1. **State the problem:** Solve the equation $$\frac{1}{2}n = 4^{n+1}$$ for $n$. 2. **Rewrite the equation:** The equation is $$\frac{1}{2}n = 4^{n+1}$$. 3. **Express the right side with base 2:** Since $4 = 2^2$, we have $$4^{n+1} = (2^2)^{n+1} = 2^{2(n+1)} = 2^{2n+2}$$. 4. **Rewrite the equation:** $$\frac{1}{2}n = 2^{2n+2}$$. 5. **Multiply both sides by 2 to clear the fraction:** $$n = 2 \times 2^{2n+2} = 2^{1} \times 2^{2n+2} = 2^{2n+3}$$. 6. **Now the equation is:** $$n = 2^{2n+3}$$. 7. **Analyze the equation:** The left side is linear in $n$, the right side is exponential in $n$. We look for integer solutions by testing small values. 8. **Test $n=0$:** Left: $0$, Right: $2^{3} = 8$ (no). 9. **Test $n=1$:** Left: $1$, Right: $2^{5} = 32$ (no). 10. **Test $n=2$:** Left: $2$, Right: $2^{7} = 128$ (no). 11. **Test $n=8$:** Left: $8$, Right: $2^{19} = 524288$ (no). 12. **Test $n= -1$:** Left: $-1$, Right: $2^{1} = 2$ (no). 13. **Conclusion:** No integer $n$ satisfies the equation. For real $n$, the equation is transcendental and requires numerical methods. 14. **Summary:** The equation $$\frac{1}{2}n = 4^{n+1}$$ simplifies to $$n = 2^{2n+3}$$, which has no simple algebraic solution. **Final answer:** No simple closed-form solution; numerical methods needed for real $n$.