The number of solutions of the equation
`x^("log"sqrt(x)^(2x)) =4` is

Options

A. 0
B. 1
C. 2
D. infinitely many

Explanation

To solve the equation \( x^{\log_{\sqrt{x}}(2x)} = 4 \), we will follow these steps:

### Step 1: Rewrite the logarithmic expression
We start with the equation:
\[ x^{\log_{\sqrt{x}}(2x)} = 4 \]
Using the change of base formula for logarithms, we can rewrite \( \log_{\sqrt{x}}(2x) \):
\[ \log_{\sqrt{x}}(2x) = \frac{\log(2x)}{\log(\sqrt{x})} \]
Since \( \log(\sqrt{x}) = \frac{1}{2} \log(x) \), we can substitute this into our equation:
\[ \log_{\sqrt{x}}(2x) = \frac{\log(2x)}{\frac{1}{2} \log(x)} = \frac{2\log(2x)}{\log(x)} \]

### Step 2: Substitute back into the equation
Now substituting back into the original equation, we have:
\[ x^{\frac{2\log(2x)}{\log(x)}} = 4 \]

### Step 3: Simplify the left-hand side
We can simplify the left-hand side:
\[ x^{\frac{2(\log(2) + \log(x))}{\log(x)}} = 4 \]
This simplifies to:
\[ x^{\frac{2\log(2)}{\log(x)} + 2} = 4 \]

### Step 4: Express 4 in terms of powers of x
We know that \( 4 = 2^2 \). Therefore, we can rewrite the equation as:
\[ x^{\frac{2\log(2)}{\log(x)} + 2} = 2^2 \]

### Step 5: Set the exponents equal
Since the bases are the same, we can set the exponents equal to each other:
\[ \frac{2\log(2)}{\log(x)} + 2 = 2 \]

### Step 6: Solve for \( x \)
Subtracting 2 from both sides gives us:
\[ \frac{2\log(2)}{\log(x)} = 0 \]
This implies that:
\[ \log(x) \to \infty \text{ (which means } x \to 1 \text{)} \]
However, we have the conditions from the original logarithmic properties that \( x > 0 \) and \( x \neq 1 \).

### Step 7: Conclusion
Since \( x = 1 \) is not a valid solution due to the logarithmic properties, and there are no other values that satisfy the equation, we conclude that the equation has no solutions.

### Final Answer
The number of solutions of the equation \( x^{\log_{\sqrt{x}}(2x)} = 4 \) is **0**.

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