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Logarithmic Functions

Convert from logarithmic to exponential form.

Logarithmic Functions

Convert between exponential and logarithmic form. (IA 10.3.1)
Evaluate logarithmic functions. (IA 10.3.2)

Try it.

Graph the exponential function \(f(x)={2}^{x}\) by making a table.

\(x\)	\(y=f(x)\)

Is it one-to-one?
Domain?
Range?
Graph the inverse of \(f(x)={2}^{x}\) on the grid above by interchanging x and y coordinates in the table.
\(x\) \(y=f(x)\)
Is the inverse one-to-one function?
Domain?
Range?

Example

Try it.

Find the inverse of \(f(x)={2}^{x}\)

Solution

Rewrite with \(y=f(x)\)	\(y={2}^{x}\)
Interchange the variables \(x\) and \(y\) .	\(x={2}^{y}\)
Solve for \(y\) .	Oops! We have no way to solve for \(y\) .

We give y a new notation:
\(y={\log }_{x}2\)
" \(y={\log }_{x}2\) " read "the logarithm, base 2 of x", means "the power to which we raise 2 to get x". The function \(y={\log }_{a}x\) is equivalent to \({a}^{y}=x\) is the logarithmic function with base \(a\), where \(a>0\), \(x>0\)

Since the equations \(y={\text{log}}_{a}x\) and \(x={a}^{y}\) are equivalent, we can go back and forth between them. This will often be the method to solve some exponential and logarithmic equations. To help with converting back and forth, let’s take a close look at the equations. Notice the positions of the exponent and base.

If we remember the logarithm is the exponent, it makes the conversion easier. You may want to repeat, “base to the exponent gives us the number.”

Example

Convert between exponential and logarithmic form.

Try it.

ⓐ Convert to logarithmic form: \({2}^{3}=8\)

Solution

Identify the base and the exponent: the base is 2 and the exponent is 3.
Then we have \(3={\text{log}}_{2}8\) .

Try it.

ⓑ Convert to exponential form: \({\text{log}}_{b}m=a\)

Solution

Identify the base and the exponent: the base is b and the exponent is a.
Then we have \({b}^{a}=m\) .

Convert between exponential and logarithmic form.

Remember these logarithmic notations to help complete the following:
Common Logarithm \(\text{log}x={\text{log}}_{x}10\)
Natural Logarithm \(\text{ln}x={\text{log}}_{x}e\)

Try it.

Convert to logarithmic form.

ⓐ \(8={2}^{x}\)
ⓑ \({10}^{-2}=0.01\)
ⓒ \({e}^{x}=40\)

Try it.

Convert to exponential form.

ⓐ \({\log }_{81}3=4\)
ⓑ \(\ln 1=0\)
ⓒ \(\log 10000=4\)

Condensed — the full section is in OpenStax Precalculus 2e.

Converting from Logarithmic to 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 \({10}^{x}=500,\) 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 \({10}^{x}=500.\) We know that \({10}^{2}=100\) and \({10}^{3}=1000,\) so it is clear that \(x\) must be some value between 2 and 3, since \(y={10}^{x}\) is increasing. We can examine a graph, as in , 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 in passes the horizontal line test. The exponential function \(y={b}^{x}\) is one-to-one, so its inverse, \(x={b}^{y}\) 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 \(y={\log }_{b}(x).\) 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 \({2}^{5}=32,\) we can write \({\log }_{2}32=5.\) 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:

\[{\log }_{b}(x)=y⇔{b}^{y}=x,\ b>0,b\ne 1\]

Note that the base \(b\) is always positive.

Because logarithm is a function, it is most correctly written as \({\log }_{b}(x),\) using parentheses to denote function evaluation, just as we would with \(f(x).\) 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 \({\log }_{b}x.\) Note that many calculators require parentheses around the \(x.\)

Condensed — the full section is in OpenStax Precalculus 2e.

Converting 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 \(x={\log }_{b}(y).\)

Example

Try it.

Write the following exponential equations in logarithmic form.

