Skip to main content

Section 2.5 Logarithmic Functions

Recall the three kinds of exponential functions \(f(x)=a^x\) depending on whether \(0\lt a\lt 1\text{,}\) \(a=1\) or \(a>1\text{:}\)

So long as \(a\neq 1\text{,}\) the function \(f(x)=a^x\) satisfies the Horizontal Line Test and therefore has an inverse. We call the inverse of \(a^x\) the logarithmic function with base a and denote it by \(\log_a\text{.}\) In particular,

\begin{equation*} \log_a x=y\iff a^y=x\text{.} \end{equation*}

The cancellation formulas for logs are:

\begin{equation*} \log_a(a^x)=x,\mbox{ for every \(x\in\R\)}\text{,} \end{equation*}
\begin{equation*} a^{\log_a(x)}=x,\mbox{ for every \(x>0\)}\text{.} \end{equation*}

Since the function \(f(x)=a^x\) for \(a\neq 1\) has domain \(\mathbb{R}\) and range \((0,\infty)\text{,}\) the logarithmic function has domain \((0,\infty)\) and range \(\mathbb{R}\text{.}\) For the most part, we only focus on logarithms with a base larger than \(1\) (i.e., \(a>1\)) as these are the most important.

Notice that every logarithm passes through the point \((1,0)\) in the same way that every exponential function passes through the point \((0,1)\text{.}\)

Some properties of logarithms are as follows.

Logarithm Properties.

Let \(A,B\) be positive numbers and \(b>0\) (\(b\neq1\)) be a base.

  • \(\ds{\log_b(AB)=\log_b A+\log_b B}\text{,}\)

  • \(\ds{\log_b\left(\frac{A}{B}\right)=\log_b A-\log_b B}\text{,}\)

  • \(\ds{\log_b(A^n)=n\log_b A}\text{,}\) where \(n\) is any real number.

Example 2.21. Compute Lorarithms.

To compute \(\log_2(24)-\log_2(3)\) we can do the following:

\begin{equation*} \log_2(24)-\log_2(3)=\log_2\left(\frac{24}{3}\right)=\log_2(8)=3\text{,} \end{equation*}

since \(2^3=8\text{.}\)

Subsection 2.5.1 The Natural Logarithm

As mentioned earlier for exponential functions, the number \(e\approx 2.71828\ldots\) is the most convenient base to use in Calculus. For this reason we give the logarithm with base \(e\) a special name: the natural logarithm. We also give it special notation:

\begin{equation*} \log_ex=\ln x\text{.} \end{equation*}

You may pronounce \(\ln\) as either: “el - en”, “lawn”, or refer to it as “natural log”. The above properties of logarithms also apply to the natural logarithm.

Often we need to turn a logarithm (in a different base) into a natural logarithm. This gives rise to the change of base formula.

Change of Base Formula.
\begin{equation*} \log_ax=\frac{\ln x}{\ln a} \end{equation*}
Example 2.22. Combine Logarithms.

Write \(\ln A+2\ln B -\ln C\) as a single logarithm.

Solution

Using properties of logarithms, we have,

\begin{align*} \ln A+2\ln B -\ln C \amp = \ln A + \ln B^2 - \ln C\\ \amp = \ln (AB^2) - \ln C\\ \amp = \ds{\ln\frac{AB^2}{C}} \end{align*}
Example 2.23. Solve Exponential Equations using Logarithms.

If \(e^{x+2}=6e^{2x}\text{,}\) then solve for \(x\text{.}\)

Solution

Taking the natural logarithm of both sides and noting the cancellation formulas (along with \(\ln e=1\)), we have:

\begin{equation*} \begin{array}{rcl} e^{x+2} \amp = \amp 6e^{2x}\\ \\ \ln e^{x+2}\amp =\amp \ln (6e^{2x})\\ \\ x+2\amp =\amp \ln 6 + \ln e^{2x}\\ \\ x+2\amp =\amp \ln 6 + 2x\\ \\ x\amp =\amp 2-\ln 6 \end{array} \end{equation*}
Example 2.24. Solve Logarithm Equations using Exponentials.

If \(\ln(2x-1)=2\ln(x)\text{,}\) then solve for \(x\text{.}\)

Solution

“Taking \(e\)” of both sides and noting the cancellation formulas, we have:

\begin{equation*} \begin{array}{rcl} e^{\ln(2x-1)} \amp = \amp e^{2\ln(x)}\\ \\ (2x-1) \amp = \amp e^{\ln(x^2)}\\ \\ 2x-1 \amp = \amp x^2\\ \\ x^2-2x+1 \amp = \amp 0\\ \\ (x-1)^2 \amp = \amp 0 \end{array} \end{equation*}

Therefore, the solution is \(x=1\text{.}\)

Exercises for Section 2.5.

Expand \(\ds\log_{10} ((x+45)^7 (x-2))\text{.}\)

Expand \(\ds\log_2 {\frac{x^3}{3x-5 +(7/x)}}\text{.}\)

Write \(\ds \log_2 3x + 17 \log_2 (x-2) - 2\log_2 (x^2 + 4x + 1)\) as a single logarithm.

Answer
\(\log_{2}\left(\dfrac{3x(x-2)^{17}}{(x^{2}+4x+1)^{2}} \right)\)
Solution

We can write the expression as a single logarithm using log-rules.

\begin{equation*} \begin{split} \log_{2}3x + 17\log_{2}(x-2) - 2\log_{2}(x^{2}+4x+1) \amp = \log_{2}3x + \log_{2}(x-2)^{17} - \log_{2}(x^{2}+4x+1)^{2} \\ \amp = \log_{2}\left(\dfrac{3x(x-2)^{17}}{(x^{2}+4x+1)^{2}} \right) \end{split} \end{equation*}

Solve \(\ds \log_2 (1+ \sqrt{x} ) = 6\) for \(x\text{.}\)

Answer
\(x = 3969\)
Solution

We solve for \(x\) as follows.

\begin{equation*} \begin{split} \amp \log_{2} \left(1+\sqrt{x}\right) = 6\\ \amp 1 + \sqrt{x} = 2^{6} \\ \amp \sqrt{x} = 2^{6}-1 \\ \amp x = \left(2^{6}-1\right)^{2} \end{split} \end{equation*}

Therefore, \(x = 3969\text{.}\)

Solve \(\ds 2^{x^2} = 8\) for \(x\text{.}\)

Solve \(\ds \log_2 (\log_3 (x) ) = 1\) for \(x\text{.}\)

Solve \(\ds a^{2x} - 4a^{x} + 4 = 0\) for \(x\) (\(a\) constant).

Answer
\(\ds x = \dfrac{\log{2}}{\log{a}}\)
login