Motivating Questions
- What is a logarithmic function, and what properties do logarithmic functions have?
- How can we use logarithmic functions to solve exponential equations?
Section 3.2 Properties of Logarithms
This section corresponds to 2.9 Derivatives of Logs in the workbook.
We defined an exponential function to be a function of the form
\begin{equation*}
P=P(t)=a(b)^t.
\end{equation*}
In this section we will discuss logarithmic functions which are inverses of exponential functions. In particular, the function
\begin{equation*}
t=\log_{b}\left(\frac{P}{a}\right)
\end{equation*}
is the inverse of the exponential function above. For \(a=1\text{,}\) the following applet illustrates this inverse relationship:
While this inverse relationship is how we will think of logarithms in practice, let’s give a formal definition:
The Logarithm.
Let \(b\neq 1 \) be a positive number, then the function
\begin{equation*}
f(t)=\log_b(t)
\end{equation*}
is called a logarithm with base \(b \text{.}\)
Upon inputting a value \(t\) the function \(\log_b(t)\) will tell you the power of \(b \) which will yield \(t \text{.}\)
Due to the relationship between logarithms and exponentials, we often say that the equations
\begin{equation*}
x=\log_b(y)\ \ \ \ \text{and}\ \ \ \ b^x=y
\end{equation*}
are equivalent.
Warning 3.2.2.
One of the more common bases of a logarithm is base \(10 \text{.}\) Since it is used so often, we have developed a short hand notation for a logarithm of base \(10\text{.}\) This short hand is shown below:
\begin{equation*}
\log_{10}(y)=\log(y).
\end{equation*}
In other words, we simply drop the subscript when referring to base \(10 \text{.}\)
Subsection 3.2.1 Properties of Logarithms
Because logarithms are actually exponents, they have several properties that can be derived from the laws of exponents. Here are the laws we will need at present.
- To multiply two powers with the same base, add the exponents and leave the base unchanged.\begin{equation*} a^m \cdot a^n = a^{m+n} \end{equation*}
- To divide two powers with the same base, subtract the exponents and leave the base unchanged.\begin{equation*} \frac{a^m}{a^n} = a^{m-n} \end{equation*}
- To raise a power to a power, keep the same base and multiply the exponents.\begin{equation*} \left(a^m\right)^n = a^{mn} \end{equation*}
Each of these laws corresponds to one of three properties of logarithms.
Properties of Logarithms.
If \(x,y,b>0\text{,}\) and \(b\neq 1\text{,}\) then
- \(\displaystyle \log_b(xy)=\log_b(x)+\log_b(y),\)
- \(\displaystyle \log_b\left(\frac{x}{y}\right)=\log_b(x)-\log_b(y),\)
- \(\displaystyle \log_b\left(x^k\right)=k\cdot \log_b(x),\)
- \(\displaystyle \log_b(b^y)=y,\)
- \(\displaystyle b^{\log_b(x)}=x.\)
We will examine the properties of logarithms closer in the Homework problems. For now, study the examples below, keeping in mind that a logarithm is is the inverse function of an exponential function.
- Property (1):\begin{equation*} \log_{2}{32}=\log_{2}{(4 \cdot 8)}=\log_2 4 +\log_2 8= 2+3=5 \end{equation*}
- Property (2):\begin{equation*} \log_{2}{\frac{16}{2}}=\log_2{16}-\log_2 2=4-1=3 \end{equation*}
- Property (3):\begin{equation*} \log_{2}{64}=\log_{2}{(4^3)}=3\log_2 4= 3 \cdot 2=6 \end{equation*}
Subsection 3.2.2 Using the Properties of Logarithms
Of course, these properties are useful not so much for computing logs but rather for simplifying expressions that contain variables. We will use them to solve exponential equations. But first, we will practice applying the properties. In Example 3.2.3., we rewrite one log in terms of simpler logs.
Example 3.2.3.
Simplify \(\log_{b}\sqrt{{xy}}\text{.}\)
Solution.
First, we write \(\sqrt{xy}\) using a fractional exponent:
\begin{equation*}
\log_{b}{\sqrt{xy}} = \log_{b}{\left((xy)^{1/2}\right)}.