\({2}^{3}=8\)
\({5}^{2}=25\)
\({10}^{-4}=\frac{1}{10,000}\)

Solution

First, identify the values of \(b,y,\text{and}x.\) Then, write the equation in the form \(x={\log }_{b}(y).\)

\({2}^{3}=8\)
Here, \(b=2,\) \(x=3,\) and \(y=8.\) Therefore, the equation \({2}^{3}=8\) is equivalent to \({\log }_{2}(8)=3.\)
\({5}^{2}=25\)
Here, \(b=5,\) \(x=2,\) and \(y=25.\) Therefore, the equation \({5}^{2}=25\) is equivalent to \({\log }_{5}(25)=2.\)
\({10}^{-4}=\frac{1}{10,000}\)
Here, \(b=10,\) \(x=-4,\) and \(y=\frac{1}{10,000}.\) Therefore, the equation \({10}^{-4}=\frac{1}{10,000}\) is equivalent to \({\text{log}}_{10}(\frac{1}{10,000})=-4.\)

Evaluating Logarithms

Knowing the squares, cubes, and roots of numbers allows us to evaluate many logarithms mentally. For example, consider \({\log }_{2}8.\) We ask, “To what exponent must \(2\) be raised in order to get 8?” Because we already know \({2}^{3}=8,\) it follows that \({\log }_{2}8=3.\)

Now consider solving \({\log }_{7}49\) and \({\log }_{3}27\) mentally.

We ask, “To what exponent must 7 be raised in order to get 49?” We know \({7}^{2}=49.\) Therefore, \({\log }_{7}49=2\)
We ask, “To what exponent must 3 be raised in order to get 27?” We know \({3}^{3}=27.\) Therefore, \({\log }_{3}27=3\)

Even some seemingly more complicated logarithms can be evaluated without a calculator. For example, let’s evaluate \({\log }_{\frac{2}{3}}\frac{4}{9}\) mentally.

We ask, “To what exponent must \(\frac{2}{3}\) be raised in order to get \(\frac{4}{9}?\) ” We know \({2}^{2}=4\) and \({3}^{2}=9,\) so \({(\frac{2}{3})}^{2}=\frac{4}{9}.\) Therefore, \({\log }_{\frac{2}{3}}(\frac{4}{9})=2.\)

Example

Try it.

Solve \(y={\log }_{4}(64)\) without using a calculator.

Solution

First we rewrite the logarithm in exponential form: \({4}^{y}=64.\) Next, we ask, “To what exponent must 4 be raised in order to get 64?”

We know

\[{4}^{3}=64\]

Therefore,

\[\log (64)4=3\]

Example

Try it.

Evaluate \(y={\log }_{3}(\frac{1}{27})\) without using a calculator.

Solution

First we rewrite the logarithm in exponential form: \({3}^{y}=\frac{1}{27}.\) Next, we ask, “To what exponent must 3 be raised in order to get \(\frac{1}{27}?\) ”

We know \({3}^{3}=27,\) but what must we do to get the reciprocal, \(\frac{1}{27}?\) Recall from working with exponents that \({b}^{-a}=\frac{1}{{b}^{a}}.\) We use this information to write

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

Therefore, \({\log }_{3}(\frac{1}{27})=-3.\)

Using Common Logarithms

Sometimes you may see a logarithm written without a base. When you see one written this way, you need to look at the expression before evaluating it. It may be that the base you use doesn't matter. If you find it in computer science, it often means \({\log }_{2}(x)\). However, in mathematics it almost always means the common logarithm of 10. In other words, the expression \(\log (x)\) often means \({\log }_{10}(x).\)

Currently, we use \({\log }_{b}(x),\text{lg}(x)\) as the common logarithm, \(\text{lb}(x)\) as the binary logarithm, and \(\ln (x)\) as the natural logarithm. Writing \(\text{lg}(x)\) without specifying a base is now considered bad form, despite being frequently found in older materials.

Example

Try it.

Evaluate \(y=\log (1000)\) without using a calculator.

