Exponential of Product

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Theorem

Let $x, y \in \R$ be real numbers.

Let $\exp x$ be the exponential of $x$.


Then:

$\map \exp {x y} = \paren {\exp y}^x$


Proof 1

Let $Y = \exp y$.

From Exponential of Natural Logarithm:

$\map \ln {\exp y} = y$

From Logarithms of Powers, we have:

$\ln Y^x = x \ln Y = x \, \map \ln {\exp y} = x y$


Thus:

$\map \exp {x y} = \map \exp {\ln Y^x} = Y^x = \paren {\exp y}^x$

$\blacksquare$


Proof 2

\(\ds \paren {\exp y}^x\) \(=\) \(\ds \map \exp {x \map \ln {\exp y} }\) Definition of Power to Real Number
\(\ds \) \(=\) \(\ds \map \exp {x y}\) Exponential of Natural Logarithm

$\blacksquare$


Proof 3

For $n \in \Z_{\ge 0}$:

\(\ds \map \exp {n y}\) \(=\) \(\ds \map \exp {\sum_{k \mathop = 1}^n y}\)
\(\ds \) \(=\) \(\ds \prod_{k \mathop = 1}^n \exp y\) Exponential of Sum of Real Numbers
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \paren {\exp y}^n\)


That is:

$\forall n \in \Z_{\ge 0}: \map \exp {n y} = \paren {\exp y}^n$


Now let $n \in \Z_{<0}$.

It follows that $-n \in \Z_{>0}$, so:

\(\ds \exp n\) \(=\) \(\ds \map \exp {-\paren {-n y} }\)
\(\ds \) \(=\) \(\ds \frac 1 {\map \exp {-n y} }\) Reciprocal of Real Exponential
\(\ds \) \(=\) \(\ds \frac 1 {\paren {\exp y}^{-n} }\) from $(1)$
\(\ds \) \(=\) \(\ds \paren {\exp y}^n\) Real Number to Negative Power: Positive Integer


Thus:

$(2): \quad \forall m \in \Z: \map \exp {m y} = \paren {\exp y}^m$


Next, for $n \in \Z_{>0}$:

\(\ds \exp y\) \(=\) \(\ds \map \exp {n \frac y n}\)
\(\ds \) \(=\) \(\ds \paren {\map \exp {\frac y n} }^n\) from $(1)$
\(\text {(3)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \sqrt [n] {\exp y}\) \(=\) \(\ds \map \exp {\frac y n}\)

So fix $r \in \Q$.


Let $r = \dfrac m n$, where $m \in \Z$ is an integer and $n \in \Z_{>0}$ is a strictly positive integer.

From the above:

\(\ds \map \exp {r y}\) \(=\) \(\ds \map \exp {\frac m n y}\)
\(\ds \) \(=\) \(\ds \sqrt [n] {\map \exp {m y} }\) from $(3$
\(\ds \) \(=\) \(\ds \sqrt [n] {\paren {\exp y}^m}\) from $(2)$
\(\ds \) \(=\) \(\ds \paren {\exp y}^r\)


Thus, from the definition of $\paren {\exp y}^x$ as the unique continuous extension of $r \mapsto \paren {\exp y}^r$ from $\Q$ to $\R$:

$\map \exp {x y} = \paren {\exp y}^x$

$\blacksquare$