Question

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

• A. $10$
• B. ${10}^{-4}$
• C. $1$
• D. $-4$

Answer 1

The correct option is A $10$

$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
Applying to the above question, we get
$T_{3+1}={}^6{C}_3{x}^{\frac{3}{2(\log x+1)}}{x}^{\frac{3}{12}}$
$=200$
$20x^{\frac{3}{2(\log x+1)}}{x}^{\frac{3}{12}}=200$
$x^{\frac{3}{2(\log x+1)}+\frac{1}{4}}=10$
Taking $\log$ to the base $10$ (${\log }_{10}$) on both sides, we get
$\frac{3}{2(\log x+1)}+\frac{1}{4}=1$
$\frac{3}{2(\log x+1)}=\frac{3}{4}$
$4=2({\log }_{10}x+1)$
$2={\log }_{10}x+1$
$1={\log }_{10}x$
Taking anti-logarithm, we get
$x=10$