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Section 3.4 The Number \(e\) and Natural Logarithm

One of the first conditions of happiness is that the link between Man and Nature shall not be broken. — Lev Nikolayevich Tolstoy, Russian writer, 1828 – 1910

Question: What is the most beautiful mathematical expression?

See Figure 3.10.

Figure 3.10. Euler's Identity

Number \(e\text{.}\)

  • \(\displaystyle e\approx 2.71828182845904523536028747135266249775724709369995\ldots\)

  • First time mentioned in 1618 by (almost certainly) William Oughtred, 1574 – 1660.

  • The notation \(e\) made its first appearance in a letter written by Leonhard Euler (1707 – 1783) in 1731.

  • What is \(e\text{?}\)

    \begin{equation*} \text{If }x\to 0\text{ then } (1+x)^{\frac{1}{x}}\to e. \end{equation*}

  • Where is \(e\text{:}\) See Figure 3.11.

    Figure 3.11. The function \(f(x)=e^x\) is the exponential function with the property that the slope of its tangent line at the point \((1,0)\) equals to 1.

  • Why \(e\text{?}\) The number \(e\) plays a role in your everyday life.

    See Figure 3.12.

    Figure 3.12. The compound interest formula and the continuous compound interest formula

Natural logarithm: The inverse function of the function \(f(x)=e^x\) is the function \(f^{-1}(x)=\log_ex\text{.}\)

We write

\begin{equation*} \log_e(x)=\ln (x), x>0. \end{equation*}

We say that, for \(x>0\text{,}\) \(\ln x\) is the \textbf{natural logarithm of } \(x\text{.}\)

See Figure 3.13.

Figure 3.13. The function \(f(x)=e^x\) its inverse function \(f^{-1}(x)=\ln (x)\text{,}\) and the line \(y=x\text{.}\)

Summary:

\begin{equation*} \begin{array}{l|l|l} \amp f(x)=e^x\amp f^{-1}(x)=\ln (x)\\ \hline \text{Domain}\amp\mathbb{R}\amp(0,\infty)\\ \hline \text{Range} \amp (0,\infty)\amp \mathbb{R}\\ \hline \text{output}=0 \amp \text{Never}\amp x=1\\ \hline \text{output}=1 \amp x=0\amp x=e\\ \hline \text{Increasing}\amp \mathbb{R}\amp (0,\infty)\\ \hline \text{Decreasing}\amp \text{Never}\amp \text{Never}\\ \hline \text{One-to-one}\amp \text{Yes}\amp \text{Yes}\\ \hline \text{Inverse}\amp \ln x\amp e^x\\ \hline \text{Odd or even}\amp \text{Neither}\amp \text{Neither}\\ \hline \text{Vertical Asymptote}\amp --\amp x=0\\ \hline \text{Horizontal Asymptote}\amp y=0\amp --\\ \end{array} \end{equation*}

All you need to know. If \(x\gt 0\) then

\begin{equation*} \ln (x) =y \Leftrightarrow x=e^y. \end{equation*}
Example 3.4.1. Logarithmic equations.

Solve the equation:

  1. \(\displaystyle \ln (x+1)=2,\)

  2. \(\ln (x+12)-\ln(x+2)=\ln (x)\text{,}\)

  3. \(\displaystyle \displaystyle \frac{\ln (x^2+12)}{\ln(x+2)}=2.\)

Solution

  1. Observe that the domain of the given equation is the set \(\{x\in \mathbb{R}:x+1\gt 0\}=(-1,\infty)\text{.}\)

    By definition, for \(x\in (-1,\infty)\text{,}\)

    \begin{equation*} \ln (x+1)=2\Leftrightarrow x+1=e^2\Leftrightarrow x=e^2-1. \end{equation*}

  2. Observe that the domain of the given equation is given by

    \begin{equation*} D=\{x:x+12\gt 0, \ x+2\gt 0,\text{ and }x\gt 0\} \end{equation*}
    \begin{equation*} =\{x:x\gt -12, \ x\gt -2,\text{ and }x\gt 0\}=\{x:x\gt 0\}=(0,\infty). \end{equation*}

    If follows that, for all \(x\gt 0\text{,}\)

    \begin{equation*} \ln (x+12)-\ln(x+2)=\ln (x)\Leftrightarrow \ln\left(\frac{x+12}{x+2}\right)=\ln (x). \end{equation*}

    Since the function \(f(x)=\ln (x)\) is one–to–one, we have that, for all \(x\gt 0\text{,}\)

    \begin{equation*} \ln\left(\frac{x+12}{x+2}\right)=\ln (x)\Leftrightarrow \frac{x+12}{x+2}=x\Leftrightarrow x+12=x^2+2x\Leftrightarrow x^2+x-12=0. \end{equation*}

    The quadratic equation \(x^2+x-12=0\) has two solutions, \(x=-4\) and \(x=3\text{.}\)

    We eliminate \(x=-4\) since it is not in the domain of the given equation.

    Hence \(x=3\) is the only solution of the given equation.

  3. Observe that the domain of the given equation is given by

    \begin{equation*} D=\{x:x^2+12>0, \ x+2>0,\text{ and }\ln (x+2)\not= 0\}=\{x:x>-2,\text{ and }x+2\not=1\} \end{equation*}
    \begin{equation*} =\{x:x>-2,\text{ and }x\not=-1\}=(-2,-1)\cup (-1,\infty). \end{equation*}

    If follows that, for all \(x\in D\text{,}\)

    \begin{equation*} \frac{\ln (x^2+12)}{\ln(x+2)}=2\Leftrightarrow \ln (x^2+12)=2\ln (x+2) \end{equation*}
    \begin{equation*} \Rightarrow \ln (x^2+12)=\ln (x+2)^2\Leftrightarrow \ln (x^2+12)=\ln (x^2+4x+4). \end{equation*}

    Since the function \(f(x)=\ln (x)\) is one–to–one, we have that, for all \(x\in D\text{,}\)

    \begin{equation*} \frac{\ln (x^2+12)}{\ln(x+2)}=2\Rightarrow x^2+12=x^2+4x+4\Leftrightarrow 4x=8\Leftrightarrow x=2. \end{equation*}

    Clearly, \(2\in D\) and we conclude that \(x=2\) is the solution of the given equation.

Reminder: From the fact that the functions \(f(x)=e^x\) and \(g(x)=\ln (x)\) are inverse to each other it follows that:

\begin{equation*} \ln \left(e^x\right)=x, x\in \mathbb{R} \end{equation*}

and

\begin{equation*} e^{\ln (x)}=x, x>0. \end{equation*}
Example 3.4.2. The meaning of \(x^x\).

  1. What is the meaning of the expression \(\displaystyle x^x\text{,}\) \(x\gt 0\text{?}\)

  2. Evaluate: \(\displaystyle \pi^\pi\)

Solution

  1. By definition, for \(x\gt 0\text{,}\)

    \begin{equation*} x^x=e^{x\ln (x)}. \end{equation*}

    We justify this definition by observing the following.

    Say that \(x\gt 0\) and recall that, for \(x\gt 0\text{,}\) \(\displaystyle x=e^{\ln (x)}\text{.}\) Then

    \begin{equation*} e^{x\ln (x)}=\left( e^{\ln (x)}\right)^x=x^x. \end{equation*}

  2. By definition,

    \begin{equation*} \pi ^\pi=e^{\pi\ln\pi}\approx 36.46215965. \end{equation*}