If the fourth term in the binomial expansion of ${(\sqrt{x^{\left(\frac{1}{1+{\log }_{10}x}\right)}}+x^{\frac{1}{12}})}^6$ is equal to 200, and $x>1$ then the value of x is

Given binomial is ${(\sqrt{x^{\left(\frac{1}{1+{\log }_{10}x}\right)}}+x^{\frac{1}{12}})}^6$ Since, the fourth term in the given expansion is 200.

$\therefore {}^6{C}_3{\left(\frac{1}{x^{1+{\log }_{10}x}}\right)}^{\frac{3}{2}}{\left(\frac{1}{x^{12}}\right)}^2=200\Rightarrow 20\times x^{[\frac{3}{2(1+{\log }_{10}x)}+\frac{1}{4}]}=200$

$\Rightarrow x^{\frac{3}{2(1+{\log }_{10}x)}+\frac{1}{4}}=10\Rightarrow [\frac{3}{2(1+{\log }_{10}x)}+\frac{1}{4}]{\log }_{10}x=1$

$\left[applying{\log }_{10}bothsides\right]$

$\Rightarrow [6+(1+{\log }_{10}x)]{\log }_{10}x=4(1+{\log }_{10}x)\Rightarrow (7+{\log }_{10}x){\log }_{10}x=4+4{\log }_{10}x$

$\Rightarrow t^2+7t=4+4t[let{\log }_{10}x=t]\Rightarrow t^2+3t-4=0\Rightarrow t=1,-4={\log }_{10}x\Rightarrow x=10,{10}^{-4}$

Since,      x > 1      x = 10