Whether $2^{38}$ or $3^{33}$ is greater without needing a calculator My question is about figuring out whether $2^{38}$ or  $3^{33}$ is greater without needing a calculator, by using the Mobius function or by other means?
 A: The difficulty with problems like this is that the smaller base has the larger exponent, which makes it hard to immediately see which is really smaller.  Hence the trick is to find an easier to understand inequality, then raise each side to a common power, making it more clear what is really going on.  In this case, the helpful hint suggests that $8 = 2^3 < 3^2 = 9$.  From this, we get
$$ 2^3 < 3^2 \implies (2^3)^{13} < (3^2)^{13} \implies 2^{39} < 3^{26}. $$
It hopefully clear that $2^{38} < 2^{39}$ and that $3^{26} < 3^{33}$, from which the desired result is obtained.
A: Yet another answer: we have $2^{38}=2^{33}2^5$, and $3^{33}=(2\cdot {3\over 2})^{33}=2^{33}({3\over 2})^{33}$, so we just need to compare $2^5$ and $({3\over 2})^{33}$. Now $({3\over 2})^2={9\over 4}>2$, so $$({3\over 2})^{33}>(({3\over 2})^2)^{16}>2^{16}>2^5.$$
A: Using lulu's hint , we have $$2^{38}<2^{39}=(2^3)^{13}<(3^2)^{13}=3^{26}<3^{33}$$
A: Binomial theorem:
$$3^{33} = (2 + 1)^{33} = 2^{33} + 31 \times 2^{32} + \binom{33}{31}2^{31} + \dots > \binom{33}{31}2^{31} = \frac{33\times 32}{2}\times2^{31}$$
$$> 32 \times 16\times 2^{31} = 2^{40} > 2^{38}.$$
A: Since$$3^{33}>2^{38}\iff\log_2(3^{33})>\log_2(2^{38})\iff\log_2(3)>\frac{38}{33}$$you will just have to prove that $\log_2(3)>\frac{38}{33}$. But $\sqrt2<\frac32$. Therefore $2^{\frac32}<3$ and this is equivalent to $\frac32<\log_23$ and so$$\log_23>\frac32>\frac{38}{33}.$$
A: Since $2^{11} \approx 3^{7} \approx 2000$, you can estimate it:
$$\begin{array} {rcl}
2^{38} & \text{vs} & 3^{33} 
\\
\left( 2^{11} \right)^{38/11} & \text{vs} & \left( 3^{7} \right)^{33/7}
\\
\frac{38}{11} & \text{vs} & \frac{33}{7}
\\
38 \times 7 & \text{vs} & 33 \times 11
\\
266 & < & 343
\end{array}$$
$38/11$ being close to a small integer is why the approximation is close enough.
