1. **State the problem:** Given the equation $$4^m = p(2^{2m-1}) + p$$, show that for $$p \neq 2$$, it can be rewritten as $$m = \frac{1}{2} \log_2 \left( \frac{2p}{2 - p} \right)$$. 2. **Rewrite the bases:** Note that $$4^m = (2^2)^m = 2^{2m}$$. Substitute this into the equation: $$2^{2m} = p \cdot 2^{2m - 1} + p$$. 3. **Isolate terms involving $$2^{2m}$$:** Rewrite the right side: $$p \cdot 2^{2m - 1} + p = p \cdot 2^{2m - 1} + p \cdot 1$$. 4. **Divide both sides by $$2^{2m - 1}$$ to simplify:** $$\frac{2^{2m}}{2^{2m - 1}} = \frac{p \cdot 2^{2m - 1}}{2^{2m - 1}} + \frac{p}{2^{2m - 1}}$$ which simplifies to $$2^{2m - (2m - 1)} = p + p \cdot 2^{-(2m - 1)}$$ $$2^1 = p + p \cdot 2^{-(2m - 1)}$$ $$2 = p + p \cdot 2^{-(2m - 1)}$$. 5. **Isolate the exponential term:** $$2 - p = p \cdot 2^{-(2m - 1)}$$ Divide both sides by $$p$$: $$\frac{2 - p}{p} = 2^{-(2m - 1)}$$. 6. **Rewrite the right side:** $$2^{-(2m - 1)} = \frac{1}{2^{2m - 1}}$$, so $$\frac{2 - p}{p} = \frac{1}{2^{2m - 1}}$$ which implies $$2^{2m - 1} = \frac{p}{2 - p}$$. 7. **Take the base-2 logarithm of both sides:** $$2m - 1 = \log_2 \left( \frac{p}{2 - p} \right)$$. 8. **Solve for $$m$$:** $$2m = 1 + \log_2 \left( \frac{p}{2 - p} \right)$$ $$m = \frac{1}{2} \left( 1 + \log_2 \left( \frac{p}{2 - p} \right) \right)$$. 9. **Rewrite the expression inside the logarithm:** Recall that $$1 = \log_2 2$$, so $$m = \frac{1}{2} \left( \log_2 2 + \log_2 \left( \frac{p}{2 - p} \right) \right) = \frac{1}{2} \log_2 \left( 2 \cdot \frac{p}{2 - p} \right) = \frac{1}{2} \log_2 \left( \frac{2p}{2 - p} \right)$$. **Final answer:** $$m = \frac{1}{2} \log_2 \left( \frac{2p}{2 - p} \right)$$, as required.