7.3: Logarithms and Logarithmic Functions
Learning Objectives
In this section you will learn to:
- Write equivalent logarithmic and exponential expressions.
- Evaluate logs.
- Use properties of logs to solve exponential equations.
Prerequisite Skills
Before you get started, take this prerequisite quiz.
1. Use patterns to simplify each exponent without the use of a calculator.
a. \(4^3\)
b. \(4^2\)
c. \(4^1\)
d. \(4^0\)
e. \(4^{-1}\)
f. \(4^{-2}\)
g. \(4^{-3}\)
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a. \(4^3=64\)
b. \(4^2=16\)
c. \(4^1=4\)
d. \(4^0=1\)
e. \(4^{-1}=\dfrac{1}{4}\)
f. \(4^{-2}=\dfrac{1}{16}\)
g. \(4^{-3}=\dfrac{1}{64}\)
If you missed this problem, review here . (Note that this will open a YouTube video in a new window.)
2. Simplify each expression.
a. \(4^\frac{1}{2}\)
b. \(9^\frac{1}{2}\)
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a. \(4^\dfrac{1}{2}=\sqrt{4}=2\)
b. \(9^\dfrac{1}{2}=\sqrt{9}=3\)
If you missed this problem, review here . (Note that this will open a different textbook in a new window.)
3. What can "easily" be determined about \(x\) in each equation?
a. \(4^x=4\)
b. \(4^x=8\)
c. \(4^x=16\)
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a. \(x=1\)
b. Since 8 is between 4 and 16, this means x must be between 1 and 2.
c. \(x=2\)
If you missed this problem, review here . (Note that this will open a different textbook in a new window.)
The Logarithm
Suppose that a population of 50 flies is expected to double every week, leading to a function of the form \(f(x) = 50(2)^x\), where \(x\) represents the number of weeks that have passed. When will this population reach 500?
Trying to solve this problem leads to
\[500 = 50(2)^x\nonumber\]
Dividing both sides by 50 to isolate the exponential leads to
\[10 = 2^x . \nonumber\]
While we have set up exponential models and used them to make predictions, you may have noticed that solving exponential equations has not yet been mentioned. The reason is simple: none of the algebraic tools discussed so far are sufficient to solve exponential equations. Consider the equation 2 x = 10 above. We know that 2 3 = 8 and 2 4 = 16, so it is clear that x must be some value between 3 and 4 since g ( x ) = 2 x is increasing. We could use technology to create a table of values or graph to better estimate the solution, but we would like to find an algebraic way to solve the equation.
We need an inverse operation to exponentiation in order to solve for the variable if the variable is in the exponent. The inverse function for an exponential function is a logarithmic function.
Logarithm
The logarithm (base b ) function, written log b ( x ), is the inverse of the exponential function (base b ), b x .
\[\mathbf{y=\log_{b}(x)} \quad \textbf{ is equivalent to } \quad \mathbf{b^y=x} \nonumber\]
In general, the statement \(b^a = c\) is equivalent to the statement \(\log_b(c) = a\).
Note: The base \(b\) must be positive: \(b>0\)
Inverse Property of Logarithms
Since the logarithm and exponential are inverses, it follows that:
\[ \log_{b}(b^x) \quad \text{ and } b^{\log_{b}(x)}=x \nonumber\]
Since log is a function, it is most correctly written as log b ( c ), using parentheses to denote function evaluation, just as we would with f(c) . However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written as log b c .
Example \(\PageIndex{1}\)
Write these exponential equations as logarithmic equations:
- 2 3 = 8
- 5 2 = 25
- \(10^{-3} = \frac{1}{1000}\)
Solution
a. 2 3 = 8 can be written as a logarithmic equation as log 2 (8) = 3b. 5 2 = 25 can be written as a logarithmic equation as log 5 (25) = 2
c. \(10^{-3} = \frac{1}{1000}\) can be written as a logarithmic equation as \(\log _{10}\left(\frac{1}{1000}\right)=-3\)
Example \(\PageIndex{2}\)
Write these logarithmic equations as exponential equations:
- \(\log _{7}(49)=2\)
- \(\log _{3}(81)=4\)
Solution
- \(\log _{7}(49)=2\) can be written as an exponential equation as \(7^2=49\)
- \(\log _{3}(81)=4\) can be written as an exponential equation as \(3^{4}=81\)
By establishing the relationship between exponential and logarithmic functions, we can now solve basic logarithmic and exponential equations by rewriting.
Example \(\PageIndex{3}\)
Use the definition of logs to solve log 4 ( x ) = 2 for x .
Solution
By rewriting this expression as an exponential, 4 2 = x , so x = 16
Example \(\PageIndex{4}\)
Use the definition of logs to solve 2 x = 10 for x .
