properties of logarithms assignment

10.4 Use the Properties of Logarithms

Learning objectives.

• Use the properties of logarithms
• Use the Change of Base Formula

Be Prepared 10.4

Before you get started, take this readiness quiz.

• Evaluate: ⓐ a 0 a 0 ⓑ a 1 . a 1 . If you missed this problem, review Example 5.14 .
• Write with a rational exponent: x 2 y 3 . x 2 y 3 . If you missed this problem, review Example 8.27 .
• Round to three decimal places: 2.5646415. If you missed this problem, review Example 1.34 .

Use the Properties of Logarithms

Now that we have learned about exponential and logarithmic functions, we can introduce some of the properties of logarithms. These will be very helpful as we continue to solve both exponential and logarithmic equations.

The first two properties derive from the definition of logarithms. Since a 0 = 1 , a 0 = 1 , we can convert this to logarithmic form and get log a 1 = 0 . log a 1 = 0 . Also, since a 1 = a , a 1 = a , we get log a a = 1 . log a a = 1 .

Properties of Logarithms

In the next example we could evaluate the logarithm by converting to exponential form, as we have done previously, but recognizing and then applying the properties saves time.

Example 10.28

Evaluate using the properties of logarithms: ⓐ log 8 1 log 8 1 and ⓑ log 6 6 . log 6 6 .

ⓐ log 8 1 Use the property, log a 1 = 0 . 0 log 8 1 = 0 log 8 1 Use the property, log a 1 = 0 . 0 log 8 1 = 0

ⓑ log 6 6 Use the property, log a a = 1 . 1 log 6 6 = 1 log 6 6 Use the property, log a a = 1 . 1 log 6 6 = 1

Try It 10.55

Evaluate using the properties of logarithms: ⓐ log 13 1 log 13 1 ⓑ log 9 9 . log 9 9 .

Try It 10.56

Evaluate using the properties of logarithms: ⓐ log 5 1 log 5 1 ⓑ log 7 7 . log 7 7 .

The next two properties can also be verified by converting them from exponential form to logarithmic form, or the reverse.

The exponential equation a log a x = x a log a x = x converts to the logarithmic equation log a x = log a x , log a x = log a x , which is a true statement for positive values for x only.

The logarithmic equation log a a x = x log a a x = x converts to the exponential equation a x = a x , a x = a x , which is also a true statement.

These two properties are called inverse properties because, when we have the same base, raising to a power “undoes” the log and taking the log “undoes” raising to a power. These two properties show the composition of functions. Both ended up with the identity function which shows again that the exponential and logarithmic functions are inverse functions.

Inverse Properties of Logarithms

For a > 0 , a > 0 , x > 0 x > 0 and a ≠ 1 , a ≠ 1 ,

In the next example, apply the inverse properties of logarithms.

Example 10.29

Evaluate using the properties of logarithms: ⓐ 4 log 4 9 4 log 4 9 and ⓑ log 3 3 5 . log 3 3 5 .

ⓐ 4 log 4 9 Use the property, a log a x = x . 9 4 log 4 9 = 9 4 log 4 9 Use the property, a log a x = x . 9 4 log 4 9 = 9

ⓑ log 3 3 5 Use the property, a log a x = x . 5 log 3 3 5 = 5 log 3 3 5 Use the property, a log a x = x . 5 log 3 3 5 = 5

Try It 10.57

Evaluate using the properties of logarithms: ⓐ 5 log 5 15 5 log 5 15 ⓑ log 7 7 4 . log 7 7 4 .

Try It 10.58

Evaluate using the properties of logarithms: ⓐ 2 log 2 8 2 log 2 8 ⓑ log 2 2 15 . log 2 2 15 .

There are three more properties of logarithms that will be useful in our work. We know exponential functions and logarithmic function are very interrelated. Our definition of logarithm shows us that a logarithm is the exponent of the equivalent exponential. The properties of exponents have related properties for exponents.

In the Product Property of Exponents, a m · a n = a m + n , a m · a n = a m + n , we see that to multiply the same base, we add the exponents. The Product Property of Logarithms , log a M · N = log a M + log a N log a M · N = log a M + log a N tells us to take the log of a product, we add the log of the factors.

