(\sqrt{\frac{1}{x^{{\log }_{10}x+1}}}\right)}^3{\left(x^{\frac{1}{12}}\right)}^3=200$

$\Rightarrow 20\times x^{\frac{3}{2\left({\log }_{10}x+1\right)}+\frac{1}{4}}=200\Rightarrow x^{(\frac{3}{2\left({\log }_{10}x+1\right)}+\frac{1}{4})}=10$

$\Rightarrow (\frac{3}{2\left({\log }_{10}x+1\right)}+\frac{1}{4})={\log }_{x}10=\frac{1}{{\log }_{10}x}$

$\Rightarrow \frac{3}{2(y+1)}+\frac{1}{4}=\frac{1}{y}$ where $y={\log }_{10}x$

$\Rightarrow y=-4$ or $y=1$ $\Rightarrow {\log }_{10}x=-4$ or $1$

$\Rightarrow x={10}^{-4}$ or $10$

$\Rightarrow x=10(\because x>1)$