Solution
By rewriting this expression as a logarithm, we get x = log 2 (10)
While this does define a solution, you may find it somewhat unsatisfying since it is difficult to compare this expression to the decimal estimate we made earlier. Also, giving an exact expression for a solution is not always useful—often we really need a decimal approximation to the solution. Luckily, this is a task that calculators and computers are quite adept at. Unluckily for us, most calculators and computers will only evaluate logarithms of two bases: base 10 and base e . Happily, this ends up not being a problem, as we’ll see soon that we can use a “change of base” formula to evaluate logarithms for other bases.
Common and Natural Logarithms
The common log is the logarithm with base 10, and is typically written \(\log (x)\) and sometimes like \(\log_{10} (x)\). If the base is not indicated in the log function, then the base b used is \(b=10\).
The natural log is the logarithm with base \(e\), and is typically written \(\ln (x)\).
Note that for any other base b, other than 10, the base must be indicated in the notation \(\log_b (x)\).
Example \(\PageIndex{5}\)
Evaluate \(\log(100)\) using the definition of the common log.
Solution
The table shows values of the common log
|
number |
number as exponential |
log(number ) |
|---|---|---|
|
1000 |
10 3 |
3 |
|
100 |
10 2 |
2 |
|
10 |
10 1 |
1 |
|
1 |
10 0 |
0 |
|
0.1 |
10 -1 |
-1 |
|
0.01 |
10 -2 |
-2 |
|
0.001 |
10 -3 |
-3 |
To evaluate log(100), we can say
\[ x = \log(100) \nonumber\]
Then rewrite the equation in exponential form using the common log base of 10
\[10^x = 100 \nonumber \]
From this, we might recognize that 100 is the square of 10, so
\[x=2 \nonumber\]
Alternatively, we can use the inverse property of logs to write
\[\log_{10}(10^2) = 2 \nonumber\]
Example \(\PageIndex{6}\)
Evaluate \(\log\left(\dfrac{1}{1,000}\right)\) using the definition of the common log.
Solution
To evaluate log(1/1,000), we can say
\[x=\log (1 / 1,000)=\log \left(1 / 10^{3}\right)=\log \left(10^{-3}\right) \nonumber\]
Then rewrite the equation in exponential form: \(10^{x}=10^{-3}\)
Therefore \(x = -3\)
Alternatively, we can use the inverse property of logs to find the answer:
\[ \log _{10}\left(10^{-3}\right)=-3 \nonumber\]
Example \(\PageIndex{7}\)
Evaluate each expression using the definition of the natural log.
- \(\ln e\)
- \(\ln e^5\)
Solution
a. To evaluate \(\ln e\), we can say
\[ x = \ln e \nonumber\]
Then rewrite into exponential form using the natural log base of e
\[ e^x = e \nonumber\]
Notice that there isn't an exponent shown on the right side, so the exponent must be 1.
\[ e^x = e^1 \nonumber\]
Therefore \(x = 1\).
It is a good idea to remember that \(\ln e = 1\).
b. To evaluate \(\ln e^5\), we can say
\[ x = \ln e^5 \nonumber\]
Then rewrite into exponential form using the natural log base of e
\[ e^x = e^5 \nonumber\]
Therefore \(x = 5\).
Alternatively, we can use the inverse property of logs to write \(\ln \left(e^{5}\right)=5\).
Example \(\PageIndex{8}\)
Evaluate the following using your calculator:
- \(\log 500\)
- \(\ln 500\)
Solution
a. Using the LOG key on the calculator to evaluate logarithms in base 10, we evaluate LOG(500)
Answer: \(\log 500 \approx 2.69897\)
b. Using the LN key on the calculator to evaluate natural logarithms , we evaluate LN(500)
Answer: \(\ln 500 \approx 6.214608\)
Using Properties of Logarithms to Solve Exponential Equations
We often need to evaluate logarithms using a base other than 10 or e . To find a way to utilize the common or natural logarithm functions to evaluate expressions like log 2 (10), we need some additional properties.
Properties of Logs: Change of Base
\[\log _{b}(A)=\frac{\log _{c}(A)}{\log _{c}(b)} \text { for any bases } b, c>0 \nonumber\]
The change of base property allows us to easily convert logarithms into base 10 or base e.
Properties of Logs: One-to-One Property of Logarithms
\[C=D \quad \text {if and only if} \quad \log _{b}(C)=\log _{b}(D) \nonumber\]
The one-to-one property of logs allows us to take the logarithm of both sides of an equation.
Properties of Logs: Power Property
\[\log _{b}\left(A^{q}\right)=q \log _{b}(A) \nonumber\]
The power property allows us to move a variable out of an exponent. When used in conjunction with the one-to-one property of logs, this becomes a convenient way to isolate the variable when it is in the exponent.
Evaluating Logarithms
Example \(\PageIndex{9}\)
Evaluate \(\log_{5}(100)\) using the change of base formula.
Solution
We can rewrite this expression using any other base.
