# Logarithmic Functions

In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes.[footnote]http://earthquake.usgs.gov/earthquakes/eqinthenews/2010/us2010rja6/#summary. Accessed 3/4/2013.[/footnote] One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings,[footnote]http://earthquake.usgs.gov/earthquakes/eqinthenews/2011/usc0001xgp/#summary. Accessed 3/4/2013.[/footnote] like those shown in the picture below. Even though both caused substantial damage, the earthquake in 2011 was 100 times stronger than the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale[footnote]http://earthquake.usgs.gov/earthquakes/eqinthenews/2010/us2010rja6/. Accessed 3/4/2013.[/footnote] whereas the Japanese earthquake registered a 9.0.[footnote]http://earthquake.usgs.gov/earthquakes/eqinthenews/2011/usc0001xgp/#details. Accessed 3/4/2013.[/footnote]## Convert Between Logarithmic And Exponential Form

In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake were 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is [latex]{10}^{x}=500[/latex], where*x*represents the difference in magnitudes on the

**Richter Scale**. How would we solve for

*x*? We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve [latex]{10}^{x}=500[/latex]. We know that [latex]{10}^{2}=100[/latex] and [latex]{10}^{3}=1000[/latex], so it is clear that

*x*must be some value between 2 and 3, since [latex]y={10}^{x}[/latex] is increasing. We can examine a graph to better estimate the solution. Estimating from a graph, however, is imprecise. To find an algebraic solution, we must introduce a new function. Observe that the graph above passes the horizontal line test. The exponential function [latex]y={b}^{x}[/latex] is

**one-to-one**, so its inverse, [latex]x={b}^{y}[/latex] is also a function. As is the case with all inverse functions, we simply interchange

*x*and

*y*and solve for

*y*to find the inverse function. To represent

*y*as a function of

*x*, we use a logarithmic function of the form [latex]y={\mathrm{log}}_{b}\left(x\right)[/latex]. The base

*b*

**logarithm**of a number is the exponent by which we must raise

*b*to get that number. We read a logarithmic expression as, "The logarithm with base

*b*of

*x*is equal to

*y*," or, simplified, "log base

*b*of

*x*is

*y*." We can also say, "

*b*raised to the power of

*y*is

*x*," because logs are exponents. For example, the base 2 logarithm of 32 is 5, because 5 is the exponent we must apply to 2 to get 32. Since [latex]{2}^{5}=32[/latex], we can write [latex]{\mathrm{log}}_{2}32=5[/latex]. We read this as "log base 2 of 32 is 5." We can express the relationship between logarithmic form and its corresponding exponential form as follows: [latex-display]{\mathrm{log}}_{b}\left(x\right)=y\Leftrightarrow {b}^{y}=x,\text{}b>0,b\ne 1[/latex-display] Note that the base

*b*is always positive. Because logarithm is a function, it is most correctly written as [latex]{\mathrm{log}}_{b}\left(x\right)[/latex], using parentheses to denote function evaluation, just as we would with [latex]f\left(x\right)[/latex]. However, when the input is a single variable or number, it is common to see the parentheses dropped and the expression written without parentheses, as [latex]{\mathrm{log}}_{b}x[/latex]. Note that many calculators require parentheses around the

*x*. We can illustrate the notation of logarithms as follows: Notice that, comparing the logarithm function and the exponential function, the input and the output are switched. This means [latex]y={\mathrm{log}}_{b}\left(x\right)[/latex] and [latex]y={b}^{x}[/latex] are inverse functions.

### A General Note: Definition of the Logarithmic Function

A**logarithm**base

*b*of a positive number

*x*satisfies the following definition. For [latex]x>0,b>0,b\ne 1[/latex], [latex-display]y={\mathrm{log}}_{b}\left(x\right)\text{ is equivalent to }{b}^{y}=x[/latex-display] where,

- we read [latex]{\mathrm{log}}_{b}\left(x\right)[/latex] as, "the logarithm with base
*b*of*x*" or the "log base*b*of*x*." - the logarithm
*y*is the exponent to which*b*must be raised to get*x*.