Product Property of Logarithms

If M > 0 , N > 0 , a > 0 M > 0 , N > 0 , a > 0 and a ≠ 1 , a ≠ 1 , then,

The logarithm of a product is the sum of the logarithms.

We use this property to write the log of a product as a sum of the logs of each factor.

Example 10.30

Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible: ⓐ log 3 7 x log 3 7 x and ⓑ log 4 64 x y . log 4 64 x y .

ⓐ log 3 7 x Use the Product Property, log a ( M · N ) = log a M + log a N . log 3 7 + log 3 x log 3 7 x = log 3 7 + log 3 x log 3 7 x Use the Product Property, log a ( M · N ) = log a M + log a N . log 3 7 + log 3 x log 3 7 x = log 3 7 + log 3 x

ⓑ log 4 64 x y Use the Product Property, log a ( M · N ) = log a M + log a N . log 4 64 + log 4 x + log 4 y Simplify by evaluating log 4 64 . 3 + log 4 x + log 4 y log 4 64 x y = 3 + log 4 x + log 4 y log 4 64 x y Use the Product Property, log a ( M · N ) = log a M + log a N . log 4 64 + log 4 x + log 4 y Simplify by evaluating log 4 64 . 3 + log 4 x + log 4 y log 4 64 x y = 3 + log 4 x + log 4 y

Try It 10.59

Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify, if possible.

ⓐ log 3 3 x log 3 3 x ⓑ log 2 8 x y log 2 8 x y

Try It 10.60

ⓐ log 9 9 x log 9 9 x ⓑ log 3 27 x y log 3 27 x y

Similarly, in the Quotient Property of Exponents, a m a n = a m − n , a m a n = a m − n , we see that to divide the same base, we subtract the exponents. The Quotient Property of Logarithms , log a M N = log a M − log a N log a M N = log a M − log a N tells us to take the log of a quotient, we subtract the log of the numerator and denominator.

Quotient Property of Logarithms

The logarithm of a quotient is the difference of the logarithms.

Note that log a M − log a N ≠ log a ( M − N ) . log a M − log a N ≠ log a ( M − N ) .

We use this property to write the log of a quotient as a difference of the logs of each factor.

Example 10.31

Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible. ⓐ log 5 5 7 log 5 5 7 and ⓑ log x 100 log x 100

ⓐ log 5 5 7 Use the Quotient Property, log a M N = log a M − log a N . log 5 5 − log 5 7 Simplify. 1 − log 5 7 log 5 5 7 = 1 − log 5 7 log 5 5 7 Use the Quotient Property, log a M N = log a M − log a N . log 5 5 − log 5 7 Simplify. 1 − log 5 7 log 5 5 7 = 1 − log 5 7

ⓑ log x 100 Use the Quotient Property, log a M N = log a M − log a N . log x − log 100 Simplify. log x − 2 log x 100 = log x − 2 log x 100 Use the Quotient Property, log a M N = log a M − log a N . log x − log 100 Simplify. log x − 2 log x 100 = log x − 2

Try It 10.61

Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.

ⓐ log 4 3 4 log 4 3 4 ⓑ log x 1000 log x 1000

Try It 10.62

ⓐ log 2 5 4 log 2 5 4 ⓑ log 10 y log 10 y

The third property of logarithms is related to the Power Property of Exponents, ( a m ) n = a m · n , ( a m ) n = a m · n , we see that to raise a power to a power, we multiply the exponents. The Power Property of Logarithms , log a M p = p log a M log a M p = p log a M tells us to take the log of a number raised to a power, we multiply the power times the log of the number.

Power Property of Logarithms

If M > 0 , a > 0 , a ≠ 1 M > 0 , a > 0 , a ≠ 1 and p p is any real number then,

The log of a number raised to a power as the product product of the power times the log of the number.

We use this property to write the log of a number raised to a power as the product of the power times the log of the number. We essentially take the exponent and throw it in front of the logarithm.