Method 1: We can use natural logarithm base e with the change of base formula
\[\log _{5}(100)=\frac{\ln (100)}{\ln (5)}=\mathrm{LN}(100) / \mathrm{LN}(5) \approx 2.861 \nonumber\]
Method 2: We can use common logarithm base 10 with the change of base formula,
\[\log _{5}(100)=\frac{\log (100)}{\log (5)}=\operatorname{LOG}(100) / \mathrm{LOG}(5) \approx 2.861 \nonumber\]
With the change of base formula, \(\log _{b}(A)=\frac{\log _{c}(A)}{\log _{c}(b)}\) for any bases \(b\), \(c >0\), we can finally find a decimal approximation to our question from the beginning of the section.
Example \(\PageIndex{10}\)
Solve \(2^x = 10\) for \(x\).
Solution
Method 1: Use the definition of logarithms, followed by the change of base formula.
\[2^x = 10 \nonumber \]
Use the definition of logarithms to rewrite the exponential equation as a logarithmic equation:
\[x=\log _{2}(10) \nonumber\]
Using the change of base formula, we can rewrite log base 2 as a logarithm of any other base. Since our calculators can evaluate natural log, one option is to choose to use the natural logarithm, which is the log base e :
Using our calculators to evaluate this, \(\frac{\ln (10)}{\ln (2)}=\mathrm{LN}(10) / \mathrm{LN}(2) \approx 3.3219\)
Method 2: Use the one-to-one property of logarithms, followed by the power property, to isolate the exponent.
\[2^x = 10 \nonumber \]
Use the one-to-one property of logarithms to take the logarithm of both sides of the equation.
\[\log(2^x) = \log(10) \nonumber \]
Apply the power property to move the exponent out of the logarithm.
\[x\log(2) = \log(10) \nonumber \]
Divide both sides by \(\log(2)\) to isolate the variable.
\[x = \frac{\log(10)}{\log(2)} \approx 3.3219 \nonumber \]
This finally allows us to answer our original question from the beginning of this section:
For the population of 50 flies that doubles every week, it will take approximately 3.32 weeks to grow to 500 flies.
Example \(\PageIndex{11}\)
Solve \(2(3)^{5x} = 24\) for \(x\) by using the one-to-one property of logs.
Solution
First, isolate the exponential expression by dividing both sides by 2.
\[2(3)^{5x} = 24 \nonumber \]
\[3^{5x} = 12 \nonumber \]
Then use the one-to-one property of logarithms to take the logarithm of both sides of the equation.
\[\log(3^{5x}) = \log(12) \nonumber \]
Apply the power property to move the exponent out of the logarithm.
\[5x\log(3) = \log(12) \nonumber \]
The x is multiplied by both the 5 and the \(\log(3)\), so divide both sides by \(5\log(3)\) to isolate the variable.
\[x = \frac{\log(12)}{5\log(3)} \approx 0.4524 \nonumber \]
We summarize the relationship between exponential and logarithmic functions
Logarithms
The logarithm (base b ) function, written log b ( x ), is the inverse of the exponential function (base b ), b x .
\[\mathbf{y=\log_{b}(x)} \quad \textbf{ is equivalent to } \quad \mathbf{b^y=x} \nonumber\]
In general, the statement \(b^a = c\) is equivalent to the statement \(\log_b(c) = a\).
Note: The base b must be positive: b>0
Inverse Property of Logarithms
Since the logarithm and exponential are inverses, it follows that:
\[ \log_{b}(b^x) \quad \text{ and } \quad b^{\log_{b}(x)}=x \nonumber\]
Properties of Logs: Exponential Property: \(\log _{b}\left(A^{q}\right)=q \log _{b}(A) \nonumber\)
Properties of Logs: Change of Base: \(\log _{b}(A)=\frac{\log _{c}(A)}{\log _{c}(b)} \text { for any base } b, c>0 \nonumber\)
The inverse, exponential, and change of base properties above will allow us to solve the equations that arise in this course.
For the sake of completeness, we state a few more properties of logarithms here.
Sum of Logs Property: \(\log _{b}(A)+\log _{b}(C)=\log _{b}(A C)\)
Difference of Logs Property: \(\log _{b}(A)-\log _{b}(C)=\log _{b}\left(\frac{A}{C}\right)\)
Power Property of Logs: \(\log_{b}(A^n)=n*\log_{b}(A)\)
Logs of Reciprocals: \(\log _{b}\left(\frac{1}{C}\right)=-\log _{b}(C)\)
Reciprocal Bases: \(\log _{1 / b} C=-\log _{b}(C)\)
Source: The material in this section of the textbook originates from David Lippman and Melonie Rasmussen, Open Text Bookstore, Precalculus: An Investigation of Functions, “ Chapter 4: Exponential and Logarithmic Functions ,” licensed under a Creative Commons CC BY-SA 3.0 license. The material here is based on material contained in that textbook but has been modified by Roberta Bloom, as permitted under this license.