Solution

First we rewrite the logarithm in exponential form: \({10}^{y}=1000.\) Next, we ask, “To what exponent must \(10\) be raised in order to get 1000?” We know

\[{10}^{3}=1000\]

Therefore, \(\log (1000)=3.\)

Example

Try it.

Evaluate \(y=\log (321)\) to four decimal places using a calculator.

Solution

Press [LOG].
Enter 321, followed by [ ) ].
Press [ENTER].

Rounding to four decimal places, \(\log (321)\approx 2.5065.\)

Condensed — the full section is in OpenStax Precalculus 2e.

Using Natural Logarithms

The most frequently used base for logarithms is \(e,\) the value of which is approximately \(2.71828\). Base \(e\) logarithms are important in calculus and some scientific applications; they are called natural logarithms. The base \(e\) logarithm, \({\log }_{e}(x),\) has its own notation, \(\ln (x).\)

Most values of \(\ln (x)\) can be found only using a calculator. The major exception is that, because the logarithm of 1 is always 0 in any base, \(\ln 1=0.\) For other natural logarithms, we can use the \(\ln\) key that can be found on most scientific calculators. We can also find the natural logarithm of any power of \(e\) using the inverse property of logarithms.

Example

Try it.

Evaluate \(y=\ln (500)\) to four decimal places using a calculator.

Solution

Press [LN].
Enter \(500,\) followed by [ ) ].
Press [ENTER].

Rounding to four decimal places, \(\ln (500)\approx 6.2146\)

Key Equations

Definition of the logarithmic function	For \(\ x>0,b>0,b\ne 1,\) \(y={\log }_{b}(x)\) if and only if \(\ {b}^{y}=x.\)
Definition of the common logarithm	For \(\ x>0,\) \(y=\log (x)\) if and only if \(\ {10}^{y}=x.\)
Definition of the natural logarithm	For \(\ x>0,\) \(y=\ln (x)\) if and only if \(\ {e}^{y}=x.\)

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

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Graph the exponential function \(f(x)={2}^{x}\) by making a table.
\(x\) \(y=f(x)\)
1. Is it one-to-one?
2. Domain?
3. Range?
4. Graph the inverse of \(f(x)={2}^{x}\) on the grid above by interchanging x and y coordinates in the table.
  \(x\) \(y=f(x)\)
5. Is the inverse one-to-one function?
6. Domain?
7. Range?
Find the inverse of \(f(x)={2}^{x}\)

Fi àwọn àgbèwọlé hàn

Rewrite with \(y=f(x)\) \(y={2}^{x}\)
Interchange the variables \(x\) and \(y\) . \(x={2}^{y}\)
Solve for \(y\) . Oops! We have no way to solve for \(y\) .
We give y a new notation:
\(y={\log }_{x}2\)
" \(y={\log }_{x}2\) " read "the logarithm, base 2 of x", means "the power to which we raise 2 to get x". The function \(y={\log }_{a}x\) is equivalent to \({a}^{y}=x\) is the logarithmic function with base \(a\), where \(a>0\), \(x>0\)
ⓐ Convert to logarithmic form: \({2}^{3}=8\)

Fi àwọn àgbèwọlé hàn

Identify the base and the exponent: the base is 2 and the exponent is 3.
Then we have \(3={\text{log}}_{2}8\) .
ⓑ Convert to exponential form: \({\text{log}}_{b}m=a\)

Fi àwọn àgbèwọlé hàn

Identify the base and the exponent: the base is b and the exponent is a.
Then we have \({b}^{a}=m\) .

