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Section 15.3 The Natural Log

Remember that in Section 14.3, we learned about the number \(e\text{.}\) In this section, we will learn all about the log that has base \(e\text{.}\)

Definition 15.10.

The Natural Logarithm is the log that has base \(e\text{.}\) It is written \(\ln(x)\text{,}\) where the first letter is a lower-case L.

Example 15.11.

Suppose \(f(x)=e^x\) and we want to find its inverse. This is an exponential function, so its inverse is the log with the same base. Since the base of this exponential is \(e\text{,}\) the inverse will the the log base \(e\text{,}\) which we just learned is the natural log. So, \(f^{-1}(x)=\ln(x)\text{.}\)

Checkpoint 15.12.

Suppose \(g(x)=\ln(x)\text{.}\) Find \(f^{-1}(x)\text{.}\)
Answer.
\(g^{-1}(x)=e^x\)
Solution.

Remember that \(\ln(x)\) is just the log function with base \(e\text{.}\) In fact, you may find it helpful to rewrite it as \(\log_e(x)\) while working on these problems. Since the inverse of a log is the exponential with the same base, we have that \(g^{-1}(x)=e^x\text{.}\)

Example 15.13.

Suppose we want to simplify each of the following.

  1. Let's simplify \(\ln\left(e^7\right)\text{.}\) Remember that \(\ln\) is just log base \(e\text{,}\) so you may find it helpful to actually rewrite it as \(\log_e\left(e^7\right)\text{.}\) Now, this looks just like the questions we did in Section 15.1. The log and exponential have the same base, so they match and we have \(\ln\left(e^7\right)=7\text{.}\)

  2. Let's simplify \(e^{\ln(2)}\text{.}\) Remember that \(\ln\) is just log base \(e\text{,}\) so you may find it helpful to actually rewrite it as \(e^{\log_e(2)}\text{.}\) Now, this looks just like the questions we did in Section 15.1. The log and exponential have the same base, so they match and we have \(e^{\ln(2)}=2\text{.}\)

  3. Let's simplify \(\ln\left(\frac{1}{e^9}\right)\text{.}\) Remember that \(\ln\) is just log base \(e\text{,}\) so you may find it helpful to actually rewrite it as \(\log_e\left(\frac{1}{e^9}\right)\text{.}\) Now, this almost looks like the questions we did in Section 15.1, except that the \(e\) is still in the denominator. So, we need to use our exponent rules to make the inside just an exponential. Remember that to move something out of the denominator, we have to make the exponent negative. So, we get

    \begin{equation*} \ln\left(\frac{1}{e^9}\right)=\log_e\left(\frac{1}{e^9}\right)=\log_e\left(e^{-9}\right) \end{equation*}

    Now, the log and exponential have the same base, so they match and we have \(\ln\left(\frac{1}{e^9}\right)=-9\text{.}\)

Checkpoint 15.14.

Simplify each of the following expressions.

  1. \(\displaystyle \ln\left(e^3\right)\)

  2. \(\displaystyle \ln\left(\sqrt[7]{e}\right)\)

  3. \(\displaystyle e^{\ln(5)}\)

  4. \(\displaystyle \ln\left(\frac{1}{e^4}\right)\)

Answer.

  1. \(\displaystyle \ln\left(e^3\right)=3\)

  2. \(\displaystyle \ln\left(\sqrt[7]{e}\right)=\frac{1}{7}\)

  3. \(\displaystyle e^{\ln(5)}=5\)

  4. \(\displaystyle \ln\left(\frac{1}{e^4}\right)=-4\)

Solution.

  1. Remember that \(\ln\) is just log base \(e\text{,}\) so you may find it helpful to actually rewrite it as \(\log_e\left(e^3\right)\text{.}\) Now, this looks just like the questions we did in Section 15.1. The log and exponential have the same base, so they match and we have \(\ln\left(e^3\right)=3\text{.}\)

  2. Remember that \(\ln\) is just log base \(e\text{,}\) so you may find it helpful to actually rewrite it as \(\log_e\left(\sqrt[7]{e}\right)\text{.}\) Now, this almost looks like the questions we did in Section 15.1, except that the \(e\) still has a root over it, instead of an exponent. So, we need to use our exponent rules to make the inside just an exponential. Remember that a root is just a fractional power. So, we get

    \begin{equation*} \ln\left(\sqrt[7]{e}\right)=\log_e\left(\sqrt[7]{e}\right)=\log_e\left(e^{1/7}\right) \end{equation*}

    Now, the log and exponential have the same base, so they match and we have \(\ln\left(\sqrt[7]{e}\right)=\frac{1}{7}\text{.}\)

  3. Remember that \(\ln\) is just log base \(e\text{,}\) so you may find it helpful to actually rewrite it as \(e^{\log_e(5)}\text{.}\) Now, this looks just like the questions we did in Section 15.1. The log and exponential have the same base, so they match and we have \(e^{\ln(5)}=5\text{.}\)

  4. Remember that \(\ln\) is just log base \(e\text{,}\) so you may find it helpful to actually rewrite it as \(\log_e\left(\frac{1}{e^4}\right)\text{.}\) Now, this almost looks like the questions we did in Section 15.1, except that the \(e\) is still in the denominator. So, we need to use our exponent rules to make the inside just an exponential. Remember that to move something out of the denominator, we have to make the exponent negative. So, we get

    \begin{equation*} \ln\left(\frac{1}{e^4}\right)=\log_e\left(\frac{1}{e^4}\right)=\log_e\left(e^{-4}\right) \end{equation*}

    Now, the log and exponential have the same base, so they match and we have \(\ln\left(\frac{1}{e^4}\right)=-4\text{.}\)

Checkpoint 15.15.

Find the domain and the equation of the asymptote for \(f(x)=-2\ln(2x-7)+5\text{.}\)

Answer.

The domain is \(\left(\frac{7}{2},\infty\right)\text{,}\) and the equation for the vertical asymptote is \(x=\frac{7}{2}\text{.}\) (You must inclue the entire equation, not just the number \(\frac{7}{2}\text{.}\))

Solution.

Remember that the domain only cares about where there is a potential problem. So, it only cares about what is inside the log. Since the inside of the log has to be greater than 0, we set \(2x-7\gt 0\) and solve.

\begin{align*} 2x-7 \amp\gt 0\\ 2x \amp\gt 7\\ x \amp \gt \frac{7}{2} \end{align*}

Now, we just have to turn that into interval notation to get our domain: \(\left(\frac{7}{2},\infty\right)\text{.}\)

To find the equation of the asymptote, we have to set the inside equal to 0 and solve for \(x\text{.}\)

\begin{align*} 2x-7 \amp= 0\\ 2xx \amp= 7\\ x \amp = \frac{7}{2} \end{align*}

Note that logs always have a vertical asymptote, but they never have a horizontal asymptote. Therefore, our answer is the entire equation \(x=\frac{7}{2}\text{,}\) not just the number \(\frac{7}{2}\text{.}\)