1. **Problem:** Find the value of $x$ that satisfies the equation $$4^{2x+1} = 32^{x-3}$$. 2. **Formula and rules:** - Express both sides with the same base if possible. - Recall that $4 = 2^2$ and $32 = 2^5$. - Use the property $a^{m} = a^{n} \Rightarrow m = n$ if bases are equal. 3. **Rewrite the equation:** $$4^{2x+1} = 32^{x-3} \Rightarrow (2^2)^{2x+1} = (2^5)^{x-3}$$ 4. **Simplify exponents:** $$2^{2(2x+1)} = 2^{5(x-3)}$$ $$2^{4x+2} = 2^{5x - 15}$$ 5. **Set exponents equal:** $$4x + 2 = 5x - 15$$ 6. **Solve for $x$:** $$4x + 2 = 5x - 15$$ $$2 + 15 = 5x - 4x$$ $$17 = x$$ **Final answer:** $x = 17$.