Comparing exponential values to find the smallest number

First, normalize all numbers with base 2 into comparable powers of 2: 4^18 = (2²)^18 = 2^36. 8^12 = (2³)^12 = 2^36. 2^40 > 2^36, so 2^40 is eliminated.

Now, compare the remaining competitors: 2^36 and 3^21. To compare bases with different exponents, align their exponents to a common multiple (like power of 3): 2^36 = (2^12)³ = 4096³. 3^21 = (3^7)³ = 2187³.

Since 2187 < 4096, it is mathematically certain that ⟨MATH⟩3^21 < 2^36⟨/MATH⟩. Therefore, 3^21 is the smallest number in the set.

Answer: (b).