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Section 1.12 Exponential functions
Definition 1.12.1 . Exponential function.
The exponential function , denoted \(\exp\text{,}\) is defined as the inverse of the natural logarithm function. Using inverse function properties, this means that we have
\begin{equation}
\exp(x)=y \iff x=\ln y \hspace{5pt} \text{ for all } x\in \R, y\in (0,\infty)\text{.}\tag{1.12.1}
\end{equation}
We also write \(e^x\) for \(\exp(x)\text{.}\)
Theorem 1.12.2 . Exponential function properties.
The exponential function is differentiable (hence also continuous) on all of \(\mathbb{R}\) and satisfies
\begin{equation*}
\frac{d}{dx}e^x=e^x\text{.}
\end{equation*}
The exponential function is increasing on \(\R\) and hence one-to-one. The graph of \(\exp\) is concave up everywhere.
We have
\begin{align}
\lim_{x\to\infty}e^x\amp =\infty\tag{1.12.2}\\
\lim_{x\to -\infty}e^x\amp =0\tag{1.12.3}
\end{align}
The domain of \(\exp\) is \(\mathbb{R}=(-\infty, \infty)\text{;}\) the range of \(\exp\) is \((0,\infty)\text{.}\)
\(e^0=1\text{.}\)
We have
\begin{align}
e^{x+y}\amp =e^xe^y \tag{1.12.4}\\
e^{x-y}\amp =e^x/e^y \tag{1.12.5}\\
e^{xy}\amp =(e^x)^y \tag{1.12.6}
\end{align}
for all \(x,y\in\mathbb{R}\text{.}\)
We have
\begin{align}
\ln(e^x)\amp =x, \text{ for all } x\in \R \tag{1.12.7}\\
e^{\ln x}\amp =x, \text{ for all } x\in (0,\infty)\text{.}\tag{1.12.8}
\end{align}
Definition 1.12.3 . Base-\(a\) exponential function.
Let \(a\) be a fixed positive number. Given any \(x\in \R\text{,}\) we define the power expression \(a^x\) as
\begin{equation}
a^x=e^{x\ln a}\text{.}\tag{1.12.9}
\end{equation}
The exponential function with base \(a\text{,}\) is the function \(f\) with domain \(\R\) defined as
\begin{equation}
f(x)=a^x=e^{x\ln a}\tag{1.12.10}
\end{equation}
for all \(x\in \R\text{.}\)
Definition 1.12.4 . Base-\(a\) logarithm.
Let \(a\) be a fixed positive number, \(a\ne 1\text{.}\) The logarithmic function with base \(a\text{,}\) denoted \(\log_a\) is defined as the inverse function of the base-\(a\) exponential function \(f(x)=a^x\text{.}\)
Theorem 1.12.5 . Logarithmic and exponential compendium.
The table below summarizes important properties of the functions
\(f(x)=\log_a x\) and
\(g(x)=a^x\) for a base
\(a\) satisfying
\(a> 1\text{.}\) Table 1.12.6. Base-\(a\) functions, \(a> 0\)
Domain
\((0,\infty)\) \(\R\)
Range
\(\R\) \((0,\infty)\)
Monotonicity
Increasing
Increasing
Limit as \(x\to\infty\)
\(\infty\) \(\infty\)
Limit as \(x\to 0^+\)
\(-\infty\) 1
Limit as \(x\to -\infty\)
NA
0
Properties
\(\log_a(xy)=\log_a x+\log_a y\) \(\log_a(x^y)=y\log_a\) \(\log_a(a^x)=x\)
\(a^{x+y}=a^xa^y\) \(a^{xy}=(a^x)^y\) \(a^{\log_a x}=x\)
Relation to \(e\)
\(\log_a x=\frac{\ln x}{\ln a}\) \(a^x=e^{x\ln a}\)
Theorem 1.12.7 . Logarithmic change of base.
Let \(a\) and \(b\) be positive numbers. We have
\begin{equation}
\log_a x=\frac{\log_b x}{\log_b a}\tag{1.12.11}
\end{equation}
for all \(x\in (0,\infty)\text{.}\)
Theorem 1.12.8 . Derivative/antiderivative compendium.
