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

Question

What is $3^{\sqrt2}$ ?

Clearly we can say that $a^m = a \times a \times ...\times a \times a$ (m times)

That is $$3^2=3 \times 3 $$

But how can we define $3^{\sqrt2}$ ? How to understand the definition of irrational Powers ? I do not need the value. I need a definition to understand this.

Answer 1

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

Answer 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 $3^\sqrt{2} = \exp(\sqrt{2}\log 3)$.

Answer 3

$3^{\sqrt2}$ is $e^{\sqrt2\cdot\ln3}$, where "$e$" is $\lim\limits_{n\to\infty}\left(1+\frac1n\right)^n$ and "$\ln$" is the natural logarithm function.

Answer 4

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