Logarithmic Functions

Section 7.3 Logarithmic Functions

Exponential functions are one-to-one and therefore have inverse functions, called logarithms. (See Section 6.9.)

Definition 7.17. Logarithmic Functions.

Suppose \(a \gt 0\) and \(a \neq 1\text{.}\) The base \(a\) exponential function \(f(x) = a^x\) has an inverse function called the base \(a\) logarithmic function denoted

\begin{equation*} f^{-1}(y) = \log_a (y), \quad y \gt 0. \end{equation*}

The domain of the logarithm is the range \((0,\infty)\) of the exponential function. As inverse functions, they satisfy \((f^{-1}\circ f )(x) = x\) and \((f\circ f^{-1})(y) = y\) which now becomes

\(\displaystyle \log_a(a^x) = x\) for all \(x\)
\(\displaystyle a^{\log_a(y)} = y\) for all \(y \gt 0\)

Alternatively, exponential equations are equivalent logarithmic equations according to the relationship

\begin{equation} y = a^x \quad \text{if and only if} \quad \log_a(y) =x.\tag{7.1} \end{equation}

Observe that the logarithm returns the exponent in the exponential equation.

Example 7.18. The Base \(2\) Logarithm.

This video discusses the case of the base 2 logarithm.

Example 7.19. Evaluating Logarithmic Expressions.

Evaluate the following expressions without the use of calculator. Video solutions follow.

\(\displaystyle \log_2(8)\)
\(\displaystyle \log_8(2)\)
\(\displaystyle \log_{\pi}\left(\frac{1}{\pi^8}\right)\)
\(\displaystyle \log_{\pi} (1)\)
\(\displaystyle \pi^{\log_{\pi}(2)}\)

Solution.

Example 7.20. Evaluating Logarithms.

Evaluate the following logarithms without the use of calculator.

\(\displaystyle \displaystyle \log_{5} (25)\)
\(\displaystyle \displaystyle \log_{5} \left(\frac{1}{5}\right)\)
\(\displaystyle \displaystyle \log_{\frac{1}{5}} \left(5\right)\)
\(\displaystyle \displaystyle \log_{5} \left(\sqrt{5}\right)\)
\(\displaystyle \displaystyle \log_{5} \left(5\right)\)
\(\displaystyle \displaystyle \log_{5} \left(1\right)\)
\(\displaystyle \displaystyle \log_{5} \left(-1\right)\)

Solution.

\(\displaystyle 5^\boxed{2} = 25 \quad\Rightarrow\quad \log_{5} (25) = \boxed{2}\)
\(\displaystyle 5^{\boxed{-1}} = \frac{1}{5} \quad\Rightarrow\quad \log_{5} \left(\frac{1}{5}\right) = \boxed{-1}\)
\(\displaystyle \left(\frac{1}{5}\right)^{\boxed{-1}} = 5 \quad\Rightarrow\quad \log_{\frac{1}{5}} \left(5\right) = \boxed{-1}\)
\(\displaystyle 5^{\boxed{1/2}} = \sqrt{5} \quad\Rightarrow\quad \log_{5} \left(\sqrt{5}\right) = \boxed{1/2}\)
\(\displaystyle 5^{\boxed{1}} = 5 \quad\Rightarrow\quad \log_{5} \left(5\right) = \boxed{1}\)
\(\displaystyle 5^{\boxed{0}} = 1 \quad\Rightarrow\quad \log_{5} \left(1\right) = \boxed{0}\)
\(5^x \gt 0\text{,}\) for all \(x\text{.}\) In particular, \(5^x \neq -1\) for any \(x\text{.}\) Thus \(\log_{5} \left(-1\right)\) is undefined.

Example 7.21. Graphing a Logarithmic Function.

To sketch the graph of \(y = \log_5 (x)\) begin by sketching the graph of its inverse function \(y = 5^x\) (dashed curve). Then reflect the graph across \(y=x\) interchanging coordinates of points as you do. The horizontal asymptote of \(y = 5^x\) will become the vertical asymptote of \(y = \log_5 (x)\) (solid curve).

Observe how the rapid exponential growth of \(y = 5^x\) becomes the slow logarithmic growth of \(y = \log_5 (x)\text{.}\)

Prev Top Next