If the fourth term in the binomial expansion of `[sqrt(x^((1)/(1+ log_(10)x)))+ x^((1)/(12))]^(6)` is equal to 200 and `x gt 1`, then the value of x is

Options

A. `10^(sqrt(2))`
B. 10
C. `10^(4)`
D. none of these

Answer

To solve the problem, we need to find the value of \( x \) such that the fourth term in the binomial expansion of

\[ \left[\sqrt{x^{\frac{1}{1+\log_{10}x}}} + x^{\frac{1}{12}}\right]^6 \]

is equal to 200, given that \( x > 1 \).

### Step 1: Identify the terms in the binomial expansion

The expression can be rewritten as:

\[ \left[x^{\frac{1}{2(1+\log_{10}x)}} + x^{\frac{1}{12}}\right]^6 \]

Let \( a = x^{\frac{1}{2(1+\log_{10}x)}} \) and \( b = x^{\frac{1}{12}} \).

### Step 2: Use the Binomial Theorem

According to the Binomial Theorem, the \( r \)-th term in the expansion of \( (a + b)^n \) is given by:

\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

For our case, we want the fourth term (\( r = 3 \)):

\[ T_4 = \binom{6}{3} a^{6-3} b^3 = \binom{6}{3} a^3 b^3 \]

### Step 3: Calculate \( \binom{6}{3} \)

\[ \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]

### Step 4: Substitute \( a \) and \( b \)

Now substituting \( a \) and \( b \):

\[ T_4 = 20 \left(x^{\frac{1}{2(1+\log_{10}x)}}\right)^3 \left(x^{\frac{1}{12}}\right)^3 \]

This simplifies to:

\[ T_4 = 20 \cdot x^{\frac{3}{2(1+\log_{10}x)}} \cdot x^{\frac{3}{12}} = 20 \cdot x^{\frac{3}{2(1+\log_{10}x)} + \frac{1}{4}} \]

### Step 5: Set the equation equal to 200

We know that \( T_4 = 200 \):

\[ 20 \cdot x^{\frac{3}{2(1+\log_{10}x)} + \frac{1}{4}} = 200 \]

Dividing both sides by 20 gives:

\[ x^{\frac{3}{2(1+\log_{10}x)} + \frac{1}{4}} = 10 \]

### Step 6: Take logarithms

Taking logarithm base 10 on both sides:

\[ \frac{3}{2(1+\log_{10}x)} + \frac{1}{4} = 1 \]

### Step 7: Solve for \( \log_{10}x \)

Rearranging gives:

\[ \frac{3}{2(1+\log_{10}x)} = 1 - \frac{1}{4} \]

Calculating the right-hand side:

\[ 1 - \frac{1}{4} = \frac{3}{4} \]

Thus:

\[ \frac{3}{2(1+\log_{10}x)} = \frac{3}{4} \]

Cross-multiplying gives:

\[ 3 \cdot 4 = 3 \cdot 2(1+\log_{10}x) \]

Simplifying:

\[ 12 = 6(1+\log_{10}x) \]

Dividing by 6:

\[ 2 = 1 + \log_{10}x \]

Subtracting 1:

\[ \log_{10}x = 1 \]

### Step 8: Solve for \( x \)

Exponentiating gives:

\[ x = 10^1 = 10 \]

### Conclusion

The value of \( x \) is \( 10 \).