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The logarithmic expression

lo g_{b} a

is read the logarithm of

a

with base

b .

Here, the base

b

is clearly written in the expression. There are two cases in which the base does not need to be written, which will be discussed in this lesson.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Explore

Mathematical Operations

The

n^{th}

root of a real number

a

expresses another real number that, when multiplied by itself

n

times, will result in

a .

Aside from the radical symbol, the notation is made up of the radicand

a

and the index

n .

However, in the case of a square root, the index can be omitted. Therefore, if in the root there is no index written, it is implied that the index is

2 .

a = 2 a

Something similar happens with the exponent of a number. An exponent is a number or expression written above a number, indicating that it is being multiplied by itself. The notation is made up of a base — the number being multiplied — and an exponent, which gives the number of times the base appears in the multiplication.

The resulting number is commonly called a power. In this example, the base

7

is multiplied by itself

4

times.

7^{4} = 4 7 \cdot 7 \cdot 7 \cdot 7

However, if the exponent of a number or expression is

1,

it does not need to be written.

a = a^{1}

In the case of logarithms, there are two bases that do not need to be written. Take a look at a calculator and try to identify them.

Discussion

Common Logarithm

A common logarithm is a logarithm of base $10 .$ For example, $lo g_{10} 1000$ is called the common logarithm of $1000 .$ It is equal to $3$ because $1 0^{3}$ is $1000 .$

Since common logarithms are used so often, the base does not need to be written.

$lo g m \equiv lo g_{10} m$

In the identity above, $m$ is a positive number. Recalling the definition of a logarithm, the common logarithm of $m$ can be defined for positive values of $m .$

$lo g m = n \Leftrightarrow 1 0^{n} = m$

Common logarithms can be evaluated using a calculator. For example, to evaluate $lo g 34,$ push $LOG,$ enter $34,$ and then hit $ENTER .$

The common logarithm of $34$ is about $1.53 .$

Pop Quiz

Evaluating Common Logarithms

Use a calculator to evaluate the common logarithms. Round the answer to two decimal places.

Example

Earthquakes and Logarithms

A well-known logarithmic scale that is used to express the amount of energy released by an earthquake is the Richter scale.

On this scale, each integer number increase indicates an intensity ten times stronger than the previous whole number. For example, an earthquake of magnitude

3

on the Richter scale is

10

times stronger than an earthquake of magnitude

2 .

An earthquake of magnitude

4

10 \times 10

100

times stronger than an earthquake of magnitude

2 .

The intensity of an earthquake is determined by a seismograph.

If two earthquakes have intensity levels

I_{1}

and

I_{2},

and magnitudes on the Richter scale

M_{1}

and

M_{2},

the following formula holds true.

lo g \frac{I _{1}}{I _{2}} = M_{1} - M_{2}

1989,

the city of San Francisco was jolted by an earthquake of magnitude

7.1

on the Richter scale. In

1906,

nearly

80

years prior, the city had suffered a more powerful earthquake that is estimated to have measured magnitude

7.9

on the same scale. Vincenzo wants to know how many times more intense the

1906

earthquake was than to the

1989

earthquake.

Hint

Substitute $7.9$ and $7.1$ for $M_{1}$ and $M_{2},$ respectively, in the given formula. Then, use the definition of a common logarithm.

Solution

To calculate how many times more intense the

1906

earthquake was compared to the one in

1989,

the value of

\frac{I _{1}}{I _{2}}

needs to be found. To do so,

7.9

and

7.1

can be substituted for

M_{1}

and

M_{2},

respectively, in the given formula.

lo g \frac{I _{1}}{I _{2}} = M_{1} - M_{2}

SubstituteII

$M_{1} = 7.9$ , $M_{2} = 7.1$

lo g \frac{I _{1}}{I _{2}} = 7.9 - 7.1

SubTerm

Subtract term

lo g \frac{I _{1}}{I _{2}} = 0.8

This equation can be rewritten by following the definition of a common logarithm.

lo g \frac{I _{1}}{I _{2}} = 0.8 \Leftrightarrow \frac{I _{1}}{I _{2}} = 1 0^{0.8}

Finally, use a calculator to evaluate

1 0^{0.8} .