*x*and

*y*values, the domain and range of the exponential function are interchanged for the logarithmic function. Therefore,

- the domain of the logarithm function with base [latex]b \text{ is} \left(0,\infty \right)[/latex].
- the range of the logarithm function with base [latex]b \text{ is} \left(-\infty ,\infty \right)[/latex].

### Q & A

#### Can we take the logarithm of a negative number?

*No. Because the base of an exponential function is always positive, no power of that base can ever be negative. We can never take the logarithm of a negative number. Also, we cannot take the logarithm of zero. Calculators may output a log of a negative number when in complex mode, but the log of a negative number is not a real number.*

### How To: Given an equation in logarithmic form [latex]{\mathrm{log}}_{b}\left(x\right)=y[/latex], convert it to exponential form.

- Examine the equation [latex]y={\mathrm{log}}_{b}x[/latex] and identify
*b*,*y*, and*x*. - Rewrite [latex]{\mathrm{log}}_{b}x=y[/latex] as [latex]{b}^{y}=x[/latex].

### Example: Converting from Logarithmic Form to Exponential Form

Write the following logarithmic equations in exponential form.- [latex]{\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}[/latex]
- [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex]

Answer:
First, identify the values of *b*, *y*, and *x*. Then, write the equation in the form [latex]{b}^{y}=x[/latex].

- [latex]{\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}[/latex] Here, [latex]b=6,y=\frac{1}{2},\text{and } x=\sqrt{6}[/latex]. Therefore, the equation [latex]{\mathrm{log}}_{6}\left(\sqrt{6}\right)=\frac{1}{2}[/latex] is equivalent to [latex]{6}^{\frac{1}{2}}=\sqrt{6}[/latex].
- [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex] Here,
*b*= 3,*y*= 2, and*x*= 9. Therefore, the equation [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex] is equivalent to [latex]{3}^{2}=9[/latex].

### Try It

Write the following logarithmic equations in exponential form.- [latex]{\mathrm{log}}_{10}\left(1,000,000\right)=6[/latex]
- [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex]

Answer:

- [latex]{\mathrm{log}}_{10}\left(1,000,000\right)=6[/latex] is equivalent to [latex]{10}^{6}=1,000,000[/latex]
- [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex] is equivalent to [latex]{5}^{2}=25[/latex]

### Convert from exponential to logarithmic form

To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base*b*, exponent

*x*, and output

*y*. Then we write [latex]x={\mathrm{log}}_{b}\left(y\right)[/latex].

### Example: Converting from Exponential Form to Logarithmic Form

Write the following exponential equations in logarithmic form.- [latex]{2}^{3}=8[/latex]
- [latex]{5}^{2}=25[/latex]
- [latex]{10}^{-4}=\frac{1}{10,000}[/latex]

Answer:
First, identify the values of *b*, *y*, and *x*. Then, write the equation in the form [latex]x={\mathrm{log}}_{b}\left(y\right)[/latex].

- [latex]{2}^{3}=8[/latex] Here,
*b*= 2,*x*= 3, and*y*= 8. Therefore, the equation [latex]{2}^{3}=8[/latex] is equivalent to [latex]{\mathrm{log}}_{2}\left(8\right)=3[/latex]. - [latex]{5}^{2}=25[/latex] Here,
*b*= 5,*x*= 2, and*y*= 25. Therefore, the equation [latex]{5}^{2}=25[/latex] is equivalent to [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex]. - [latex]{10}^{-4}=\frac{1}{10,000}[/latex] Here,
*b*= 10,*x*= –4, and [latex]y=\frac{1}{10,000}[/latex]. Therefore, the equation [latex]{10}^{-4}=\frac{1}{10,000}[/latex] is equivalent to [latex]{\text{log}}_{10}\left(\frac{1}{10,000}\right)=-4[/latex].

