Dear Uncle Colin,

How do I compare two numbers with irrational exponents? I want to know whether $2^{\sqrt{3}}$ is larger or smaller than $3^{\sqrt{2}}$.

- How’d You Power A Thing Irrationally? Aaargh!

Hi, HYPATIA, and thanks for your message! There’s a sensible way to do this and an alternative that I rather like.

The sensible way

Let $x = 2^{\sqrt{3}}$ and $y = 3^{\sqrt{2}}$.

Then $\ln(x) = \sqrt{3} \ln(2)$ and $\ln(y) = \sqrt{2} \ln(3)$.

These are both positive, so squaring them will keep them in the same order.

$(\ln(x))^2 = 3 (\ln(2))^2$ and $(\ln(y))^2 = 2(\ln(3))^2$.

Now, $\ln(3)> 1$, so the second of these is greater than 2.

Also, $\ln(2) < \sqrt{\frac{1}{2}}$, so the second of these is smaller than $\frac{3}{2}$. Therefore $y$ is larger.

Another way

$\sqrt{2} > 1.4$ so $3^{\sqrt{2}} > 3^{7/5}$.

$\sqrt{3} < 1.75$, so $2^{\sqrt{3}} < 2^{7/4}$.

Now, $3^4 > 2^5$, so $3^{7/5} > 2^{7/4}$

Putting it all together, $3^{\sqrt{2}} > 3^{7/5} > 2^{7/4} > 2^{\sqrt{3}}$

Woosh

“Ahem. $2^{\sqrt{3}} < 4$. $3^{\sqrt{2}} > 3^{4/3} > 4$.”

“How do y…”

“$81 > 64$ so $3^4 > 4^3$ and $3^{4/3} > 4$. As you were.”

“Thank you, sensei.”

Hope that helps!

- Uncle Colin