Which number amongst $2^40$, $3^21$, $4^18$ and $8^12$ is the smallest?

Correct Answer: (B) 3^21

Expressing the powers of 4 and 8 in base 2: $4^{18} = (2^2)^{18} = 2^{36}$ and $8^{12} = (2^3)^{12} = 2^{36}$. Since $2^{36} < 2^{40}$, we compare $3^{21}$ and $2^{36}$. Writing both with exponent 3 gives $3^{21} = (3^7)^3 = 2187^3$, whereas $2^{36} = (2^{12})^3 = 4096^3$. Because $2187 < 4096$, it follows that $3^{21} < 2^{36}$. Therefore, $3^{21}$ is strictly the smallest number.