lf the 6th term in the expansion of ${\left(\frac{1}{x^{8/3}}+x^2{\log }_{10}x\right)}^8$ is

5600, then $x$ equals

It is given that 6th term in the expansion of 

$\begin{array}{l}{\left(\frac{1}{x^{8/3}}+x^2{\log }_{10}x\right)}^8\text{ is }5600\\ \therefore {}^8{C}_5{\left(x^2{\log }_{10}x\right)}^5{\left(\frac{1}{x^{8/3}}\right)}^3=5600\\ \Rightarrow 56x^{10}{\left({\log }_{10}x\right)}^5\cdot \frac{1}{x^8}=5600\\ \Rightarrow x^2{\left({\log }_{10}x\right)}^5=100\end{array}$

$\begin{array}{l}\Rightarrow x^2{\left({\log }_{10}x\right)}^5={10}^2{\left({\log }_{10}10\right)}^5\\ \Rightarrow x=10\end{array}$