Example 10.32

Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible. ⓐ log 5 4 3 log 5 4 3 and ⓑ log x 10 log x 10

ⓐ log 5 4 3 Use the Power Property, log a M p = p log a M . 3 log 5 4 log 5 4 3 = 3 log 5 4 log 5 4 3 Use the Power Property, log a M p = p log a M . 3 log 5 4 log 5 4 3 = 3 log 5 4

ⓑ log x 10 Use the Power Property, log a M p = p log a M . 10 log x log x 10 = 10 log x log x 10 Use the Power Property, log a M p = p log a M . 10 log x log x 10 = 10 log x

Try It 10.63

Use the Power Property of Logarithms to write each logarithm as a product of logarithms. Simplify, if possible.

ⓐ log 7 5 4 log 7 5 4 ⓑ log x 100 log x 100

Try It 10.64

ⓐ log 2 3 7 log 2 3 7 ⓑ log x 20 log x 20

We summarize the Properties of Logarithms here for easy reference. While the natural logarithms are a special case of these properties, it is often helpful to also show the natural logarithm version of each property.

If M > 0 , N > 0 , a > 0 , a ≠ 1 M > 0 , N > 0 , a > 0 , a ≠ 1 and p p is any real number then,

Now that we have the properties we can use them to “expand” a logarithmic expression. This means to write the logarithm as a sum or difference and without any powers.

We generally apply the Product and Quotient Properties before we apply the Power Property.

Example 10.33

Use the Properties of Logarithms to expand the logarithm log 4 ( 2 x 3 y 2 ) log 4 ( 2 x 3 y 2 ) . Simplify, if possible.

log 4 ( 2 x 3 y 2 ) Use the Product Property, log a M · N = log a M + log a N . log 4 2 + log 4 x 3 + log 4 y 2 Use the Power Property, log a M p = p log a M , on the last two terms. log 4 2 + 3 log 4 x + 2 log 4 y Simplify. 1 2 + 3 log 4 x + 2 log 4 y log 4 ( 2 x 3 y 2 ) = 1 2 + 3 log 4 x + 2 log 4 y log 4 ( 2 x 3 y 2 ) Use the Product Property, log a M · N = log a M + log a N . log 4 2 + log 4 x 3 + log 4 y 2 Use the Power Property, log a M p = p log a M , on the last two terms. log 4 2 + 3 log 4 x + 2 log 4 y Simplify. 1 2 + 3 log 4 x + 2 log 4 y log 4 ( 2 x 3 y 2 ) = 1 2 + 3 log 4 x + 2 log 4 y

Try It 10.65

Use the Properties of Logarithms to expand the logarithm log 2 ( 5 x 4 y 2 ) log 2 ( 5 x 4 y 2 ) . Simplify, if possible.

Try It 10.66

Use the Properties of Logarithms to expand the logarithm log 3 ( 7 x 5 y 3 ) log 3 ( 7 x 5 y 3 ) . Simplify, if possible.

When we have a radical in the logarithmic expression, it is helpful to first write its radicand as a rational exponent.

Example 10.34

Use the Properties of Logarithms to expand the logarithm log 2 x 3 3 y 2 z 4 log 2 x 3 3 y 2 z 4 . Simplify, if possible.

log 2 x 3 3 y 2 z 4 Rewrite the radical with a rational exponent. log 2 ( x 3 3 y 2 z ) 1 4 Use the Power Property, log a M p = p log a M . 1 4 log 2 ( x 3 3 y 2 z ) Use the Quotient Property, log a M · N = log a M − log a N . 1 4 ( log 2 ( x 3 ) − log 2 ( 3 y 2 z ) ) Use the Product Property, log a M · N = log a M + log a N , in the second term. 1 4 ( log 2 ( x 3 ) − ( log 2 3 + log 2 y 2 + log 2 z ) ) Use the Power Property, log a M p = p log a M , inside the parentheses. 1 4 ( 3 log 2 x − ( log 2 3 + 2 log 2 y + log 2 z ) ) Simplify by distributing. 1 4 ( 3 log 2 x − log 2 3 − 2 log 2 y − log 2 z ) log 2 x 3 3 y 2 z 4 = 1 4 ( 3 log 2 x − log 2 3 − 2 log 2 y − log 2 z ) log 2 x 3 3 y 2 z 4 Rewrite the radical with a rational exponent. log 2 ( x 3 3 y 2 z ) 1 4 Use the Power Property, log a M p = p log a M . 1 4 log 2 ( x 3 3 y 2 z ) Use the Quotient Property, log a M · N = log a M − log a N . 1 4 ( log 2 ( x 3 ) − log 2 ( 3 y 2 z ) ) Use the Product Property, log a M · N = log a M + log a N , in the second term. 1 4 ( log 2 ( x 3 ) − ( log 2 3 + log 2 y 2 + log 2 z ) ) Use the Power Property, log a M p = p log a M , inside the parentheses. 1 4 ( 3 log 2 x − ( log 2 3 + 2 log 2 y + log 2 z ) ) Simplify by distributing. 1 4 ( 3 log 2 x − log 2 3 − 2 log 2 y − log 2 z ) log 2 x 3 3 y 2 z 4 = 1 4 ( 3 log 2 x − log 2 3 − 2 log 2 y − log 2 z )

