1. **Problem Statement:** Find the value of $x$ given the logarithmic equation $$\log_2 (2x-1) - 2 \log_4 (x) = 0$$ and the relation $$\log_4 A = \frac{1}{2} \log_2 A$$. 2. **Formula and Rules:** Recall the change of base and logarithm power rules: - $$\log_a b = \frac{\log_c b}{\log_c a}$$ for any base $c$. - $$n \log_a b = \log_a b^n$$. - Given $$\log_4 A = \frac{1}{2} \log_2 A$$, we can rewrite logarithms base 4 in terms of base 2. 3. **Step-by-step Solution:** - Rewrite $$2 \log_4 (x)$$ using the given relation: $$2 \log_4 (x) = 2 \times \frac{1}{2} \log_2 (x) = \log_2 (x)$$. - Substitute into the original equation: $$\log_2 (2x - 1) - \log_2 (x) = 0$$. - Use the logarithm subtraction rule: $$\log_2 \left( \frac{2x - 1}{x} \right) = 0$$. - Since $$\log_2 (y) = 0$$ implies $$y = 1$$, set: $$\frac{2x - 1}{x} = 1$$. - Multiply both sides by $x$: $$2x - 1 = x$$. - Solve for $x$: $$2x - x = 1 \Rightarrow x = 1$$. 4. **Check domain restrictions:** - Argument of logarithms must be positive: - $$2x - 1 > 0 \Rightarrow x > \frac{1}{2}$$. - $$x > 0$$. - Since $x=1$ satisfies both, it is valid. **Final answer:** $$x = 1$$