1 0^{0.8}

UseCalc

Use a calculator

6.309573 \dots

RoundDec

Round to $2$ decimal place(s)

6.31

The

1906

earthquake was about

6.31

times more intense than the

1989

earthquake.

Example

Using the Properties of Common Logarithms

After studying the relationship between earthquakes and logarithms, Vincenzo became more interested in this fascinating math topic.

Now he wants to pair the logarithmic expressions that involve common logarithms with their corresponding simplified expression or number. Help him do this!

Hint

Use the properties of logarithms and the definition of a common logarithm.

Solution

First, simplify the expressions on the left. They can then be paired with their corresponding expressions on the right.

Expression $1$

To simplify this expression, the Quotient Property of Logarithms and the definition of a common logarithm will be used.

lo g 10 x - lo g x

$lo g (m) - lo g (n) = lo g (\frac{m}{n})$

lo g \frac{1 0 x}{x}

CancelCommonFac

Cancel out common factors

lo g \frac{1 0 x}{x}

SimpQuot

Simplify quotient

lo g 10

$lo g (10) = 1$

1

The expression

lo g 10 x - lo g x

should be paired with

1 .

Expression $2$

To simplify this expression, the Power Property of Logarithms, the Product Property of Logarithms, and the definition of a common logarithm will be used.

2 lo g x + lo g 10 x

$m \cdot lo g (a) = lo g (a^{m})$

lo g x^{2} + lo g 10 x

$lo g (m n) = lo g (m) + lo g (n)$

lo g x^{2} + lo g 10 + lo g x

$lo g (10) = 1$

lo g x^{2} + 1 + lo g x

CommutativePropAdd

\CommutativePropAdd

1 + lo g x + lo g x^{2}

$lo g (m) + lo g (n) = lo g (m n)$

1 + lo g (x \cdot x^{2})

MultBasePow

$a \cdot a^{m} = a^{1 + m}$

1 + lo g x^{3}

The expression

2 lo g x + lo g 10 x

should be paired with

1 + lo g x^{3} .

Expression $3$

To simplify the third expression, the Product Property of Logarithms, the Quotient Property of Logarithms, a logarithm identity, and the definition of a common logarithm will be used.

lo g 0.1 x

$lo g (m n) = lo g (m) + lo g (n)$

lo g 0.1 + lo g x

NumberToFrac

$a = \frac{1 0 \cdot a}{1 0}$

lo g \frac{1 0 \cdot 0 . 1}{1 0} + lo g x

Multiply

lo g \frac{1}{1 0} + lo g x

$lo g (\frac{m}{n}) = lo g (m) - lo g (n)$

lo g 1 - lo g 10 + lo g x

$lo g (1) = 0$

0 - lo g 10 + lo g x

$lo g (10) = 1$

0 - 1 + lo g x

SubTerm

Subtract term

- 1 + lo g x

CommutativePropAdd

\CommutativePropAdd

lo g x - 1

The expression

lo g 0.1 x

should be paired with

lo g x - 1 .

Expression $4$

To simplify the second to last expression, the Quotient Property of Logarithms, the Power Property of Logarithms, and the definition of a common logarithm will be used.

lo g \frac{1 0 0}{x} + lo g x^{2}

$lo g (\frac{m}{n}) = lo g (m) - lo g (n)$

lo g 100 - lo g x + lo g x^{2}

WritePow

Write as a power

lo g 1 0^{2} - lo g x + lo g x^{2}

$lo g (a^{m}) = m \cdot lo g (a)$

2 lo g 10 - lo g x + 2 lo g x

$lo g (10) = 1$

2 (1) - lo g x + 2 lo g x

IdPropMult

\IdPropMult

2 - lo g x + 2 lo g x

AddTerms

Add terms

2 + lo g x

The expression

lo g \frac{1 0 0}{x} + lo g x^{2}

should be paired with

2 + lo g x .

