If the third term in the binomial expansion of `(1+x^(log_(2)x))^(5)` equals 2560, then a possible value of x is:

Options

A. `1/4`
B. `1/8`
C. `4sqrt2`
D. `2sqrt2`

Explanation

To solve the problem, we need to find the possible value of \( x \) given that the third term in the binomial expansion of \( (1 + x^{\log_2 x})^5 \) equals 2560.

### Step-by-Step Solution:

1. **Identify the Third Term**:
The third term in the binomial expansion can be expressed as:
\[ T_3 = \binom{n}{r} a^{n-r} b^r \]
where \( n = 5 \), \( r = 2 \), \( a = 1 \), and \( b = x^{\log_2 x} \). Therefore, the third term is:
\[ T_3 = \binom{5}{2} (1)^{5-2} (x^{\log_2 x})^2 \]
Simplifying this gives:
\[ T_3 = \binom{5}{2} (x^{\log_2 x})^2 = 10 (x^{\log_2 x})^2 \]

2. **Set the Equation**:
We know from the problem statement that this term equals 2560:
\[ 10 (x^{\log_2 x})^2 = 2560 \]

3. **Solve for \( x^{\log_2 x} \)**:
Dividing both sides by 10:
\[ (x^{\log_2 x})^2 = \frac{2560}{10} = 256 \]
Taking the square root of both sides:
\[ x^{\log_2 x} = 16 \]

4. **Express 16 in Terms of Powers of 2**:
We can express 16 as \( 2^4 \):
\[ x^{\log_2 x} = 2^4 \]

5. **Take Logarithm on Both Sides**:
Taking logarithm base 2:
\[ \log_2(x^{\log_2 x}) = \log_2(2^4) \]
This simplifies to:
\[ \log_2 x \cdot \log_2 x = 4 \]

6. **Set Up the Equation**:
Let \( y = \log_2 x \). Then we have:
\[ y^2 = 4 \]

7. **Solve for \( y \)**:
Taking the square root:
\[ y = 2 \quad \text{or} \quad y = -2 \]

8. **Convert Back to \( x \)**:
Now, substituting back for \( x \):
- If \( y = 2 \):
\[ \log_2 x = 2 \implies x = 2^2 = 4 \]
- If \( y = -2 \):
\[ \log_2 x = -2 \implies x = 2^{-2} = \frac{1}{4} \]

9. **Conclusion**:
Therefore, the possible values of \( x \) are \( 4 \) and \( \frac{1}{4} \). Since the question asks for a possible value of \( x \), we can conclude that:
\[ x = \frac{1}{4} \]