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Step-by-Step Solution
Step 1: Identify the binomial expression and the required term
We are given the binomial
$ \bigl(\sqrt{x^{\tfrac{1}{1 + \log_{10} x}}} + x^{\tfrac{1}{12}}\bigr)^6 $
and told that its fourth term (i.e., $T_4$) is equal to 200, with $x > 1$.
Step 2: Write the general term and substitute for the 4th term
In the expansion of $(a + b)^6$, the $(k+1)$-th term is given by
$ {}^6C_{k} \, a^{6-k} \, b^{k} $.
For the fourth term, $k = 3$, so
$ T_4 = {}^6C_3 \, \bigl(\sqrt{x^{\tfrac{1}{1 + \log_{10} x}}}\bigr)^{6-3} \, \bigl(x^{\tfrac{1}{12}}\bigr)^3 .$
Since ${}^6C_3 = 20$, we have
$ T_4 = 20 \,\bigl(\sqrt{x^{\tfrac{1}{1 + \log_{10} x}}}\bigr)^{3} \,\bigl(x^{\tfrac{1}{12}}\bigr)^3.$
Step 3: Simplify the expression for $T_4$
Note that $\bigl(\sqrt{x^{p}}\bigr)^3 = x^{\tfrac{3p}{2}}$ and $(x^{\tfrac{1}{12}})^3 = x^{\tfrac{3}{12}} = x^{\tfrac{1}{4}}.$
Let $p = \tfrac{1}{1 + \log_{10} x}$. Then:
$ T_4 = 20 \times x^{\tfrac{3p}{2}} \times x^{\tfrac{1}{4}}
= 20 \times x^{\bigl(\tfrac{3p}{2} + \tfrac{1}{4}\bigr)}. $
Substituting $ T_4 = 200 $ gives
$ 20 \times x^{\left(\tfrac{3}{2(1 + \log_{10} x)} + \tfrac{1}{4}\right)} = 200. $
Simplifying this yields
$ x^{\left(\tfrac{3}{2(1 + \log_{10} x)} + \tfrac{1}{4}\right)} = 10. $
Step 4: Take logarithm and define a new variable
Take $\log_{10}$ on both sides. Let $t = \log_{10} x$. Then the equation becomes:
$ \Bigl(\tfrac{3}{2(1 + t)} + \tfrac{1}{4}\Bigr)\,t = 1. $
Step 5: Manipulate the equation in terms of $t$
Rewrite the fraction:
$ \tfrac{3}{2(1 + t)} = \tfrac{3}{2 + 2t}, $
so
$ \tfrac{3}{2(1 + t)} + \tfrac{1}{4}
= \tfrac{3 + \tfrac{(1 + t)}{2}}{2(1 + t)} $
after multiplying a suitable factor. A simpler approach is to find a common denominator:
$ \tfrac{3}{2(1 + t)} + \tfrac{1}{4}
= \tfrac{3 \cdot 2 + (1 + t)}{4(1 + t)}
= \tfrac{6 + (1 + t)}{4(1 + t)}
= \tfrac{(1 + t) + 6}{4(1 + t)}
= \tfrac{1 + t + 6}{4(1 + t)}
= \tfrac{t + 7}{4(1 + t)}. $
Hence the equation becomes:
$ \left(\tfrac{t + 7}{4(1 + t)}\right) t = 1. $
Multiply through by $4(1+t)$:
$ (t + 7)\,t = 4(1 + t). $
This expands to:
$ t^2 + 7t = 4 + 4t. $
Rearranging:
$ t^2 + 3t - 4 = 0. $
Step 6: Solve for $t$ and then for $x$
Factor the quadratic:
$ t^2 + 3t - 4 = (t + 4)(t - 1) = 0, $
yielding the solutions
$ t = -4 $ or $ t = 1 $.
Since $t = \log_{10} x$, we have:
1) If $ t = 1 $, then $ \log_{10} x = 1 \implies x = 10. $
2) If $ t = -4 $, then $ \log_{10} x = -4 \implies x = 10^{-4}. $
Because we are given that $ x > 1 $, we discard $ x = 10^{-4} $.
Hence the valid solution is $ x = 10 $.
Step 7: Conclude the value of $x$
Therefore, the value of $x$ that satisfies the given conditions and makes the fourth term 200 is
$ \boxed{10}. $
