If the $4^{th}$ term of ${(\sqrt{x^{\frac{1}{{\log }_{10}^x+1}}}+x^{\frac{1}{12}})}^6$ is 200 then find ‘x’ when  $x>1$

$$\begin{gathered} T_4=200 \\ \Rightarrow x^{\frac{3}{2}\left(\frac{1}{1+{\log }_{10}^x}\right)+\frac{1}{4}}=10 \\ \Rightarrow \left(\frac{3}{2}\left(\frac{1}{1+{\log }_{10}^x}\right)+\frac{1}{4}\right){\log }_{10}^x=1 \\ {\log }_{10}^x=t \\ \Rightarrow t=-4,1 \\ \therefore x=10 \end{gathered}$$