We collect here the new derivative formulas obtained via logarithms and exponential functions, along with their equivalent antiderivative formulas.
\begin{align*}
\frac{d}{dx}\ln\vert x\vert=\frac{1}{x} \amp \iff \int\frac{1}{x} \, dx=\ln\vert x\vert+C\\
\frac{d}{dx}\ln\vert \cos x \vert=-\tan x \amp \iff \int\tan x \, dx=-\ln\vert \cos x \vert+C=\ln\vert\sec x\vert+C\\
\frac{d}{dx}\ln\vert \sin x\vert=\cot x \amp \iff \int\cot x \, dx=\ln\vert \sin x\vert+C\\
\frac{d}{dx}\ln\vert \sec x+\tan x\vert=\sec x \amp \iff \int\sec x \, dx=\ln\vert \sec x+\tan x\vert+C\\
\frac{d}{dx}\ln\vert \csc x+\cot x\vert=-\csc x \amp \iff \int\csc x \, dx=-\ln\vert \csc x+\cot x\vert+C\\
\frac{d}{dx}e^x=e^x \amp \iff \int e^x \, dx=e^x+C\\
\frac{d}{dx}a^x=(\ln a)a^x \amp \iff \int a^x \, dx=\frac{1}{\ln a}a^x+C\\
\frac{d}{dx}\log_a \vert x\vert =\frac{1}{(\ln a)\, x} \amp \iff \int \frac{1}{(\ln a)\, x} \, dx=\log_a\vert x\vert+C
\end{align*}
Example 1.12.9 . Solving exponential equations.
Find all \(t\) satisfying \(\displaystyle 2^{-t^2}=\frac{1}{16}\text{.}\) Simplify your answer as much as possible.
Solution .
We have
\begin{align*}
2^{-t^2}=\frac{1}{16}\amp \iff \ln\left(2^{-t^2} \right)= \ln(1/16) \\
\amp \iff -t^2\ln 2=-\ln 16\\
\amp\iff t^2=\frac{\ln 16}{\ln 2} \\
\amp \iff t^2=\log_2 16 \amp (\log_2 16=\ln 16/\ln 2)\\
\amp \iff t^2=4 \amp (2^4=16\iff \log_2 16=4)\\
\amp \iff t=\pm 2\text{.}
\end{align*}
Alternatively, using the fact that the function \(f(x)=2^x\) is one-to-one and \(1/16=2^{-4}\text{,}\) we see that
\begin{align*}
2^{-t^2}=\frac{1}{16}\amp \iff 2^{-t^2}=2^{-1/4} \\
\amp \iff -t^2=-4\\
\amp \iff t=\pm 2\text{.}
\end{align*}
Example 1.12.10 . Exponential and logarithmic derivatives.
Compute \(f'(x)\) for each of the following functions.
\(\displaystyle f(x)=\ln(\sin x)e^{\cos x}\)
\(\displaystyle f(x)=\log_3(2^x+3^{x^2})\)
Solution .
We have
\begin{align*}
f'(x) \amp = (\ln \sin x)'e^{\cos x}+\ln\sin x(e^{\cos x})'\\
\amp = \frac{\cos x}{\sin x}e^{\cos x}+\ln\sin x \cdot e^{\cos x}(-\sin x)\\
\amp = e^{\cos x}(\cot x-\sin x\ln \sin x)\text{.}
\end{align*}
We have
\begin{align*}
f'(x) \amp = \frac{(2^x+3^{x^2})'}{\ln 3(2^x+3^{x^2})}\\
\amp = \frac{\ln 2\cdot 2^x+2\ln 3 \cdot x \cdot 3^{x^2}}{\ln 3(2^x+3^{x^2})}\text{.}
\end{align*}
Example 1.12.11 . Exponential and logarithmic integrals.
Compute the following definite and indefinite integrals.
\(\displaystyle \displaystyle\int (e^t)^2\sin(e^{2t})\, dt\)
\(\displaystyle \displaystyle\int_0^\pi \sin(2^x)2^{\cos(2^x)+x}\, dx\)
Solution .
We have
\begin{align*}
\int (e^t)^2\sin(e^{2t})\, dt \amp \int u \sin (u^2)\, du \amp (u=e^t, du=e^t\, dt\\
\amp = -\frac{1}{2}\cos u^2 +C\\
\amp = -\frac{1}{2}\cos (e^{2t})+C\text{.}
\end{align*}
Consider the somewhat tricky substitution: \(u=2^{\cos 2^x}\text{,}\) \(du=-(\ln 2)^2\sin 2^x\cdot 2^x\cdot 2^{\cos 2^x} dx\text{.}\) We then have
\begin{align*}
\int_{0}^\pi \sin(2^x)2^{\cos(2^x)+x}\, dx \amp = -\frac{1}{(\ln 2)^2}\int_{2^{\cos 1}}^{2^{\cos 2^\pi}} du \\
\amp = -\frac{1}{(\ln 2)^2}u\Bigr\vert_{2^{\cos 1}}^{2^{\cos 2^\pi}}\\
\amp = \frac{1}{(\ln 2)^2}(2^{\cos 1}-2^{\cos 2^\pi})\text{.}
\end{align*}