Try It 10.67

Use the Properties of Logarithms to expand the logarithm log 4 x 4 2 y 3 z 2 5 log 4 x 4 2 y 3 z 2 5 . Simplify, if possible.

Try It 10.68

Use the Properties of Logarithms to expand the logarithm log 3 x 2 5 y z 3 log 3 x 2 5 y z 3 . Simplify, if possible.

The opposite of expanding a logarithm is to condense a sum or difference of logarithms that have the same base into a single logarithm. We again use the properties of logarithms to help us, but in reverse.

To condense logarithmic expressions with the same base into one logarithm, we start by using the Power Property to get the coefficients of the log terms to be one and then the Product and Quotient Properties as needed.

Example 10.35

Use the Properties of Logarithms to condense the logarithm log 4 3 + log 4 x − log 4 y log 4 3 + log 4 x − log 4 y . Simplify, if possible.

The log expressions all have the same base, 4. log 4 3 + log 4 x − log 4 y The first two terms are added, so we use the Product Property, log a M + log a N = log a M · N . log 4 3 x − log 4 y Since the logs are subtracted, we use the Quotient Property, log a M − log a N = log a M N . log 4 3 x y log 4 3 + log 4 x − log 4 y = log 4 3 x y The log expressions all have the same base, 4. log 4 3 + log 4 x − log 4 y The first two terms are added, so we use the Product Property, log a M + log a N = log a M · N . log 4 3 x − log 4 y Since the logs are subtracted, we use the Quotient Property, log a M − log a N = log a M N . log 4 3 x y log 4 3 + log 4 x − log 4 y = log 4 3 x y

Try It 10.69

Use the Properties of Logarithms to condense the logarithm log 2 5 + log 2 x − log 2 y log 2 5 + log 2 x − log 2 y . Simplify, if possible.

Try It 10.70

Use the Properties of Logarithms to condense the logarithm log 3 6 − log 3 x − log 3 y log 3 6 − log 3 x − log 3 y . Simplify, if possible.

Example 10.36

Use the Properties of Logarithms to condense the logarithm 2 log 3 x + 4 log 3 ( x + 1 ) 2 log 3 x + 4 log 3 ( x + 1 ) . Simplify, if possible.

The log expressions have the same base, 3. 2 log 3 x + 4 log 3 ( x + 1 ) Use the Power Property, log a M + log a N = log a M · N . log 3 x 2 + log 3 ( x + 1 ) 4 The terms are added, so we use the Product Property, log a M + log a N = log a M · N . log 3 x 2 ( x + 1 ) 4 2 log 3 x + 4 log 3 ( x + 1 ) = log 3 x 2 ( x + 1 ) 4 The log expressions have the same base, 3. 2 log 3 x + 4 log 3 ( x + 1 ) Use the Power Property, log a M + log a N = log a M · N . log 3 x 2 + log 3 ( x + 1 ) 4 The terms are added, so we use the Product Property, log a M + log a N = log a M · N . log 3 x 2 ( x + 1 ) 4 2 log 3 x + 4 log 3 ( x + 1 ) = log 3 x 2 ( x + 1 ) 4

Try It 10.71

Use the Properties of Logarithms to condense the logarithm 3 log 2 x + 2 log 2 ( x − 1 ) 3 log 2 x + 2 log 2 ( x − 1 ) . Simplify, if possible.

