Integrals, Exponential Functions, and Logarithms: Learn It 2

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Integrals, Exponential Functions, and Logarithms: Learn It 2

Defining the Number $e$

Now that we have the natural logarithm defined, we can use that function to define the number $e.$

the number $e$

The number $e$ is defined to be the real number such that

ln e = 1

$\text{ln}e=1$

To put it another way, the area under the curve $y=\frac{1}{t}$ between $t=1$ and $t=e$ is $1$ (Figure 3).

This figure is a graph. It is the curve y=1/t. It is decreasing and in the first quadrant. Under the curve is a shaded area. The area is bounded to the left at x=1 and to the right at x=e. The area is labeled “area=1”. — Figure 3. The area under the curve from 1 to $e$ is equal to one.

The proof that such a number exists and is unique is left to you. (Hint: Use the Intermediate Value Theorem to prove existence and the fact that $\text{ln}x$ is increasing to prove uniqueness.)

The number $e$ can be shown to be irrational, although we won’t do so here. Its approximate value is given by

e \approx 2.71828182846

$e\approx 2.71828182846$

The Exponential Function

We now turn our attention to the function ${e}^{x}.$ Note that the natural logarithm is one-to-one and therefore has an inverse function. For now, we denote this inverse function by $\text{exp }x.$ Then,

exp (ln x) = x for x > 0 and ln (exp x) = x for all x .

$\text{exp}(\text{ln }x)=x\text{ for }x>0\text{ and }\text{ln}(\text{exp }x)=x\text{ for all }x.$

The following figure shows the graphs of $\text{exp }x$ and $\text{ln }x.$

This figure is a graph. It has three curves. The first curve is labeled exp x. It is an increasing curve with the x-axis as a horizontal asymptote. It intersects the y-axis at y=1. The second curve is a diagonal line through the origin. The third curve is labeled lnx. It is an increasing curve with the y-axis as an vertical axis. It intersects the x-axis at x=1. — Figure 4. The graphs of $\text{ln}x$ and $\text{exp}x.$

We hypothesize that $\text{exp }x={e}^{x}.$

For rational values of $x,$ this is easy to show. If $x$ is rational, then we have $\text{ln }({e}^{x})=x\text{ln }e=x.$ Thus, when $x$ is rational, ${e}^{x}=\text{exp }x.$

For irrational values of $x,$ we simply define ${e}^{x}$ as the inverse function of $\text{ln }x.$

defining the exponential function

For any real number $x,$ define $y={e}^{x}$ to be the number for which

ln y = ln (e^{x}) = x

$\text{ln}y=\text{ln}({e}^{x})=x$

Then we have ${e}^{x}=\text{exp}(x)$ for all $x,$ and thus

e^{ln x} = x for x > 0 and ln (e^{x}) = x

${e}^{\text{ln}x}=x\text{ for }x>0\text{ and }\text{ln}({e}^{x})=x$

for all $x.$

Properties of the Exponential Function

Since the exponential function was defined in terms of an inverse function, and not in terms of a power of $e,$ we must verify that the usual laws of exponents hold for the function ${e}^{x}.$

properties of the exponential function

If $p$ and $q$ are any real numbers and $r$ is a rational number, then

${e}^{p}{e}^{q}={e}^{p+q}$
$\frac{{e}^{p}}{{e}^{q}}={e}^{p-q}$
${({e}^{p})}^{r}={e}^{pr}$

Proof

Note that if $p$ and $q$ are rational, the properties hold. However, if $p$ or $q$ are irrational, we must apply the inverse function definition of ${e}^{x}$ and verify the properties. Only the first property is verified here; the other two are left to you. We have

ln (e^{p} e^{q}) = ln (e^{p}) + ln (e^{q}) = p + q = ln (e^{p + q}) .

$\text{ln}({e}^{p}{e}^{q})=\text{ln}({e}^{p})+\text{ln}({e}^{q})=p+q=\text{ln}({e}^{p+q}).$

Since $\text{ln}x$ is one-to-one, then

e^{p} e^{q} = e^{p + q} .

${e}^{p}{e}^{q}={e}^{p+q}.$

$_\blacksquare$

As with part iv. of the logarithm properties, we can extend property iii. to irrational values of $r,$ and we do so by the end of the section.

We also want to verify the differentiation formula for the function $y={e}^{x}.$

To do this, we need to use implicit differentiation. Let $y={e}^{x}.$ Then,

\begin{matrix} ln y & = & x \frac{d}{d x} ln y & = & \frac{d}{d x} x \frac{1}{y} \frac{d y}{d x} & = & 1 \frac{d y}{d x} & = & y . \end{matrix}

$\begin{array}{ccc}\hfill \text{ln}y& =\hfill & x\hfill \\ \hfill \frac{d}{dx}\text{ln}y& =\hfill & \frac{d}{dx}x\hfill \\ \hfill \frac{1}{y}\frac{dy}{dx}& =\hfill & 1\hfill \\ \hfill \frac{dy}{dx}& =\hfill & y.\hfill \end{array}$

Thus, we see

\frac{d}{d x} e^{x} = e^{x}

$\frac{d}{dx}{e}^{x}={e}^{x}$

as desired, which leads immediately to the integration formula

\int e^{x} d x = e^{x} + C

$\displaystyle\int {e}^{x}dx={e}^{x}+C$

We apply these formulas in the following examples.

Evaluate the following derivatives:

$\frac{d}{dt}{e}^{3t}{e}^{{t}^{2}}$
$\frac{d}{dx}{e}^{3{x}^{2}}$

We apply the chain rule as necessary.

$\frac{d}{dt}{e}^{3t}{e}^{{t}^{2}}=\frac{d}{dt}{e}^{3t+{t}^{2}}={e}^{3t+{t}^{2}}(3+2t)$
$\frac{d}{dx}{e}^{3{x}^{2}}={e}^{3{x}^{2}}6x$

Evaluate the following integral: $\displaystyle\int \frac{4}{{e}^{3x}}dx.$

$\displaystyle\int \frac{4}{{e}^{3x}}dx=-\frac{4}{3}{e}^{-3x}+C$

Watch the following video to see the worked solution to this example.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “6.7 Try It Problems” here (opens in new window).

Integrals, Exponential Functions, and Logarithms: Learn It 2

Defining the Number ee

the number ee

The Exponential Function

defining the exponential function

Properties of the Exponential Function

properties of the exponential function

Defining the Number $e$

the number $e$