a) Show that \( \log_{r^2} x = \frac{1}{2} \log_r x \):

Method 1:
Use change of base: \( \log_{r^2} x = \frac{\log_r x}{\log_r r^2} \). [1]
Since \( \log_r r^2 = 2 \log_r r = 2 \), we get \( \frac{\log_r x}{2} = \frac{1}{2} \log_r x \). [1]
Thus: \( \log_{r^2} x = \frac{1}{2} \log_r x \). [2]

Method 2:
Use change of base: \( \log_{r^2} x = \frac{1}{\log_x r^2} \). [1]
Since \( \log_x r^2 = 2 \log_x r \), we get \( \frac{1}{2 \log_x r} = \frac{\log_r x}{2} \). [1]
Thus: \( \log_{r^2} x = \frac{1}{2} \log_r x \). [2]

b) Express \( y \) in terms of \( x \):

Method 1:
Given: \( \log_2 y + \log_4 x + \log_4 2x = 0 \).
Combine: \( \log_4 (x \cdot 2x) = \log_4 2x^2 \), so \( \log_2 y + \log_4 2x^2 = 0 \). [1]
Change base: \( \log_4 2x^2 = \frac{1}{2} \log_2 2x^2 \), so \( \log_2 y + \frac{1}{2} \log_2 2x^2 = 0 \). [1]
Simplify: \( \log_2 y = -\frac{1}{2} \log_2 2x^2 \). [1]
Rewrite: \( \log_2 y = \log_2 \left( \frac{1}{\sqrt{2x^2}} \right) = \log_2 \left( \frac{1}{\sqrt{2} x} \right) \). [1]
Thus: \( y = \frac{1}{\sqrt{2}} x^{-1} \). [1]
Note: \( y \) is in the form \( p x^q \) where \( p = \frac{1}{\sqrt{2}}, q = -1 \). [5]

Method 2:
Given: \( \log_2 y + \log_4 x + \log_4 2x = 0 \).
Change base: \( \log_4 x = \frac{1}{2} \log_2 x \), \( \log_4 2x = \frac{1}{2} \log_2 2x \). [1]
Rewrite: \( \log_2 y + \frac{1}{2} \log_2 x + \frac{1}{2} \log_2 2x = 0 \). [1]
Combine: \( \log_2 y + \log_2 x^{1/2} + \log_2 (2x)^{1/2} = \log_2 \left( y \cdot x^{1/2} \cdot (2x)^{1/2} \right) = \log_2 \sqrt{2xy} = 0 \). [1]
Solve: \( \sqrt{2xy} = 1 \). [1]
Thus: \( y = \frac{1}{\sqrt{2}} x^{-1} \). [1]
Note: \( y \) is in the form \( p x^q \) where \( p = \frac{1}{\sqrt{2}}, q = -1 \). [5]

c) Find the value of \( \alpha \):
From part (b): \( y = \frac{1}{\sqrt{2}} x^{-1} \).
Area of \( R \): \( \int_1^\alpha \frac{1}{\sqrt{2}} x^{-1} \, dx \). [1]
Integrate: \( \left[ \frac{1}{\sqrt{2}} \ln x \right]_1^\alpha \). [1]
Evaluate: \( \frac{1}{\sqrt{2}} \ln \alpha – \frac{1}{\sqrt{2}} \ln 1 = \frac{1}{\sqrt{2}} \ln \alpha \). [1]
Set equal to area: \( \frac{1}{\sqrt{2}} \ln \alpha = \sqrt{2} \). [1]
Solve: \( \ln \alpha = 2 \), so \( \alpha = e^2 \). [1]
Note: Follow through from part (b) if \( y \) is in the form \( p x^q \). [5]