Which of the following are correct ?

Options

A. graph of `y=2^(log_(2)x)` is
B. graph of `y=x^(log_(x)2)` is
C. graph of `y=2^(log_(2)x^(2)` is
D. graph of `y=2^(2log_(2)x)` is

Answer

To solve the question of determining which of the given graphs are correct, we will analyze each option step by step using logarithmic properties.

### Step-by-Step Solution:

1. **Option 1: \( y = 2 \log x \)**
- Using the property \( a^{\log_m n} = n^{\log_m a} \), we can rewrite:
\[ y = 2 \log x = \log(x^2) \]
- This means \( y = \log(x^2) \) which implies \( x^2 > 0 \). Therefore, \( x > 0 \).
- The graph of \( y = x \) is a straight line through the origin, but since \( x \) must be greater than 0, the graph is valid for \( x > 0 \).
- **Conclusion**: This graph is correct.

2. **Option 2: \( y = x^{\log_2 x} \)**
- We can rewrite this using the property \( \log_m n^p = p \log_m n \):
\[ y = x^{\log_2 x} = 2^{\log_2 x} = 2 \]
- Here, \( x \) must not equal 0 or 1 (since \( \log_2 x \) is undefined at these points).
- The graph of \( y = 2 \) is a horizontal line, which is valid except at \( x = 0 \) and \( x = 1 \).
- **Conclusion**: This graph is correct.

3. **Option 3: \( y = 2 \log_2 (x^2) \)**
- Using the logarithmic property, we can simplify:
\[ y = 2 \log_2 (x^2) = 2 \cdot 2 \log_2 x = 4 \log_2 x \]
- The graph of \( y = x^2 \) is a parabola that opens upwards. The domain for \( x^2 \) is all real numbers except \( x = 0 \).
- **Conclusion**: This graph is correct.

4. **Option 4: \( y = 2^{2 \log_2 x} \)**
- We can simplify this using the property of exponents:
\[ y = 2^{2 \log_2 x} = (2^{\log_2 x})^2 = x^2 \]
- The graph of \( y = x^2 \) is a parabola, and since \( x \) must be greater than 0 for the logarithm to be defined, the domain is \( x > 0 \).
- **Conclusion**: This graph is correct.

### Final Conclusion:
All options provided are correct.