### Try It

Write the following exponential equations in logarithmic form.- [latex]{3}^{2}=9[/latex]
- [latex]{5}^{3}=125[/latex]
- [latex]{2}^{-1}=\frac{1}{2}[/latex]

Answer:

- [latex]{3}^{2}=9[/latex] is equivalent to [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex]
- [latex]{5}^{3}=125[/latex] is equivalent to [latex]{\mathrm{log}}_{5}\left(125\right)=3[/latex]
- [latex]{2}^{-1}=\frac{1}{2}[/latex] is equivalent to [latex]{\text{log}}_{2}\left(\frac{1}{2}\right)=-1[/latex]

## Evaluate Logarithms

Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider [latex]{\mathrm{log}}_{2}8[/latex]. We ask, "To what exponent must 2 be raised in order to get 8?" Because we already know [latex]{2}^{3}=8[/latex], it follows that [latex]{\mathrm{log}}_{2}8=3[/latex]. Now consider solving [latex]{\mathrm{log}}_{7}49[/latex] and [latex]{\mathrm{log}}_{3}27[/latex] mentally.- We ask, "To what exponent must 7 be raised in order to get 49?" We know [latex]{7}^{2}=49[/latex]. Therefore, [latex]{\mathrm{log}}_{7}49=2[/latex]
- We ask, "To what exponent must 3 be raised in order to get 27?" We know [latex]{3}^{3}=27[/latex]. Therefore, [latex]{\mathrm{log}}_{3}27=3[/latex]

- We ask, "To what exponent must [latex]\frac{2}{3}[/latex] be raised in order to get [latex]\frac{4}{9}[/latex]? " We know [latex]{2}^{2}=4[/latex] and [latex]{3}^{2}=9[/latex], so [latex]{\left(\frac{2}{3}\right)}^{2}=\frac{4}{9}[/latex]. Therefore, [latex]{\mathrm{log}}_{\frac{2}{3}}\left(\frac{4}{9}\right)=2[/latex].

### How To: Given a logarithm of the form [latex]y={\mathrm{log}}_{b}\left(x\right)[/latex], evaluate it mentally.

- Rewrite the argument
*x*as a power of*b*: [latex]{b}^{y}=x[/latex]. - Use previous knowledge of powers of
*b*identify*y*by asking, "To what exponent should*b*be raised in order to get*x*?"

### Example: Solving Logarithms Mentally

Solve [latex]y={\mathrm{log}}_{4}\left(64\right)[/latex] without using a calculator.Answer: First we rewrite the logarithm in exponential form: [latex]{4}^{y}=64[/latex]. Next, we ask, "To what exponent must 4 be raised in order to get 64?" We know [latex]{4}^{3}=64[/latex] Therefore, [latex-display]\mathrm{log}{}_{4}\left(64\right)=3[/latex-display]

### Try It

Solve [latex]y={\mathrm{log}}_{121}\left(11\right)[/latex] without using a calculator.Answer: [latex]{\mathrm{log}}_{121}\left(11\right)=\frac{1}{2}[/latex] (recalling that [latex]\sqrt{121}={\left(121\right)}^{\frac{1}{2}}=11[/latex] )

### Example: Evaluating the Logarithm of a Reciprocal

Evaluate [latex]y={\mathrm{log}}_{3}\left(\frac{1}{27}\right)[/latex] without using a calculator.Answer: First we rewrite the logarithm in exponential form: [latex]{3}^{y}=\frac{1}{27}[/latex]. Next, we ask, "To what exponent must 3 be raised in order to get [latex]\frac{1}{27}[/latex]"? We know [latex]{3}^{3}=27[/latex], but what must we do to get the reciprocal, [latex]\frac{1}{27}[/latex]? Recall from working with exponents that [latex]{b}^{-a}=\frac{1}{{b}^{a}}[/latex]. We use this information to write

[latex]\begin{array}{l}{3}^{-3}=\frac{1}{{3}^{3}}\hfill \\ =\frac{1}{27}\hfill \end{array}[/latex]

Therefore, [latex]{\mathrm{log}}_{3}\left(\frac{1}{27}\right)=-3[/latex].### Try It

Evaluate [latex]y={\mathrm{log}}_{2}\left(\frac{1}{32}\right)[/latex] without using a calculator.Answer: [latex]{\mathrm{log}}_{2}\left(\frac{1}{32}\right)=-5[/latex]

### Use common logarithms

To convert from exponents to logarithms, we follow the same steps in reverse. We identify the base*b*, exponent

*x*, and output

*y*. Then we write [latex]x={\mathrm{log}}_{b}\left(y\right)[/latex].