Expression $5$

Finally, to simplify the last expression, the Product Property of Logarithms and the Power Property of Logarithms will be used.

lo g x + lo g x^{2} - 3 lo g x

$lo g (m) + lo g (n) = lo g (m n)$

lo g (x \cdot x^{2}) - 3 lo g x

MultBasePow

$a \cdot a^{m} = a^{1 + m}$

lo g x^{3} - 3 lo g x

$m \cdot lo g (a) = lo g (a^{m})$

lo g x^{3} - lo g x^{3}

SubTerm

Subtract term

0

The expression

lo g x + lo g x^{2} - 3 lo g x

should be paired with

0 .

Explore

Calculating the Value of an Expression for Bigger Values of $n$

Consider the following algebraic expression.

(1 + \frac{1}{n})^{n}

This expression can be evaluated for different values of

n .

Pay close attention to the value of the expression as

n

increases.

What conclusion can be made?

Discussion

The Natural Base

Logarithms and exponents are closely related. This part of the lesson will explore a particular type of exponential expression. Recall the formula for compound interest.

Here,

r

is the interest rate in decimal form. If the interest rate is

100 %,

which is extremely profitable, the value of

r

equals

1 .

A = P (1 + \frac{1}{n})^{n t}

The number

n

is the amount of times the interest is compounded each year. This is how often the accrued interest is added to the balance. The more often the interest is compounded, the higher the profit will be each year. What happens as

n

increases? To examine this, the formula will be rewritten using the the Power of a Power Property.

A = P (1 + \frac{1}{n})^{n t} ⇕ A = P ((1 + \frac{1}{n})^{n})^{t}

In the revised formula, the expression in the outer parentheses is a constant that depends only on

n .

To analyze what happens as

n

increases, larger and larger values will be substituted into this expression.

As the values of

n

increase higher and higher, the expression seems to approach a value of about

2.718 .

In fact, as

n

goes to infinity, the value of the expression approaches a constant that is called the natural base, or

e .

Concept

The Number $e$

The number $e$ — commonly called the natural base — is an irrational mathematical constant named by the mathematician Leonhard Euler.

$e = 2.7182818284 \dots$

Euler's number

e

appears in several areas of mathematics and has multiple uses. For example, it is often used as the base of exponential and logarithmic functions.

Discussion

The Natural Logarithm

A natural logarithm is a logarithm with base $e$ . Although it is correct to write $lo g_{e},$ the natural logarithm is more commonly written as $ln .$

$ln m = lo g_{e} m$

This means that $ln m$ equals the exponent to which $e$ must be raised to equal $m .$

$ln m = n \Leftrightarrow m = e^{n}$

Natural logarithms can be evaluated using a calculator. For example, to evaluate $ln 21,$ push $LN,$ input $21,$ and then hit $ENTER .$

The natural logarithm of

21

is about

3.04 .

Pop Quiz

Evaluating Natural Logarithms

Use a calculator to evaluate the natural logarithms. Round the answer to two decimal places.

Example

Using the Properties of Natural Logarithms

Vincenzo already knows how to use the properties of logarithms when dealing with common logarithms. Now he wants to use the properties when natural logarithms are involved.

Properties for common and natural logarithms

To do so, Vincenzo asked his math teacher for some extra practice. She asked him to identify the only expression that is not equivalent to the others. Help him do this!

Hint

Simplify all the expressions using the properties of logarithms.

Solution

Simplify the expressions so that the expression that is not equivalent with the others can be easily identified.

Expression $1$

To simplify this expression, the Product Property of Logarithms, the Quotient Property of Logarithms, and the Power Property of Logarithms will be used.

ln x^{2} + ln \frac{y}{x} + ln z

$ln (a^{b}) = b \cdot ln (a)$

2 ln x + ln \frac{y}{x} + ln z

$ln (\frac{a}{b}) = ln (a) - ln (b)$

2 ln x + ln y - ln x + ln z

SubTerm

Subtract term

ln x + ln y + ln z

$ln (a) + ln (b) = ln (a b)$

ln x y + ln z

$ln (a) + ln (b) = ln (a b)$

ln x y z

The expression

ln x^{2} + ln \frac{y}{x} + ln z

is equivalent to

ln x y z .

