Question

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

• A. $10\sqrt{2}$
• B. $10$
• C. ${10}^4$
• D. $10/\sqrt{2}$

Answer 1

The correct option is B $10$

${(\sqrt{x^{\left(\frac{1}{1+\log x}\right)}}+\sqrt[12]{12x})}^6$
$={(x^{\frac{1}{2(1+\log x)}}+x^{\frac{1}{12}})}^6$
Now, $T_4=T_{3+1}{=}^6{C}_3{x}^{\frac{3}{2(1+\log x)}}{x}^{\frac{1}{4}}$
$\Rightarrow 200=20x^{\frac{3}{2(1+\log x)}+\frac{1}{4}}$
$\Rightarrow 10=x^{\frac{3}{2(1+\log x)}+\frac{1}{4}}$
Taking log on both sides,
$\Rightarrow 1=\left[\frac{3}{2(1+\log x)}+\frac{1}{4}\right]lo{g}_{10}x$
$\Rightarrow \frac{1}{logx}-\frac{1}{4}=\frac{3}{2(1+\log x)}$
$\Rightarrow (\log x{)}^2+3\log x-4=0$
$\Rightarrow (\log x+4)(\log x-1)=0$
$\Rightarrow {\log }_{10}x=1$ ($\because \log x=-4$, not possible )
$\Rightarrow x=10$