### Example: Converting from Exponential Form to Logarithmic Form

Write the following exponential equations in logarithmic form.- [latex]{2}^{3}=8[/latex]
- [latex]{5}^{2}=25[/latex]
- [latex]{10}^{-4}=\frac{1}{10,000}[/latex]

Answer: First, identify the values of *b*, *y*, and *x*. Then, write the equation in the form [latex]x={\mathrm{log}}_{b}\left(y\right)[/latex].

- [latex]{2}^{3}=8[/latex]Here,
*b*= 2,*x*= 3, and*y*= 8. Therefore, the equation [latex]{2}^{3}=8[/latex] is equivalent to [latex]{\mathrm{log}}_{2}\left(8\right)=3[/latex]. - [latex]{5}^{2}=25[/latex]Here,
*b*= 5,*x*= 2, and*y*= 25. Therefore, the equation [latex]{5}^{2}=25[/latex] is equivalent to [latex]{\mathrm{log}}_{5}\left(25\right)=2[/latex]. - [latex]{10}^{-4}=\frac{1}{10,000}[/latex]Here,
*b*= 10,*x*= –4, and [latex]y=\frac{1}{10,000}[/latex]. Therefore, the equation [latex]{10}^{-4}=\frac{1}{10,000}[/latex] is equivalent to [latex]{\text{log}}_{10}\left(\frac{1}{10,000}\right)=-4[/latex].

### Try It

Write the following exponential equations in logarithmic form.- [latex]{3}^{2}=9[/latex]
- [latex]{5}^{3}=125[/latex]
- [latex]{2}^{-1}=\frac{1}{2}[/latex]

Answer:

- [latex]{3}^{2}=9[/latex] is equivalent to [latex]{\mathrm{log}}_{3}\left(9\right)=2[/latex]
- [latex]{5}^{3}=125[/latex] is equivalent to [latex]{\mathrm{log}}_{5}\left(125\right)=3[/latex]
- [latex]{2}^{-1}=\frac{1}{2}[/latex] is equivalent to [latex]{\text{log}}_{2}\left(\frac{1}{2}\right)=-1[/latex]

## Key Equations

Definition of the logarithmic function | For [latex]\text{ } x>0,b>0,b\ne 1[/latex],[latex]y={\mathrm{log}}_{b}\left(x\right)[/latex] if and only if [latex]\text{ }{b}^{y}=x[/latex]. |

Definition of the common logarithm | For [latex]\text{ }x>0[/latex], [latex]y=\mathrm{log}\left(x\right)[/latex] if and only if [latex]\text{ }{10}^{y}=x[/latex]. |

Definition of the natural logarithm | For [latex]\text{ }x>0[/latex], [latex]y=\mathrm{ln}\left(x\right)[/latex] if and only if [latex]\text{ }{e}^{y}=x[/latex]. |

## Key Concepts

- The inverse of an exponential function is a logarithmic function, and the inverse of a logarithmic function is an exponential function.
- Logarithmic equations can be written in an equivalent exponential form, using the definition of a logarithm.
- Exponential equations can be written in their equivalent logarithmic form using the definition of a logarithm.
- Logarithmic functions with base
*b*can be evaluated mentally using previous knowledge of powers of*b*. - Common logarithms can be evaluated mentally using previous knowledge of powers of 10.
- When common logarithms cannot be evaluated mentally, a calculator can be used.
- Real-world exponential problems with base 10 can be rewritten as a common logarithm and then evaluated using a calculator.
- Natural logarithms can be evaluated using a calculator.

## Glossary

**common logarithm**the exponent to which 10 must be raised to get

*x*; [latex]{\mathrm{log}}_{10}\left(x\right)[/latex] is written simply as [latex]\mathrm{log}\left(x\right)[/latex].

**logarithm**the exponent to which

*b*must be raised to get

*x*; written [latex]y={\mathrm{log}}_{b}\left(x\right)[/latex]

**natural logarithm**the exponent to which the number

*e*must be raised to get

*x*; [latex]{\mathrm{log}}_{e}\left(x\right)[/latex] is written as [latex]\mathrm{ln}\left(x\right)[/latex].

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