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I'd never encountered a problem like "show that $2^{50}<3^{33}$" but I think I ended up solving it after doing some weird stuff with logs and Maclaurin series:

$$ \begin{align*} 2^{50}&<3^{33}\\ \ln2^{50}&<\ln3^{33}\\ 50\ln2&<33\ln3\\ \frac{50}{33}&<\frac{\ln3}{\ln2}\\ \end{align*} $$

When I got to this point I thought I might be able to use the result $\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\cdots$ to rewrite the right-hand side into something like a fraction. $$ \begin{align*} \frac{50}{33}&<\frac{\ln3}{\ln2}\\ &<\frac{\ln(1+2)}{\ln(1+1)}\\ &<\frac{2-\frac{2^2}2+\frac{2^3}3-\cdots}{1-\frac12+\frac13-\cdots}\\ \end{align*} $$

Here I noticed that actually, the bottom is clearly greater than a half, because the sum of all the terms is greater than the sum of the first two, $1-\frac12+\frac13-\cdots>1-\frac12$.

So I actually went back a step:

$$ \begin{align*} \frac{50}{33}&<\frac{\ln3}{\frac12}\\ \frac{50}{33}&<2\ln3\\ \end{align*} $$

Here I think I've proved that $2^{50}<3^{33}$, because the above inequality ($\frac{50}{33}<2\ln3$) is clearly true:

$$ \ln3>1\implies 2\ln3>2.\\ \frac{50}{33}<2\implies \frac{50}{33}<2\ln3. $$

However, I'm not sure if I've done this the "right" way. I'm pretty sure there must be alternative, simpler, "more elegant" ways to prove $2^{50}<3^{33}$.

Can people show me how they would prove this result?

minseong
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3 Answers3

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$$3^{33}=\frac{3^{35}}{9}=\frac{(3^7)^5}{9}=\frac{2187^5}{9}>\frac{2048^5}{9}=\frac{2^{55}}{9}=2^{50}\cdot\frac{32}{9}>2^{50}.$$

Michael Rozenberg
  • 194,933
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One possible method is to manipulate several smaller inequalities to get the original one. I hope you are happy with how index laws work.

We have:

\begin{align} (2^{11} = 2048) & \lt (2187 = 3^7) \tag{1}\\ (2^{6} = 64) & \lt(243 = 3^5) \tag{2} \\ \end{align}

Multiply $4\cdot(1)$ and $(2)$,

\begin{align} 2^{44+6} & \lt 3^{28+5} \\ 2^{50} & \lt 3^{33}\\ \end{align} And we are finished.

Landuros
  • 2,781
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Here's a much easier way to approach the problem:

$$2^{50} \ ? \ 3^{33}$$ $$\Rightarrow 2^{150} \ ? \ 3^{100} > 3^{99}$$ $$\Rightarrow (2^3)^{50} \ ? \ (3^2)^{50}$$

Since $2^3 < 3^2$, then $2^{150} < 3^{100}$ which means that $2^{50} < 3^{33}$. While there might be a problem with line $2$, since $3^2$ is $\frac{9}{8}$ of $2^3$, $3^{100}$ is $(\frac{9}{8})^{50}$ of $2^{150}$. However, $(\frac{9}{8})^{50} > 3$ which means that changing from $3^{99}$ to $3^{100}$ won't affect the sign.

Toby Mak
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