\end{equation*}
Then we apply Property (3) to rewrite the exponent as a coefficient:
\begin{equation*}
\log_{b}{\left((xy)^{1/2}\right)} = \frac{1}{2}\log_{b}{(xy)}.
\end{equation*}
Finally, by Property (1) we write the log of a product as a sum of logs:
\begin{equation*}
\frac{1}{2}\left(\log_{b}{xy}\right) = \frac{1}{2}\left(\log_{b}{x} + \log_{b}{y}\right).
\end{equation*}
Thus, \(\log_{b}\sqrt{xy} = \frac{1}{2}\left(\log_{b}{x} + \log_{b}{y}\right)\text{.}\)
Example 3.2.4.
Simplify \(\log_{b}{xy^2}\text{.}\)
Answer.
\(\log_b x + 2\log_b y\)
Warning 3.2.5.
Be careful when using the properties of logarithms! Compare the statements below:
-
\(\log_{b}{(2x)} = \log_{b}{2} + \log_{b}{x} \ \ \ \text{ by Property 1,}\)but\(\log_{b}{(2 + x)} \ne \log_{b}{2} + \log_{b}{x}.\)
-
\(\log_{b}{\left(\dfrac{x}{5}\right)}= \log_b x - \log_b 5 \ \ \ \text{ by Property 2,}\)but\(\log_{b}{\left(\dfrac{x}{5}\right)} \ne \dfrac{\log_b x}{\log_b 5}.\)
We can also use the properties of logarithms to combine sums and differences of logarithms into one logarithm.
Example 3.2.6.
Express \(3(\log_b x - \log_b y)\) as a single logarithm with a coefficient of \(1\text{.}\)
Solution.
We begin by applying Property (2) to combine the logs.
\begin{equation*}
3(\log_b x - \log_b y) = 3 \log_{b}\left(\frac{x}{y}\right)
\end{equation*}
Then, using Property (3), we replace the coefficient \(3\) by an exponent \(3\text{.}\)
\begin{equation*}
3 \log_{b}\left(\frac{x}{y}\right)=\log_{b}\left(\left(\frac{x}{y}\right)^3\right)
\end{equation*}
Example 3.2.7.
Express \(2\log_b x + 4\log_{b}{(x + 3)}\) as a single logarithm with a coefficient of \(1\text{.}\)
Answer.
\(\log_b \left(x^2(x+3)^4\right) \)
Subsection 3.2.3 The Natural Exponential Function
There is another base for logarithms and exponential functions that is often used in applications. This base is an irrational number called \(e\text{,}\) where
\begin{equation*}
e \approx 2.71828182845.
\end{equation*}
The number \(e\) is essential for many advanced topics, and it is often called the natural base.
The base \(e\) logarithm of a number \(x\text{,}\) or \(\log_ e x\text{,}\) is called the natural logarithm of \(x\) and is denoted by \(\ln x\text{.}\)
The Natural Logarithm.
The natural logarithm is the logarithm base \(e\text{.}\)
\begin{equation*}
\ln x = \log_{e}{x}, ~~~~ x\gt 0
\end{equation*}
We use natural logarithms in the same way that we use logs to other bases. The properties of logarithms that we studied above also apply to logarithms base \(e\text{.}\)
Subsection 3.2.4 Solving Exponential Equations
Suppose we want to solve the equation
\begin{equation*}
5^x = 7.
\end{equation*}
We could rewrite the equation in logarithmic form to obtain the exact solution
\begin{equation*}
x = \log_{5}{7}.
\end{equation*}
However, sometimes we are stuck in a situation where we cannot evaluate \(\log_{5}{7}\text{.}\) For example, some calculators do not have a log base \(5\) button. So, if we want a decimal approximation for the solution, we begin by taking the base \(10\) logarithm of both sides of the original equation, even though the base of the power is not \(10\text{.}\) This gives us
\begin{equation*}
\log_{10}{(5^x)} = \log_{10}{7}.
\end{equation*}
Then we use Property (3) to rewrite the left side as
\begin{equation*}
x \log_{10}{5} = \log_{10}{7}.
