If the fourth term in the expansion of `{sqrt(1/(""_"x"log^(x+1))+1/(_x 12)}` is equal to `200a n dx >1,` then find `xdot`

Explanation

To solve the problem step by step, we need to find the value of \( x \) given that the fourth term in the expansion of \( \left( \sqrt{x^{\frac{1}{\log(x+1)}}} + \frac{1}{x^{\frac{1}{12}}} \right)^6 \) is equal to 200.

### Step 1: Identify the fourth term in the binomial expansion
The fourth term \( T_4 \) in the expansion of \( (a + b)^n \) is given by the formula:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our case, \( n = 6 \), \( a = \sqrt{x^{\frac{1}{\log(x+1)}}} \), and \( b = \frac{1}{x^{\frac{1}{12}}} \). The fourth term corresponds to \( r = 3 \).
...