1. **State the problem:** Solve for $x$ in the equation $$\log_a(4x) - \log_a(2x+1) = \frac{1}{2} \log_4 4.$$\n\n2. **Recall logarithm properties:**\n- The difference of logarithms with the same base is the logarithm of the quotient: $$\log_a M - \log_a N = \log_a \left(\frac{M}{N}\right).$$\n- The power rule for logarithms: $$c \log_b M = \log_b M^c.$$\n- Also, note that $$\log_4 4 = 1$$ because any log base of itself is 1.\n\n3. **Apply the difference rule on the left side:**\n$$\log_a \left(\frac{4x}{2x+1}\right) = \frac{1}{2} \log_4 4.$$\n\n4. **Simplify the right side:**\nSince $$\log_4 4 = 1,$$ then $$\frac{1}{2} \log_4 4 = \frac{1}{2} \times 1 = \frac{1}{2}.$$\nSo the equation becomes:\n$$\log_a \left(\frac{4x}{2x+1}\right) = \frac{1}{2}.$$\n\n5. **Rewrite the equation in exponential form:**\nBy definition, $$\log_a y = b \iff y = a^b.$$\nSo,\n$$\frac{4x}{2x+1} = a^{\frac{1}{2}} = \sqrt{a}.$$\n\n6. **Solve for $x$:**\nMultiply both sides by $2x+1$:\n$$4x = \sqrt{a} (2x + 1).$$\n\n7. **Distribute $\sqrt{a}$:**\n$$4x = 2x \sqrt{a} + \sqrt{a}.$$\n\n8. **Group $x$ terms on one side:**\n$$4x - 2x \sqrt{a} = \sqrt{a}.$$\n\n9. **Factor out $x$:**\n$$x(4 - 2 \sqrt{a}) = \sqrt{a}.$$\n\n10. **Divide both sides by $(4 - 2 \sqrt{a})$:**\n$$x = \frac{\sqrt{a}}{4 - 2 \sqrt{a}}.$$\n\n11. **Simplify the denominator by factoring out 2:**\n$$x = \frac{\sqrt{a}}{2(2 - \sqrt{a})}.$$\n\n12. **Rationalize the denominator:**\nMultiply numerator and denominator by the conjugate $(2 + \sqrt{a})$:\n$$x = \frac{\sqrt{a} (2 + \sqrt{a})}{2 (2 - \sqrt{a})(2 + \sqrt{a})}.$$\n\n13. **Simplify the denominator using difference of squares:**\n$$(2 - \sqrt{a})(2 + \sqrt{a}) = 2^2 - (\sqrt{a})^2 = 4 - a.$$\n\n14. **Expand numerator:**\n$$\sqrt{a} \times 2 + \sqrt{a} \times \sqrt{a} = 2 \sqrt{a} + a.$$\n\n15. **Final expression for $x$:**\n$$x = \frac{2 \sqrt{a} + a}{2 (4 - a)}.$$\n\n**Answer:** $$\boxed{x = \frac{2 \sqrt{a} + a}{2 (4 - a)}}.$$