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 value of \( x \) such that the third term in the binomial expansion of \( (1 + x^{\log_2 x})^5 \) equals 2560.

### Step-by-step Solution:

1. **Identify the Binomial Expansion**:
The binomial expansion of \( (1 + a)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^r \]
Here, \( n = 5 \) and \( a = x^{\log_2 x} \).

2. **Find the Third Term**:
The third term \( T_3 \) corresponds to \( r = 2 \):
\[ T_3 = \binom{5}{2} (x^{\log_2 x})^2 \]

3. **Calculate the Binomial Coefficient**:
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]
Therefore, the third term becomes:
\[ T_3 = 10 (x^{\log_2 x})^2 \]

4. **Set the Third Term Equal to 2560**:
We have:
\[ 10 (x^{\log_2 x})^2 = 2560 \]
Dividing both sides by 10:
\[ (x^{\log_2 x})^2 = 256 \]

5. **Take the Square Root**:
Taking the square root of both sides gives:
\[ x^{\log_2 x} = 16 \quad \text{(since } \sqrt{256} = 16\text{)} \]

6. **Express 16 in Terms of Powers of 2**:
We know that \( 16 = 2^4 \), so we can write:
\[ x^{\log_2 x} = 2^4 \]

7. **Use Logarithmic Properties**:
Taking logarithm base 2 of both sides:
\[ \log_2(x^{\log_2 x}) = \log_2(2^4) \]
This simplifies to:
\[ \log_2 x \cdot \log_2 x = 4 \]
Let \( y = \log_2 x \), then:
\[ y^2 = 4 \]

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

9. **Convert Back to \( x \)**:
- If \( y = 2 \):
\[ \log_2 x = 2 \implies x = 2^2 = 4 \]
- If \( y = -2 \):
\[ \log_2 x = -2 \implies x = 2^{-2} = \frac{1}{4} \]

### Possible Values of \( x \):
Thus, the possible values of \( x \) are:
\[ x = 4 \quad \text{or} \quad x = \frac{1}{4} \]