We are given the expression: \[ 6 + \log_{\frac{1}{4}} \left( \frac{1}{\sqrt{2}} \sqrt{1 - \frac{1}{\sqrt{2}} \sqrt{1 - \frac{1}{\sqrt{2}} \sqrt{1 - \frac{1}{\sqrt{2}} \sqrt{1 - \frac{1}{\sqrt{2}} \dots \infty}}}} \right) \] Step 1: Evaluating the Infinite Nested Radical
Consider the infinite nested radical: \[ x = \sqrt{1 - \frac{1}{\sqrt{2}} x} \] Squaring both sides, \[ x^2 = 1 - \frac{1}{\sqrt{2}} x \] Rearrange into a quadratic equation: \[ x^2 + \frac{1}{\sqrt{2}} x - 1 = 0 \] Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1 \), \( b = \frac{1}{\sqrt{2}} \), and \( c = -1 \): \[ x = \frac{-\frac{1}{\sqrt{2}} \pm \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + 4}}{2} \] \[ x = \frac{-\frac{1}{\sqrt{2}} \pm \sqrt{\frac{1}{2} + 4}}{2} \] \[ x = \frac{-\frac{1}{\sqrt{2}} \pm \sqrt{\frac{9}{2}}}{2} \] \[ x = \frac{-\frac{1}{\sqrt{2}} \pm \frac{3}{\sqrt{2}}}{2} \] \[ x = \frac{-1 + 3}{2\sqrt{2}} \text{ or } x = \frac{-1 - 3}{2\sqrt{2}} \] \[ x = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}}, \quad \text{(choosing the positive root)} \] Thus, \[ x = \frac{1}{\sqrt{2}} \] Step 2: Evaluating the Logarithm
We now substitute \( x = \frac{1}{\sqrt{2}} \): \[ \log_{\frac{1}{4}} \left( \frac{1}{\sqrt{2}} x \right) = \log_{\frac{1}{4}} \left( \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \right) \] \[ = \log_{\frac{1}{4}} \left( \frac{1}{2} \right) \] Since \( \frac{1}{4} = 4^{-1} \), we convert the logarithm: \[ \log_{\frac{1}{4}} \frac{1}{2} = \frac{\log 1/2}{\log 1/4} \] \[ = \frac{\log_4 2^{-1}}{\log_4 4^{-1}} \] \[ = \frac{-\log_4 2}{-1} = \log_4 2 \] Since \( 4 = 2^2 \), we use \( \log_{2^2} 2 = \frac{1}{2} \): \[ \log_4 2 = \frac{1}{2} \] Step 3: Final Calculation \[ 6 + \frac{1}{2} = \frac{13}{2} \]