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

$$2.2^{3.3}\gtrless3.3^{2.2}$$ $$(2.2^3)^{1.1}\gtrless (3.3^2)^{1.1}$$ $$2.2^3\gtrless 3.3^2$$ $$(2\cdot 1.1)^3\gtrless (3\cdot 1.1)^2$$ $$2^3\cdot 1.1\gtrless 3^2$$ $$8\cdot 1.1\gtrless 9$$ $$8.8<9.$$

So, $$2.2^{3.3}<3.3^{2.2}.$$

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:

$$2.2^{3.3}\gtrless3.3^{2.2}$$ $$3.3 \log 2.2 \gtrless 2.2 \log 3.3$$ $$1.5 \log 2.2 \gtrless \log 3.3$$ $$2.2^{1.5} \gtrless 3.3$$ $$2.2 \cdot 2.2^{0.5} \gtrless 3.3$$ $$2.2^{0.5} \gtrless 1.5$$ $$(2.2^{0.5})^2 \gtrless (1.5)^2$$ $$2.2 < 2.25$$ $$\Rightarrow 2.2^{3.3}\lt3.3^{2.2}$$