Try It 10.72

Use the Properties of Logarithms to condense the logarithm 2 log x + 2 log ( x + 1 ) 2 log x + 2 log ( x + 1 ) . Simplify, if possible.

Use the Change-of-Base Formula

To evaluate a logarithm with any other base, we can use the Change-of-Base Formula . We will show how this is derived.

Suppose we want to evaluate log a M . log a M Let y = log a M . y = log a M Rewrite the expression in exponential form. a y = M Take the log b of each side. log b a y = log b M Use the Power Property. y log b a = log b M Solve for y . y = log b M log b a Substitute y = log a M . log a M = log b M log b a Suppose we want to evaluate log a M . log a M Let y = log a M . y = log a M Rewrite the expression in exponential form. a y = M Take the log b of each side. log b a y = log b M Use the Power Property. y log b a = log b M Solve for y . y = log b M log b a Substitute y = log a M . log a M = log b M log b a

The Change-of-Base Formula introduces a new base b . b . This can be any base b we want where b > 0 , b ≠ 1 . b > 0 , b ≠ 1 . Because our calculators have keys for logarithms base 10 and base e , we will rewrite the Change-of-Base Formula with the new base as 10 or e .

Change-of-Base Formula

For any logarithmic bases a , b a , b and M > 0 , M > 0 ,

When we use a calculator to find the logarithm value, we usually round to three decimal places. This gives us an approximate value and so we use the approximately equal symbol (≈) (≈) .

Example 10.37

Rounding to three decimal places, approximate log 4 35 . log 4 35 .

Try It 10.73

Rounding to three decimal places, approximate log 3 42 . log 3 42 .

Try It 10.74

Rounding to three decimal places, approximate log 5 46 . log 5 46 .

Access these online resources for additional instruction and practice with using the properties of logarithms.

• Using Properties of Logarithms to Expand Logs
• Using Properties of Logarithms to Condense Logs
• Change of Base

Section 10.4 Exercises

Practice makes perfect.

In the following exercises, use the properties of logarithms to evaluate.

ⓐ log 4 1 log 4 1 ⓑ log 8 8 log 8 8

ⓐ log 12 1 log 12 1 ⓑ ln e ln e

ⓐ 3 log 3 6 3 log 3 6 ⓑ log 2 2 7 log 2 2 7

ⓐ 5 log 5 10 5 log 5 10 ⓑ log 4 4 10 log 4 4 10

ⓐ 8 log 8 7 8 log 8 7 ⓑ log 6 6 −2 log 6 6 −2

ⓐ 6 log 6 15 6 log 6 15 ⓑ log 8 8 −4 log 8 8 −4

ⓐ 10 log 5 10 log 5 ⓑ log 10 −2 log 10 −2

ⓐ 10 log 3 10 log 3 ⓑ log 10 −1 log 10 −1

ⓐ e ln 4 e ln 4 ⓑ ln e 2 ln e 2

ⓐ e ln 3 e ln 3 ⓑ ln e 7 ln e 7

In the following exercises, use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify if possible.

log 4 6 x log 4 6 x

log 5 8 y log 5 8 y

log 2 32 x y log 2 32 x y

log 3 81 x y log 3 81 x y

log 100 x log 100 x

log 1000 y log 1000 y

In the following exercises, use the Quotient Property of Logarithms to write each logarithm as a sum of logarithms. Simplify if possible.

log 3 3 8 log 3 3 8

log 6 5 6 log 6 5 6

log 4 16 y log 4 16 y

log 5 125 x log 5 125 x

log x 10 log x 10

log 10,000 y log 10,000 y

ln e 3 3 ln e 3 3

ln e 4 16 ln e 4 16

In the following exercises, use the Power Property of Logarithms to expand each. Simplify if possible.

