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Precalculus JumpStart

Anthony van Groningen

Section 7.4 The Natural Logarithm

The base \(e\) situation is a convenient choice of base in calculus and deserves its own notation.
The base \(e\) exponential function \(f(x) = e^x\) is one-to-one has an inverse function \(f^{-1}\) called the natural logarithm function denoted by \(f^{-1}(y) = \ln(y) = \log_e (y)\text{.}\)
As for any other logarithm, the inverse properties are
  1. \(\displaystyle \ln(e^x) = x\) for all \(x\)
  2. \(\displaystyle e^{\ln(y)} = y\) for all \(y \gt 0\)
Alternatively, exponential equations are equivalent logarithmic equations according to
\begin{equation*} y = e^x \quad \text{if and only if} \quad \ln(y) =x. \end{equation*}
Observe that the natural logarithm returns the exponent in the base \(e\) exponential equation. Below is the graph of the natural logarithm (solid curve).
Figure 7.22.

Example 7.23. Using Technology.

According to my calculator,
\begin{equation*} \ln(10) \approx 2.303. \end{equation*}
Interpreting this in exponential form, we find
\begin{equation*} e^{2.303} \approx 10. \end{equation*}

Example 7.24. Evaluating Logarithms.

Evaluate the following logarithms without the use of calculator.
  1. \(\displaystyle \displaystyle \ln \left(\sqrt[3]{e}\right)\)
  2. \(\displaystyle \displaystyle \ln \left(\frac{1}{e^3}\right)\)
  3. \(\displaystyle \displaystyle \ln \left(e\right)\)
  4. \(\displaystyle \displaystyle \ln \left(1\right)\)
  5. \(\displaystyle \displaystyle \ln \left(0\right)\)
Solution.
  1. \(\displaystyle e^\boxed{1/3} = \sqrt[3]{e} \quad\Rightarrow\quad \ln \left(\sqrt[3]{e}\right) = \boxed{1/3}\)
  2. \(\displaystyle e^{\boxed{-3}} = \frac{1}{e^3} \quad\Rightarrow\quad \ln \left(\frac{1}{e^3}\right) = \boxed{-3}\)
  3. \(\displaystyle e^{\boxed{1}} = e \quad\Rightarrow\quad \ln \left(e\right) = \boxed{1}\)
  4. \(\displaystyle e^{\boxed{0}} = 1 \quad\Rightarrow\quad \ln \left(1\right) = \boxed{0}\)
  5. \(e^x \gt 0\text{,}\) for all \(x\text{.}\) In particular, \(e^x \neq 0\text{,}\) for any \(x\text{.}\) Thus \(\ln \left(0\right)\) is undefined.

Example 7.25. Graphing Logarithmic Functions.

Find the domain and sketch the graph of \(y = \ln(x+1)\text{.}\)
Solution.
We require that the input into the logarithm is positive. In this case, \(x+1 \gt 0\) or equivalently, \(x \gt -1\text{.}\) The domain is then the interval \((-1,+\infty)\text{.}\) Adding one to the input has the effect of translating the graph one unit to the left so that the vertical asymptote of the logarithm is now \(x=-1\text{.}\)
Figure 7.26. The graph of \(y = \ln(x+1)\) (solid) compared to \(y = \ln(x)\) (dashed).
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