Find the value of x: ⓐ \({\text{log}}_{x}36=2,\) ⓑ \({\text{log}}_{4}x=3,\) and ⓒ \({\text{log}}_{\frac{1}{2}}\frac{1}{8}=x.\)

Fi àwọn àgbèwọlé hàn

ⓐ

	\(\ {\text{log}}_{x}36\ =\ 2\)
Convert to exponential form.	\(\ {x}^{2}\ =\ 36\)
Solve the quadratic.	\(\ x=6,\ x=-6\)
The base of a logarithmic function must be positive, so we eliminate \(x=-6\) .	\(\ x\ =\ 6\ \text{Therefore,}\ {\text{log}}_{6}36=2.\)

ⓑ

	\(\ {\text{log}}_{4}x\ =\ 3\)
Convert to exponential form.	\(\ {4}^{3}\ =\ x\)
Simplify.	\(\ x\ =\ 64\ \text{Therefore,}\ {\text{log}}_{4}64\ =\ 3.\)

ⓒ

	\({\text{log}}_{\frac{1}{2}}\frac{1}{8}\ =\ x\)
Convert to exponential form.	\(\ {(\frac{1}{2})}^{x}\ =\ \frac{1}{8}\)
Rewrite \(\frac{1}{8}\) as \({(\frac{1}{2})}^{3}\) .	\(\ {(\frac{1}{2})}^{x}\ =\ {(\frac{1}{2})}^{3}\)
With the same base, the exponents must be equal.	\(\ x\ =\ 3\ \text{Therefore,}\ {\text{log}}_{\frac{1}{2}}\frac{1}{8}=3\)

Evaluate each of the following.
1. ⓐ \({\log }_{100}10\)
2. ⓑ \(\log 0.1\)
3. ⓒ \({\log }_{2}4\)
4. ⓓ \({\log }_{1}20\)
5. ⓔ \({\log }_{4}4\)
6. ⓕ \({\log }_{\frac{1}{9}}3\)
7. ⓖ \({\log }_{\sqrt{2}}2\)
8. ⓗ \(\ln {e}^{-5}\)
Write the following logarithmic equations in exponential form.
1. ⓐ \({\log }_{6}(\sqrt{6})=\frac{1}{2}\)
2. ⓑ \({\log }_{3}(9)=2\)
Fi àwọn àgbèwọlé hàn
First, identify the values of \(b,y,\text{and }x.\) Then, write the equation in the form \({b}^{y}=x.\)
- ⓐ \({\log }_{6}(\sqrt{6})=\frac{1}{2}\)
  Here, \(b=6,y=\frac{1}{2},\text{and}x=\sqrt{6.}\) Therefore, the equation \({\log }_{6}(\sqrt{6})=\frac{1}{2}\) is equivalent to \[{6}^{\frac{1}{2}}=\sqrt{6}.\]
- ⓑ \({\log }_{3}(9)=2\)
  Here, \(b=3,y=2,\text{and}x=9.\) Therefore, the equation \({\log }_{3}(9)=2\) is equivalent to \({3}^{2}=9.\)
Write the following logarithmic equations in exponential form.
1. ⓐ\({\log }_{10}(1,000,000)=6\)
2. ⓑ\({\log }_{5}(25)=2\)
Fi àwọn àgbèwọlé hàn
1. ⓐ\({\log }_{10}(1,000,000)=6\) is equivalent to \({10}^{6}=1,000,000\)
2. ⓑ\({\log }_{5}(25)=2\) is equivalent to \({5}^{2}=25\)
Write the following exponential equations in logarithmic form.
1. \({2}^{3}=8\)
2. \({5}^{2}=25\)
3. \({10}^{-4}=\frac{1}{10,000}\)
Fi àwọn àgbèwọlé hàn
First, identify the values of \(b,y,\text{and}x.\) Then, write the equation in the form \(x={\log }_{b}(y).\)
1. \({2}^{3}=8\)
  Here, \(b=2,\) \(x=3,\) and \(y=8.\) Therefore, the equation \({2}^{3}=8\) is equivalent to \({\log }_{2}(8)=3.\)
2. \({5}^{2}=25\)
  Here, \(b=5,\) \(x=2,\) and \(y=25.\) Therefore, the equation \({5}^{2}=25\) is equivalent to \({\log }_{5}(25)=2.\)
3. \({10}^{-4}=\frac{1}{10,000}\)
  Here, \(b=10,\) \(x=-4,\) and \(y=\frac{1}{10,000}.\) Therefore, the equation \({10}^{-4}=\frac{1}{10,000}\) is equivalent to \({\text{log}}_{10}(\frac{1}{10,000})=-4.\)
Write the following exponential equations in logarithmic form.
1. ⓐ \({3}^{2}=9\)
2. ⓑ \({5}^{3}=125\)
3. ⓒ \({2}^{-1}=\frac{1}{2}\)
Fi àwọn àgbèwọlé hàn
1. ⓐ \({3}^{2}=9\) is equivalent to \({\log }_{3}(9)=2\)
2. ⓑ \({5}^{3}=125\) is equivalent to \({\log }_{5}(125)=3\)
3. ⓒ \({2}^{-1}=\frac{1}{2}\) is equivalent to \({\text{log}}_{2}(\frac{1}{2})=-1\)
Solve \(y={\log }_{4}(64)\) without using a calculator.