\end{equation*}
Note how using Property (3) allows us to solve the equation: The variable, \(x\text{,}\) is no longer in the exponent, and it is multiplied by a constant, \(\log_{10}{5}\text{.}\) To finish the solution, we divide both sides by \(\log_{10}{5}\) to get
\begin{equation*}
x = \frac{\log_{10}{7}}{\log_{10}{5}}.
\end{equation*}
On your calculator, enter the sequence
LOG( \(7\) ) \(\div\) LOG ( \(5\) ) ENTER
to find that \(x \approx 1.2091\text{.}\)
Warning 3.2.8.
Do not confuse the expression \(\dfrac{\log_{10}{7}}{\log_{10}{5}}\) with \(\log_{10}{\left(\dfrac{7}{5}\right)}\text{;}\) they are not the same! Property (2) allows us to simplify \(\log{\left(\dfrac{x}{y}\right)}\text{,}\) but not \(\dfrac{\log x}{\log y}\text{.}\) We cannot rewrite \(\dfrac{\log_{10}{7}}{\log_{10}{5}}\text{,}\) so we must evaluate it as \((\log 7)/(\log 5)\text{.}\) You can check on your calculator that
\begin{equation*}
\dfrac{\log_{10}{7}}{\log_{10}{5}}\ne \log_{10}{\left(\dfrac{7}{5}\right)}= \log_{10}{1.4}\text{.}
\end{equation*}
Example 3.2.9.
Solve \(1640 = 80 \cdot 6^{0.03x}\text{.}\)
Solution.
First we divide both sides by \(80\) to obtain
\begin{equation*}
20.5 = 6^{0.03x}.
\end{equation*}
Next, we take the base \(10\) logarithm of both sides of the equation and use Property (3) of logarithms to get
\begin{equation*}
\log_{10}{20.5} = \log_{10}{(6^{0.03x})}= 0.03x \log_{10}{6}.
\end{equation*}
On the right side of the equation, \(x\) is multiplied by two constants, \(0.03\) and \(\log_{10}{6}\text{.}\) So, to solve for \(x\) we must divide both sides of the equation by \(0.03 \log_{10}{6}\text{.}\) We use a calculator to evaluate the answer:
\begin{equation*}
x = \frac{\log_{10}{20.5}}{0.03 \log_{10}{6}}\approx 56.19.
\end{equation*}
(On your calculator, remember to enclose the denominator, \(0.03 \log_{10}{6}\text{,}\) in parentheses.)
Warning 3.2.10.
In Example 3.2.9, do not try to simplify
\begin{equation*}
80 \cdot 6^{0.03x} \rightarrow 480^{0.03x}~~ \text{ Incorrect!}
\end{equation*}
Remember that the order of operations tells us to compute the power \(6^{0.03x}\) before multiplying by \(80\text{.}\)
Example 3.2.11.
Solve \(5(1.2)^{2.5x} = 77\text{.}\)
Hint.
Divide both sides by 5.
Take the log of both sides.
Apply Property (3) to simplify the left side.
Solve for \(x\text{.}\)
Answer.
\(x=\dfrac{\log 15.4}{2.5\log 1.2}\approx 5.999 \)
Subsection 3.2.5 Summary
- A logarithmic function is the inverse of an exponential function, and the function \(f(t) = \log_b(t)\) will give the power of \(b\) which yields \(t\text{.}\)
- Logarithms have many important properties based on the properties of exponents, which can be used to expand or contract logarithmic expressions, and solve equations involving exponential or logarithmic functions.
- There are two shorthand notations for logarithms: writing just \(\log(x)\) means \(\log_{10}(x)\text{,}\) and writing \(\ln(x)\) means \(\log_e(x)\text{.}\)
Exercises 3.2.6 Exercises
1. Logarithms to Exponents.
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2. Definition of Logarithms.
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3. Exponents to Logarithms.
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4. Evaluating Logarithms.
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5. Evaluating Logarithms.
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6. Evaluating Logarithms.
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7. Evaluating Logarithms.
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8. Applications of Logarithms.
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9. Applications of Logarithms.
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10. Properties of Logarithms.
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11. Solving Exponents with Logarithms.
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