What is $3^{\sqrt2}$? Definition of irrational Powers

I assume you're comfortable with exponentiation with rational exponents.

Let $(x_n)$ be a sequence of rational numbers such that $\lim x_n = \sqrt 2$. Then $3^{\sqrt2} = \lim 3^{x_n}$.

One such sequence is $$ x_{n+1} = \frac12\left(x_n + \frac2{x_n}\right), \quad x_0=2. $$

You can always define $x^y$ for $x>0$ by taking the formal logarithm and defining it in order to maintain the functional property of the logarithm:

$$\log x^y = y\log x$$

so you define:

$$x^y := \exp(y\log x)$$

In your case, it gives $3^\sqrt{2} = \exp(\sqrt{2}\log 3)$.

$3^{\sqrt{2}}= sup\{3^q \,| \, q \in \mathbb{Q} \wedge q^2<2 \}$