If `sqrt((log)_2x)-0. 5=(log)_2sqrt(x ,)` then `x` equals odd integer (b) prime number composite number (d) irrational

Options

A. odd integer
B. prime number
C. composite number
D. irrational

Explanation

To solve the equation \( \sqrt{\log_2 x} - 0.5 = \log_2 \sqrt{x} \), we will follow these steps:

### Step-by-step Solution:

1. **Rewrite the logarithmic expression:**
We know that \( \log_2 \sqrt{x} = \log_2 (x^{1/2}) \). By using the property of logarithms \( \log_b (a^n) = n \cdot \log_b a \), we can rewrite this as:
\[ \log_2 \sqrt{x} = \frac{1}{2} \log_2 x \]

2. **Substituting into the equation:**
Substitute \( \log_2 \sqrt{x} \) back into the original equation:
\[ \sqrt{\log_2 x} - 0.5 = \frac{1}{2} \log_2 x \]

3. **Let \( t = \sqrt{\log_2 x} \):**
To simplify the equation, let \( t = \sqrt{\log_2 x} \). Therefore, \( \log_2 x = t^2 \). Substitute these into the equation:
\[ t - 0.5 = \frac{1}{2} t^2 \]

4. **Rearranging the equation:**
Rearranging gives:
\[ \frac{1}{2} t^2 - t + 0.5 = 0 \]
To eliminate the fraction, multiply the entire equation by 2:
\[ t^2 - 2t + 1 = 0 \]

5. **Factoring the quadratic equation:**
The equation can be factored as:
\[ (t - 1)^2 = 0 \]
This implies:
\[ t - 1 = 0 \quad \Rightarrow \quad t = 1 \]

6. **Finding \( \log_2 x \):**
Recall that \( t = \sqrt{\log_2 x} \). Thus:
\[ \sqrt{\log_2 x} = 1 \quad \Rightarrow \quad \log_2 x = 1^2 = 1 \]

7. **Solving for \( x \):**
Now, we can solve for \( x \):
\[ \log_2 x = 1 \quad \Rightarrow \quad x = 2^1 = 2 \]

### Conclusion:
The value of \( x \) is \( 2 \).

### Options Analysis:
- (a) Odd integer: **False** (2 is not odd)
- (b) Prime number: **True** (2 is a prime number)
- (c) Composite number: **False** (2 is not composite)
- (d) Irrational: **False** (2 is not irrational)

Thus, the correct option is (b) Prime number.