1. We are asked to solve the equation $$1^n + 2^n + 3^n + 4^n = 7^{n-1} + 11$$ for $n$. 2. Notice that $1^n = 1$ for any integer $n$, so the left side simplifies to $$1 + 2^n + 3^n + 4^n.$$ 3. We can test some integer values of $n$ to check if equality holds. 4. For $n=1$: $$1 + 2^1 + 3^1 + 4^1 = 1 + 2 + 3 + 4 = 10$$ $$7^{1-1} + 11 = 7^0 + 11 = 1 + 11 = 12$$ So, left side $10 \neq 12$. 5. For $n=2$: $$1 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16 = 30$$ $$7^{2-1} + 11 = 7^{1} + 11 = 7 + 11 = 18$$ Left side $30 \neq 18$. 6. For $n=3$: $$1 + 2^3 + 3^3 + 4^3 = 1 + 8 + 27 + 64 = 100$$ $$7^{3-1} + 11 = 7^{2} + 11 = 49 + 11 = 60$$ Left side $100 \neq 60$. 7. For $n=4$: $$1 + 2^4 + 3^4 + 4^4 = 1 + 16 + 81 + 256 = 354$$ $$7^{4-1} + 11 = 7^{3} + 11 = 343 + 11 = 354$$ Left side equals right side. 8. Thus, $n=4$ satisfies the equation. Final answer: $$n = 4.$$