“Here’s a quick one,” suggested a fellow tutor. “Prove that $2^{50} < 3^{33}$.”

Easy, I thought: but I knew better than to say it aloud.

First approach

“I know that $9 > 8$,” I said, checking on my fingers. “So if $2^3 < 3^2$, then $2^{150} < 3^{100}$ and $2^{50} < 3^{\frac{100}{3}}$.”

I was mid-boom when I realised I hadn’t proved it at all! The left-hand side could still be between $3^{33}$ and $3^{\frac{100}{3}}$. I was going to need to tighten things up!

Logs! They’re all made out of logs!

Comparing $2^{50}$ with $3^{33}$ is the same as comparing $50\ln(2)$ with $33\ln(3)$ - or comparing $\frac{50}{33}\ln(2)$ with $\ln(3)$.

Now, $\ln(2) < 0.69315$, so $\frac{\ln(2)}{3} < 0.23105$ and $\frac{\ln(2)}{33} < 0.0211$.

Multiplying that by 50 gives 1.05, which is smaller than $\ln(3)$. But that feels a bit unsatisfactory.

Let’s bring out the binomial

I don’t just know that $\frac{9}{8} > 1$, I know that $\frac{9}{8} = 1 + \frac{1}{8}$. I also know that $\br{ 1 + x}^8 > 1 + 8x$ if $x>0$, so $\br {\frac{9}{8} }^8 > 2$.

I can rewrite that as $\frac{3^{16}}{2^{24}} > 2$, so $3^{16} > 2^{25}$, which implies that $3^{32} > 2^{50}$ (and hence that $3^{33} > 2^{50}$.

I might have known you’d have something to say, sensei

$\ln\br{\frac{9}{8}} \gt \frac{2}{17}$.

So $34 \ln(3) - 51\ln(2) \gt 2 \gt \ln(2)+\ln(3)$.

So $33\ln(3) \gt 52 \ln(2)$, and $2^{52} \lt 3^{33}$.”

Another route

Amelia points out that $3^{7} > 2^{11}$, so $3^{28} > 2^{44}$. Further, $3^{4} > 2^{6}$, so we can say $3^{32} > 2^{50}$ as before.

Have you got a better way?