1. **State the problem:** We need to find the value of $x$ such that $$9^{24} \div 9^{3} \times 9^{4} = 9^{x}.$$\n\n2. **Recall the exponent rules:** When dividing powers with the same base, subtract the exponents: $$a^{m} \div a^{n} = a^{m-n}.$$ When multiplying powers with the same base, add the exponents: $$a^{m} \times a^{n} = a^{m+n}.$$\n\n3. **Apply the division rule first:** $$9^{24} \div 9^{3} = 9^{24-3} = 9^{21}.$$\n\n4. **Now multiply by $9^{4}$:** $$9^{21} \times 9^{4} = 9^{21+4} = 9^{25}.$$\n\n5. **Therefore, the equation becomes:** $$9^{25} = 9^{x}.$$\n\n6. **Since the bases are equal and nonzero, the exponents must be equal:** $$x = 25.$$\n\n**Final answer:** $x = 25$