Expression $2$

To simplify the second expression, the Product Property of Logarithms, the Power Property of Logarithms, and the definition of a natural logarithm will be used.

ln e x + ln y^{2} z - (ln y + 1)

$ln (a b) = ln (a) + ln (b)$

ln e + ln x + ln y^{2} + ln z - (ln y + 1)

$ln (e) = 1$

1 + ln x + ln y^{2} + ln z - (ln y + 1)

$ln (a^{b}) = b \cdot ln (a)$

1 + ln x + 2 ln y + ln z - (ln y + 1)

Distr

Distribute $- 1$

1 + ln x + 2 ln y + ln z - ln y - 1

SubTerms

Subtract terms

ln x + ln y + ln z

$ln (a) + ln (b) = ln (a b)$

ln x y + ln z

$ln (a) + ln (b) = ln (a b)$

ln x y z

The expression

ln e x + ln y^{2} z - (ln y + 1)

is also equivalent to

ln x y z .

Expression $3$

To simplify the third expression, the Quotient Property of Logarithms and the definition of a natural logarithm will be used.

- ln \frac{e}{x y z} + 1

$ln (\frac{a}{b}) = ln (a) - ln (b)$

- (ln e - ln x y z) + 1

$ln (e) = 1$

- (1 - ln x y z) + 1

Distr

Distribute $- 1$

- 1 + ln x y z + 1

AddTerms

Add terms

ln x y z

The expression

- ln \frac{e}{x y z} + 1

is also equivalent to

ln x y z .

Expression $4$

To simplify the second to last expression, the Product Property of Logarithms and the Quotient Property of Logarithms will be used.

ln x y + ln z - ln y

$ln (a) + ln (b) = ln (a b)$

ln x y z - ln y

$ln (a) - ln (b) = ln (\frac{a}{b})$

ln \frac{x y z}{y}

CancelCommonFac

Cancel out common factors

ln \frac{x y z}{y}

SimpQuot

Simplify quotient

ln x z

The expression

ln x y + ln z - ln y

is equivalent to

ln x z .

Therefore, this expression is not equivalent to the previous expressions.

Expression $5$

The last expression will be simplified by using the Power Property of Logarithms and the Product Property of Logarithms.

2 ln x y + ln z^{2} - ln z

SqrtToPowD

$a = a^{\frac{1}{2}}$

2 ln (x y)^{\frac{1}{2}} + ln z^{2} - ln z

$ln (a^{b}) = b \cdot ln (a)$

2 (\frac{1}{2}) ln x y + 2 ln z - ln z

DenomMultFracToNumber

$2 \cdot \frac{a}{2} = a$

1 ln x y + 2 ln z - ln z

IdPropMult

\IdPropMult

ln x y + 2 ln z - ln z

SubTerm

Subtract term

ln x y + ln z

$ln (a) + ln (b) = ln (a b)$

ln x y z

The last expression,

2 ln x y + ln z^{2} - ln z,

is also equivalent to

ln x y z .

Therefore, the expression that is not equivalent to the others is the fourth expression.

Discussion

Change of Base Formula

Most calculators only calculate common and natural logarithms. These are logarithms with base $10$ or $e .$ Luckily, there is a formula that allows any logarithm to be written in terms of common or natural logarithms.

Rule

Change of Base Formula

A logarithm of arbitrary base can be rewritten as the quotient of two logarithms with the same base by using the change of base formula.

$lo g_{c} a = \frac{lo g _{b} a}{lo g _{b} c}$

This rule is valid for positive values of $a, b,$ and $c,$ where $b$ and $c$ are different than $1 .$

Proof

Let

x = lo g_{c} a .