Fi àwọn àgbèwọlé hàn

First we rewrite the logarithm in exponential form: \({4}^{y}=64.\) Next, we ask, “To what exponent must 4 be raised in order to get 64?”
We know
\[{4}^{3}=64\]
Therefore,
\[\log (64)4=3\]
Solve \(y={\log }_{121}(11)\) without using a calculator.

Fi àwọn àgbèwọlé hàn

\({\log }_{121}(11)=\frac{1}{2}\) (recalling that \(\sqrt{121}={(121)}^{\frac{1}{2}}=11\) )
Evaluate \(y={\log }_{3}(\frac{1}{27})\) without using a calculator.

Fi àwọn àgbèwọlé hàn

First we rewrite the logarithm in exponential form: \({3}^{y}=\frac{1}{27}.\) Next, we ask, “To what exponent must 3 be raised in order to get \(\frac{1}{27}?\) ”
We know \({3}^{3}=27,\) but what must we do to get the reciprocal, \(\frac{1}{27}?\) Recall from working with exponents that \({b}^{-a}=\frac{1}{{b}^{a}}.\) We use this information to write
\[\begin{array}{l}{3}^{-3}=\frac{1}{{3}^{3}} \\ =\frac{1}{27}\end{array}\]
Therefore, \({\log }_{3}(\frac{1}{27})=-3.\)
Evaluate \(y={\log }_{2}(\frac{1}{32})\) without using a calculator.

Fi àwọn àgbèwọlé hàn

\({\log }_{2}(\frac{1}{32})=-5\)
Evaluate \(y=\log (1000)\) without using a calculator.

Fi àwọn àgbèwọlé hàn

First we rewrite the logarithm in exponential form: \({10}^{y}=1000.\) Next, we ask, “To what exponent must \(10\) be raised in order to get 1000?” We know
\[{10}^{3}=1000\]
Therefore, \(\log (1000)=3.\)
Evaluate \(y=\log (1,000,000).\)

Fi àwọn àgbèwọlé hàn

\(\log (1,000,000)=6\)
Evaluate \(y=\log (321)\) to four decimal places using a calculator.
Fi àwọn àgbèwọlé hàn
- Press [LOG].
- Enter 321, followed by [ ) ].
- Press [ENTER].
Rounding to four decimal places, \(\log (321)\approx 2.5065.\)
Evaluate \(y=\log (123)\) to four decimal places using a calculator.

Fi àwọn àgbèwọlé hàn

\(\log (123)\approx 2.0899\)
The amount of energy released from one earthquake was 500 times greater than the amount of energy released from another. The equation \({10}^{x}=500\) represents this situation, where \(x\) is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?
Fi àwọn àgbèwọlé hàn
We begin by rewriting the exponential equation in logarithmic form.
\[\begin{array}{lll}{10}^{x} & =500 & \\ \log (500) & =x & \text{Use the definition of the common log}\text{.}\end{array}\]
Next we evaluate the logarithm using a calculator:
- Press [LOG].
- Enter \(500,\) followed by [ ) ].
- Press [ENTER].
- To the nearest thousandth, \(\log (500)\approx 2.699.\)
The difference in magnitudes was about \(2.699.\)
The amount of energy released from one earthquake was \(\text{8,500}\) times greater than the amount of energy released from another. The equation \({10}^{x}=8500\) represents this situation, where \(x\) is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?

