The number of real solutions of the equation `2log_2log_2x+log_(1//2)log_2(2sqrt2x)=1` is:

Options

A. 1
B. 2
C. 3
D. more then 3

Answer

To solve the equation \( 2 \log_2 \log_2 x + \log_{\frac{1}{2}} \log_2 (2\sqrt{2}x) = 1 \), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the original equation:
\[ 2 \log_2 \log_2 x + \log_{\frac{1}{2}} \log_2 (2\sqrt{2}x) = 1 \]
We know that \( \log_{\frac{1}{2}} a = -\log_2 a \), so we can rewrite the second term:
\[ 2 \log_2 \log_2 x - \log_2 \log_2 (2\sqrt{2}x) = 1 \]

### Step 2: Simplify the Logarithmic Terms
Next, we can express \( \log_2 (2\sqrt{2}x) \):
\[ \log_2 (2\sqrt{2}x) = \log_2 (2) + \log_2 (\sqrt{2}) + \log_2 (x) = 1 + \frac{1}{2} + \log_2 (x) = \frac{3}{2} + \log_2 (x) \]
Substituting this back into the equation gives us:
\[ 2 \log_2 \log_2 x - \log_2 \left( \frac{3}{2} + \log_2 x \right) = 1 \]

### Step 3: Let \( y = \log_2 \log_2 x \)
Let \( y = \log_2 \log_2 x \). Then, we can rewrite the equation as:
\[ 2y - \log_2 \left( \frac{3}{2} + 2^y \right) = 1 \]
Rearranging gives us:
\[ 2y - 1 = \log_2 \left( \frac{3}{2} + 2^y \right) \]

### Step 4: Exponentiate Both Sides
Exponentiating both sides to eliminate the logarithm:
\[ 2^{2y - 1} = \frac{3}{2} + 2^y \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ 2^{2y - 1} - 2^y - \frac{3}{2} = 0 \]
Let \( z = 2^y \). Then, \( 2^{2y - 1} = \frac{z^2}{2} \), and we can rewrite the equation as:
\[ \frac{z^2}{2} - z - \frac{3}{2} = 0 \]
Multiplying through by 2 to eliminate the fraction gives:
\[ z^2 - 2z - 3 = 0 \]

### Step 6: Solve the Quadratic Equation
Now, we solve the quadratic equation using the quadratic formula:
\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm 4}{2} \]
This gives us:
\[ z = 3 \quad \text{or} \quad z = -1 \]

### Step 7: Back Substitute for \( y \)
Since \( z = 2^y \), we only consider \( z = 3 \) (as \( z = -1 \) is not valid for logarithmic functions):
\[ 2^y = 3 \implies y = \log_2 3 \]

### Step 8: Find \( x \)
Now, substituting back to find \( x \):
\[ \log_2 \log_2 x = \log_2 3 \implies \log_2 x = 3 \implies x = 2^3 = 8 \]

### Conclusion: Number of Real Solutions
Since we derived a valid \( x \) from our calculations, we conclude that there is **1 real solution** to the original equation.