Therefore, by the definition of a logarithm, it is known that

a = c^{x} .

x = lo g_{c} a \Leftrightarrow a = c^{x}

By the Reflexive Property of Equality,

lo g_{b} a

is equal to itself.

lo g_{b} a = lo g_{b} a

Substitute

$a = c^{x}$

lo g_{b} a = lo g_{b} c^{x}

$lo g_{b} (a^{m}) = m \cdot lo g_{b} (a)$

lo g_{b} a = x lo g_{b} c

Substitute

$x = l o g_{c} a$

lo g_{b} a = l o g_{c} a lo g_{b} c

DivEqn

$LHS / lo g_{b} c = RHS / lo g_{b} c$

\frac{lo g _{b} a}{lo g _{b} c} = lo g_{c} a

RearrangeEqn

Rearrange equation

lo g_{c} a = \frac{lo g _{b} a}{lo g _{b} c} ✓

Note that this formula is helpful to calculate any logarithm using a calculator, since the new base can be any positive number different than $1 .$ This means that the new base can be $10$ or $e .$

lo g_{c} a = \frac{lo g a}{lo g c} lo g_{c} a = \frac{ln a}{ln c}

Example

Using the Change of Base Formula

Vincenzo has discovered that he can use the Change of Base Formula to calculate the value of any logarithm.

Vincenzo at the computer with a thought bubble where the Change of Base Formula is written

Help him answer the following questions!

a Use common logarithms and a calculator to find the value of

lo g_{4} 23 .

Round the answer to two decimal places.

b Use natural logarithms and a calculator to find the value of

lo g_{7} 56 .

Round the answer to two decimal places.

Hint

a By the Change of Base Formula,

lo g_{4} 23 = \frac{l o g 2 3}{l o g 4} .

b By the Change of Base Formula,

lo g_{7} 56 = \frac{l n 5 6}{l n 7} .

Solution

a Recall the Change of Base Formula.

lo g_{b} a = \frac{lo g _{c} a}{lo g _{c} b}

Here,

a,

b,

and

c

are positive numbers, with

b

and

c

different than

1 .

While the base

c

can be any number, in this case it might be most helpful if it was

10 .

Then the logarithm of

23

with base

4

could be expressed in terms of common logarithms.

lo g_{4} 23 = \frac{lo g 2 3}{lo g 4}

The value of the numeric expression on the right-hand side of the above equation can be found by using a calculator.

Therefore, the value of $lo g_{4} 23$ rounded to the nearest hundredth is $2.26 .$

b Once again, recall the Change of Base Formula, this time with natural logarithms.

ln_{b} a = \frac{lo g _{c} a}{lo g _{c} b}

Here,

a,

b,

and

c

are positive numbers, with

b

and

c

different than

1 .

For this exercise, let

c

e .

Then the logarithm of

56

with base

7

can be expressed in terms of natural logarithms.

lo g_{7} 56 = \frac{ln 5 6}{ln 7}

The value of the numeric expression on the right-hand side of the above equation can be found by using a calculator.

Therefore, the value of $lo g_{7} 56$ rounded to the nearest hundredth is $2.07 .$

Closure

Leonhard Euler and His Contributions

In this lesson, the Swiss mathematician Leonhard Euler was mentioned. Euler was also a physicist, astronomer, geographer, logician, and engineer. During his life, Euler came up with principles that set the foundations for most of the mathematics used nowadays. He was a revolutionary thinker in the fields of geometry, calculus, trigonometry, differential equations, and number theory.

Among other important contributions, he introduced most of the mathematical notations that are used today. Euler was the first person to use the letter

e

to denote the base of a natural logarithm. Furthermore, although he was not the first to use it, Euler popularized the use of the Greek letter

π

to indicate the ratio of the circumference of a circle to its diameter.

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Catch-Up and Review

Hint

Solution

Hint

Solution

Expression 1

Expression 2

Expression 3

Expression 4

Expression 5

Hint

Solution

Expression 1

Expression 2

Expression 3

Expression 4

Expression 5

Proof

Hint

Solution

Expression $1$

Expression $2$

Expression $3$

Expression $4$

Expression $5$

Expression $1$

Expression $2$

Expression $3$

Expression $4$

Expression $5$