The solution set of the equation
`"log"_(x)2 xx "log"_(2x)2 = "log"_(4x) 2,` is

Options

A. `{2^(-sqrt(2)),2^sqrt(2)}`
B. `{(1)/(2),2}`
C. `{(1)/4,2^2}`
D. None of these

Answer

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

### Step 1: Rewrite the logarithms
We can rewrite the logarithms using the change of base formula:
\[ \log_x 2 = \frac{\log_2 2}{\log_2 x}, \quad \log_{2x} 2 = \frac{\log_2 2}{\log_2 (2x)}, \quad \log_{4x} 2 = \frac{\log_2 2}{\log_2 (4x)} \]
Since \( \log_2 2 = 1 \), we can simplify:
\[ \log_x 2 = \frac{1}{\log_2 x}, \quad \log_{2x} 2 = \frac{1}{\log_2 (2) + \log_2 (x)} = \frac{1}{1 + \log_2 x}, \quad \log_{4x} 2 = \frac{1}{\log_2 (4) + \log_2 (x)} = \frac{1}{2 + \log_2 x} \]

### Step 2: Substitute into the equation
Substituting these into the original equation gives:
\[ \frac{1}{\log_2 x} \cdot \frac{1}{1 + \log_2 x} = \frac{1}{2 + \log_2 x} \]

### Step 3: Cross-multiply
Cross-multiplying leads to:
\[ 1 = \log_2 x \cdot (1 + \log_2 x) \cdot \frac{1}{2 + \log_2 x} \]
This simplifies to:
\[ (2 + \log_2 x) = \log_2 x + (\log_2 x)^2 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ (\log_2 x)^2 + \log_2 x - 2 = 0 \]

### Step 5: Factor the quadratic equation
This is a standard quadratic equation which can be factored:
\[ (\log_2 x - 1)(\log_2 x + 2) = 0 \]
Setting each factor to zero gives:
\[ \log_2 x - 1 = 0 \quad \text{or} \quad \log_2 x + 2 = 0 \]

### Step 6: Solve for \( x \)
From \( \log_2 x - 1 = 0 \):
\[ \log_2 x = 1 \implies x = 2^1 = 2 \]
From \( \log_2 x + 2 = 0 \):
\[ \log_2 x = -2 \implies x = 2^{-2} = \frac{1}{4} \]

### Final Solution
The solution set is:
\[ \{ 2, \frac{1}{4} \} \]