Which number amongst $2^{40}$, $3^{21}$, $4^{18}$ and $8^{12}$ is the smallest?
Which number amongst $2^{40}$, $3^{21}$, $4^{18}$ and $8^{12}$ is the smallest?
$2^{40}$
$3^{21}$
$4^{18}$
$8^{12}$
Solution
Rewrite in base 2 where possible: $4^{18} = 2^{36}$ and $8^{12} = 2^{36}$. So $2^{40}$, $2^{36}$, $2^{36}$ and $3^{21}$ are to be compared. Comparing $2^{36}$ with $3^{21}$: $2^{36} = (2^{12})^3 = 4096^3$ and $3^{21} = (3^{7})^3 = 2187^3$. Since $4096 > 2187$, $3^{21}$ is the smallest.