How to find out which number is larger without a calculator?

So I have a question which is:

Which is larger?

$$2.2^{3.3} \text{ or } 3.3^{2.2} $$

Now I need to find out with using a calculator but the answer is $3.3^{2.2}$.

The only thing I could think of is rounding.

So you know:

$2^3=8$ and $3^2=9$

I’m interested in seeing if there are other ways just because there might be the possiblity of being asked:

Which is larger?

$$2.5^{3.5} \text{ or } 3.5^{2.5} $$

So if I use the rounding idea, would I just round normally. I would get:

$$ 3^4 \text{ or } 4^3 $$

which shows $3^4$ is larger.

This is the only way I could think of, are there any other ways without using a calculator to determine which is larger?

Solutions Collecting From Web of "How to find out which number is larger without a calculator?"

$$(2.2^3)^{1.1}\gtrless (3.3^2)^{1.1}$$
$$2.2^3\gtrless 3.3^2$$
$$(2\cdot 1.1)^3\gtrless (3\cdot 1.1)^2$$
$$2^3\cdot 1.1\gtrless 3^2$$
$$8\cdot 1.1\gtrless 9$$


Can you find now which number is larger: $2.25^{3.375}$ or $3.375^{2.25}$ 😉 ?

One nice approach is the calculus approach. In particular:
a^b > b^a \iff a^{1/a} > b^{1/b}
In order to compare $a^{1/a}$ to $b^{1/b}$, we could consider the function $f(x) = x^{1/x}$ (or its logarithm, $\frac{\ln(x)}{x}$). On which interval is the function increasing? On which interval is it decreasing?

Note that this will only work, however, if both values are either greater than $e \approx 2.718$, or if both are less than $e$ (so, if $a = 3$ and $b=4$). Otherwise, it’s not immediately clear.

Another method is to use the fact that $\log a^b=b\log a$, so that you can reduce the exponents even if they don’t have any obvious common divisors:

$$3.3 \log 2.2 \gtrless 2.2 \log 3.3$$
$$1.5 \log 2.2 \gtrless \log 3.3$$
$$2.2^{1.5} \gtrless 3.3$$
$$2.2 \cdot 2.2^{0.5} \gtrless 3.3$$
$$2.2^{0.5} \gtrless 1.5$$
$$(2.2^{0.5})^2 \gtrless (1.5)^2$$
$$2.2 < 2.25$$
$$\Rightarrow 2.2^{3.3}\lt3.3^{2.2}$$

Let’s calculate the quotient:
\frac {3.3^{2.2}}{2.2^{3.3}} = \frac {3.3^{2.2}}{2.2^{2.2} \cdot 2.2^{\fbox{1.1}}} =
\frac {(3/2)^{2.2}}{2.2^{1.1}} > \frac {1.5^2}{2.2^{1.1}} = \dots ?
EDIT: I edited out the shameful mistake, but now this calculation isn’t useful. Oleg fixed it nicely, see below.