If the sixth term in the expansion of ${\left(\frac{1}{x^{\frac{8}{3}}}+x^2{\log }_{10}x\right)}^8$ is $5600,$ then the value of $x$ is

• A

  $2$
• B

  $\sqrt{5}$
• C

  $\sqrt{10}$
• Correct answer
  D

  $10$

Solution

Note that for ${\log }_{10}x$ to be defined, $x>0.$
We have, $T_6=T_{5+1}={\mathrm{}}^8\mathrm{}{C}_5{\left(\frac{1}{x^{\frac{8}{3}}}\right)}^{8-5}{\left(x^2{\log }_{10}x\right)}^5$
$\Rightarrow 5600=\frac{8\times 7\times 6}{3\times 2}\left(\frac{1}{x^8}\right)x^{10}{\left({\log }_{10}x\right)}^5$
$\Rightarrow 100=x^2{\left({\log }_{10}x\right)}^5$
$\Rightarrow \frac{100}{x^2}={\left({\log }_{10}x\right)}^5$
The curves $y=\frac{100}{x^2}$ and $y={\left({\log }_{10}x\right)}^5$ intersect in just one point $\left(\mathrm{10,1}\right)$

$\Rightarrow x=10$

Hence, the correct answer is option: 'D'.