Find x, if `log_(2x)sqrtx+log_(2sqrtx)x=0`:

Options

A. `1,2^(-5/6)`
B. `1,2^(-6//5)`
C. `4,-2`
D. none of these

Answer

To solve the equation \( \log_{2x} \sqrt{x} + \log_{2\sqrt{x}} x = 0 \), we will follow these steps:

### Step 1: Rewrite the logarithms using the change of base formula
Using the change of base formula, we can express the logarithms in terms of base 2:
\[ \log_{2x} \sqrt{x} = \frac{\log_2 \sqrt{x}}{\log_2 (2x)} \quad \text{and} \quad \log_{2\sqrt{x}} x = \frac{\log_2 x}{\log_2 (2\sqrt{x})} \]

### Step 2: Simplify the logarithms
Now, simplify the logarithms:
- For \( \log_2 \sqrt{x} \):
\[ \log_2 \sqrt{x} = \log_2 x^{1/2} = \frac{1}{2} \log_2 x \]
- For \( \log_2 (2x) \):
\[ \log_2 (2x) = \log_2 2 + \log_2 x = 1 + \log_2 x \]
- For \( \log_2 (2\sqrt{x}) \):
\[ \log_2 (2\sqrt{x}) = \log_2 2 + \log_2 \sqrt{x} = 1 + \frac{1}{2} \log_2 x \]

### Step 3: Substitute back into the equation
Now substitute these back into the equation:
\[ \frac{\frac{1}{2} \log_2 x}{1 + \log_2 x} + \frac{\log_2 x}{1 + \frac{1}{2} \log_2 x} = 0 \]

### Step 4: Set a variable for simplification
Let \( y = \log_2 x \). Then the equation becomes:
\[ \frac{\frac{1}{2} y}{1 + y} + \frac{y}{1 + \frac{1}{2} y} = 0 \]

### Step 5: Find a common denominator and simplify
The common denominator for the left-hand side is \( (1 + y)(1 + \frac{1}{2} y) \):
\[ \frac{\frac{1}{2} y (1 + \frac{1}{2} y) + y(1 + y)}{(1 + y)(1 + \frac{1}{2} y)} = 0 \]
This simplifies to:
\[ \frac{\frac{1}{2} y + \frac{1}{4} y^2 + y + y^2}{(1 + y)(1 + \frac{1}{2} y)} = 0 \]
Combine like terms:
\[ \frac{\frac{5}{4} y^2 + \frac{3}{2} y}{(1 + y)(1 + \frac{1}{2} y)} = 0 \]

### Step 6: Set the numerator to zero
Set the numerator equal to zero:
\[ \frac{5}{4} y^2 + \frac{3}{2} y = 0 \]
Factor out \( y \):
\[ y \left( \frac{5}{4} y + \frac{3}{2} \right) = 0 \]
This gives us two solutions:
1. \( y = 0 \)
2. \( \frac{5}{4} y + \frac{3}{2} = 0 \)

### Step 7: Solve for \( y \)
From \( y = 0 \):
\[ \log_2 x = 0 \implies x = 2^0 = 1 \]

From \( \frac{5}{4} y + \frac{3}{2} = 0 \):
\[ \frac{5}{4} y = -\frac{3}{2} \implies y = -\frac{3/2}{5/4} = -\frac{3 \cdot 4}{2 \cdot 5} = -\frac{6}{5} \]
Thus,
\[ \log_2 x = -\frac{6}{5} \implies x = 2^{-\frac{6}{5}} = \frac{1}{2^{6/5}} = \frac{1}{\sqrt[5]{64}} = \frac{1}{4\sqrt[5]{4}} \]

### Final Solutions
The solutions for \( x \) are:
1. \( x = 1 \)
2. \( x = \frac{1}{\sqrt[5]{64}} \)