# 8.5: The Exponential Function

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## Definition

We call the inverse of the logarithm function the exponential function. We denote the value of the exponential function at a real number $$x$$ by $$\exp (x)$$.

## Proposition $$\PageIndex{1}$$

The exponential function has domain $$\mathrm{R}$$ and range $$(0,+\infty)$$. Moreover, the exponential function is increasing and differentiable on $$\mathbb{R}$$. If $$f(x)=\exp (x),$$ then $$f^{\prime}(x)=\exp (x)$$.

Proof

Only the final statement of the proposition requires proof. If we let $$g(x)=\log (x),$$ then

$f^{\prime}(x)=\frac{1}{g^{\prime}(\exp (x))}=\exp (x).$

Q.E.D.

## Proposition $$\PageIndex{2}$$

For any real numbers $$x$$ and $$y$$,

$\exp (x+y)=\exp (x) \exp (y).$

Proof

The result follows from

$\log (\exp (x) \exp (y))=\log (\exp (x))+\log (\exp (y))=x+y.$

Q.E.D.

## Proposition $$\PageIndex{3}$$

For any real number $$x$$,

$\exp (-x)=\frac{1}{\exp (x)}.$

Proof

The result follows from

$\log \left(\frac{1}{\exp (x)}\right)=-\log (\exp (x))=-x.$

Q.E.D.

## Exercise $$\PageIndex{1}$$

Use Thylor's theorem to show that

$\exp (1)=e=\sum_{n=0}^{\infty} \frac{1}{n !}.$

## Proposition $$\PageIndex{4}$$

For any rational number $$\alpha$$,

$\exp (\alpha)=e^{\alpha}.$

Proof

Since $$\log (e)=1,$$ we have

$\log \left(e^{\alpha}\right)=\alpha \log (e)=\alpha.$

Q.E.D.

## Definition

If $$\alpha$$ is an irrational number, we define

$e^{\alpha}=\exp (\alpha).$

Note that for any real numbers $$x$$ and $$y$$,

$e^{x+y}=e^{x} e^{y}$

and

$e^{-x}=\frac{1}{e^{x}}.$

Moreover, $$\log \left(e^{x}\right)=x$$ and, if $$x>0, e^{\log (x)}=x$$.

## Definition

If $$x$$ and $$a$$ are real numbers with $$a>0,$$ we define

$a^{x}=e^{x \log (a)}.$

## Exercise $$\PageIndex{2}$$

Define $$f:(0,+\infty) \rightarrow \mathbb{R}$$ by $$f(x)=x^{a},$$ where $$a \in \mathbb{R}, a \neq 0$$. Show that $$f^{\prime}(x)=a x^{a-1}$$.

## Exercise $$\PageIndex{3}$$

Suppose $$a$$ is a positive real number and $$f: \mathbb{R} \rightarrow \mathbb{R}$$ is defined by $$f(x)=a^{x} .$$ Show that $$f^{\prime}(x)=a^{x} \log (a)$$.

## Proposition $$\PageIndex{5}$$

For any real number $$\alpha>0$$,

$\lim _{x \rightarrow+\infty} x^{\alpha} e^{-x}=0.$

Proof

We know that

$\lim _{y \rightarrow+\infty} \frac{\log (y)}{y^{\frac{1}{2}}}=0.$

Hence

$\lim _{y \rightarrow+\infty} \frac{(\log (y))^{\alpha}}{y}=0.$

Letting $$y=e^{x},$$ we have

$\lim _{x \rightarrow+\infty} \frac{x^{\alpha}}{e^{x}}=0.$

Q.E.D.

## Proposition $$\PageIndex{6}$$

For any real number $$\alpha$$,

$\lim _{x \rightarrow+\infty}\left(1+\frac{\alpha}{x}\right)^{x}=e^{\alpha}.$

Proof

First note that, letting $$x=\frac{1}{h}$$,

$\lim _{x \rightarrow+\infty}\left(1+\frac{\alpha}{x}\right)^{x}=\lim _{h \rightarrow 0+}(1+\alpha h)^{\frac{1}{k}}=\lim _{h \rightarrow 0^{+}} e^{\frac{1}{h} \log (1+\alpha h)}.$

Using I'Hópital's rule, we have

$\lim _{h \rightarrow 0^{+}} \frac{\log (1+\alpha h)}{h}=\lim _{h \rightarrow 0^{+}} \frac{\alpha}{1+\alpha h}=\alpha,$

and the result follows from the continuity of the exponential function. $$\quad$$ Q.E.D.

## Definition

We define the hyperbolic sine and hyperbolic cosine functions by

$\sinh (x)=\frac{e^{x}-e^{-x}}{2}$

and

$\cosh (x)=\frac{e^{x}+e^{-x}}{2},$

respectively.

## Exercise $$\PageIndex{4}$$

Show that for any real numbers $$x$$ and $$y$$,

$\sinh (x+y)=\sinh (x) \cosh (y)+\sinh (y) \cosh (x)$

and

$\cosh (x+y)=\cosh (x) \cosh (y)+\sinh (x) \sinh (y).$

## Exercise $$\PageIndex{5}$$

Show that for any real number $$x$$,

$\cosh ^{2}(x)-\sinh ^{2}(x)=1.$

## Exercise $$\PageIndex{6}$$

If $$f(x)=\sinh (x)$$ and $$g(x)=\cosh (x),$$ show that

$f^{\prime}(x)=\cosh (x)$

and

$g^{\prime}(x)=\sinh (x).$

This page titled 8.5: The Exponential Function is shared under a CC BY-NC-SA 1.0 license and was authored, remixed, and/or curated by Dan Sloughter via source content that was edited to the style and standards of the LibreTexts platform.