Fi àwọn àgbèwọlé hàn

The difference in magnitudes was about \(3.929.\)
Evaluate \(y=\ln (500)\) to four decimal places using a calculator.
Fi àwọn àgbèwọlé hàn
- Press [LN].
- Enter \(500,\) followed by [ ) ].
- Press [ENTER].
Rounding to four decimal places, \(\ln (500)\approx 6.2146\)
Evaluate \(\ln (-500).\)

Fi àwọn àgbèwọlé hàn

It is not possible to take the logarithm of a negative number in the set of real numbers.
What is a base \(b\) logarithm? Discuss the meaning by interpreting each part of the equivalent equations \({b}^{y}=x\) and \({\log }_{b}x=y\) for \(b>0,b\ne 1.\)

Fi àwọn àgbèwọlé hàn

A logarithm is an exponent. Specifically, it is the exponent to which a base \(b\) is raised to produce a given value. In the expressions given, the base \(b\) has the same value. The exponent, \(y,\) in the expression \({b}^{y}\) can also be written as the logarithm, \({\log }_{b}x,\) and the value of \(x\) is the result of raising \(b\) to the power of \(y.\)
How is the logarithmic function \(f(x)={\log }_{b}x\) related to the exponential function \(g(x)={b}^{x}?\) What is the result of composing these two functions?
How can the logarithmic equation \({\log }_{b}x=y\) be solved for \(x\) using the properties of exponents?

Fi àwọn àgbèwọlé hàn

Since the equation of a logarithm is equivalent to an exponential equation, the logarithm can be converted to the exponential equation \({b}^{y}=x,\) and then properties of exponents can be applied to solve for \(x.\)
Discuss the meaning of the common logarithm. What is its relationship to a logarithm with base \(b,\) and how does the notation differ?
Discuss the meaning of the natural logarithm. What is its relationship to a logarithm with base \(b,\) and how does the notation differ?

Fi àwọn àgbèwọlé hàn

The natural logarithm is a special case of the logarithm with base \(b\) in that the natural log always has base \(e.\) Rather than notating the natural logarithm as \({\log }_{e}(x),\) the notation used is \(\ln (x).\)
\({\text{log}}_{4}(q)=m\)
\({\text{log}}_{a}(b)=c\)

Fi àwọn àgbèwọlé hàn

\({a}^{c}=b\)
\({\log }_{16}(y)=x\)
\({\log }_{x}(64)=y\)

Fi àwọn àgbèwọlé hàn

\({x}^{y}=64\)
\({\log }_{y}(x)=-11\)
\({\log }_{15}(a)=b\)

Fi àwọn àgbèwọlé hàn

\({15}^{b}=a\)
\({\log }_{y}(137)=x\)
\({\log }_{13}(142)=a\)

Fi àwọn àgbèwọlé hàn

\({13}^{a}=142\)
\(\text{log}(v)=t\)
\(\text{ln}(w)=n\)

Fi àwọn àgbèwọlé hàn

\({e}^{n}=w\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\approx

approximately equal
Equal to the precision shown, not exactly.

\neq

not equal
The two sides are different.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

imaginary unit
i² = −1.

\theta

theta
The usual name for an angle.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

How to: Logarithmic Functions

Convert from logarithmic to exponential form.
Convert from exponential to logarithmic form.
Evaluate logarithms.
Use common logarithms.
Use natural logarithms.
Convert between exponential and logarithmic form. (IA 10.3.1)
Evaluate logarithmic functions. (IA 10.3.2)
Is it one-to-one?

Questions people ask

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Wárá

Parts of this page are adapted from OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Diẹ̀ nínú Precalculus

Complex numbers Polynomial functions Rational functions Sequences and series The binomial theorem Conic sections Vectors Exponential and logarithmic functions Polynomial division and the remainder theorem Parametric equations and polar coordinates