Solve for \( x \). Leave answer in radical form. \( \log _{4} x+\log _{4} 2 x= \)

To solve the equation \( \log_{4} x + \log_{4} (2x) = 0 \), we can use the properties of logarithms. 1. **Combine the logarithms**: Using the property \( \log_b a + \log_b c = \log_b (a \cdot c) \), we can combine the two logarithms: \[ \log_{4} x + \log_{4} (2x) = \log_{4} (x \cdot 2x) = \log_{4} (2x^2) \] 2. **Set the equation**: Now we have: \[ \log_{4} (2x^2) = 0 \] 3. **Convert the logarithmic equation to exponential form**: The equation \( \log_{4} (2x^2) = 0 \) can be rewritten in exponential form: \[ 2x^2 = 4^0 \] Since \( 4^0 = 1 \), we have: \[ 2x^2 = 1 \] 4. **Solve for \( x^2 \)**: Dividing both sides by 2 gives: \[ x^2 = \frac{1}{2} \] 5. **Take the square root**: Taking the square root of both sides, we find: \[ x = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2} \] Since \( x \) must be positive in the context of logarithms, we discard the negative solution. Thus, the final answer is: \[ x = \frac{\sqrt{2}}{2} \]