log 3 x 2 log 3 x 2

log 2 x 5 log 2 x 5

log x −2 log x −2

log x −3 log x −3

log 4 x log 4 x

log 5 x 3 log 5 x 3

ln x 3 ln x 3

ln x 4 3 ln x 4 3

In the following exercises, use the Properties of Logarithms to expand the logarithm. Simplify if possible.

log 5 ( 4 x 6 y 4 ) log 5 ( 4 x 6 y 4 )

log 2 ( 3 x 5 y 3 ) log 2 ( 3 x 5 y 3 )

log 3 ( 2 x 2 ) log 3 ( 2 x 2 )

log 5 ( 21 4 y 3 ) log 5 ( 21 4 y 3 )

log 3 x y 2 z 2 log 3 x y 2 z 2

log 5 4 a b 3 c 4 d 2 log 5 4 a b 3 c 4 d 2

log 4 x 16 y 4 log 4 x 16 y 4

log 3 x 2 3 27 y 4 log 3 x 2 3 27 y 4

log 2 2 x + y 2 z 2 log 2 2 x + y 2 z 2

log 3 3 x + 2 y 2 5 z 2 log 3 3 x + 2 y 2 5 z 2

log 2 5 x 3 2 y 2 z 4 4 log 2 5 x 3 2 y 2 z 4 4

log 5 3 x 2 4 y 3 z 3 log 5 3 x 2 4 y 3 z 3

In the following exercises, use the Properties of Logarithms to condense the logarithm. Simplify if possible.

log 6 4 + log 6 9 log 6 4 + log 6 9

log 4 + log 25 log 4 + log 25

log 2 80 − log 2 5 log 2 80 − log 2 5

log 3 36 − log 3 4 log 3 36 − log 3 4

log 3 4 + log 3 ( x + 1 ) log 3 4 + log 3 ( x + 1 )

log 2 5 − log 2 ( x − 1 ) log 2 5 − log 2 ( x − 1 )

log 7 3 + log 7 x − log 7 y log 7 3 + log 7 x − log 7 y

log 5 2 − log 5 x − log 5 y log 5 2 − log 5 x − log 5 y

4 log 2 x + 6 log 2 y 4 log 2 x + 6 log 2 y

6 log 3 x + 9 log 3 y 6 log 3 x + 9 log 3 y

log 3 ( x 2 − 1 ) − 2 log 3 ( x − 1 ) log 3 ( x 2 − 1 ) − 2 log 3 ( x − 1 )

log ( x 2 + 2 x + 1 ) − 2 log ( x + 1 ) log ( x 2 + 2 x + 1 ) − 2 log ( x + 1 )

4 log x − 2 log y − 3 log z 4 log x − 2 log y − 3 log z

3 ln x + 4 ln y − 2 ln z 3 ln x + 4 ln y − 2 ln z

1 3 log x − 3 log ( x + 1 ) 1 3 log x − 3 log ( x + 1 )

2 log ( 2 x + 3 ) + 1 2 log ( x + 1 ) 2 log ( 2 x + 3 ) + 1 2 log ( x + 1 )

In the following exercises, use the Change-of-Base Formula, rounding to three decimal places, to approximate each logarithm.

log 3 42 log 3 42

log 5 46 log 5 46

log 12 87 log 12 87

log 15 93 log 15 93

log 2 17 log 2 17

log 3 21 log 3 21

Study Guides > Intermediate Algebra

Properties of logarithms, learning objectives.

• Define properties of logarithms, and use them to solve equations
• Define the product rule for logarithms, and use it to solve equations
• Define the quotient and power rules for logarithms
• Use the quotient and power rules for logarithms to simplify logarithmic expressions
• Combine product, power and quotient rules to simplify logarithmic expressions
• Expand logarithmic expressions that have negative or fractional exponents
• Condense logarithmic expressions
• Use properties of logarithms to define the change of base formula
• Change the base of logarithmic expressions into base 10, or base e

Recall that we can express the relationship between logarithmic form and its corresponding exponential form as follows:

Note that the base b  is always positive and that the logarithmic and exponential functions "undo" each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here.

Zero and Identity Exponent Rule for Logarithms and Exponentials

Use the the fact that exponentials and logarithms are inverses to prove the zero and identity exponent rule for the following:

1.[latex]{\mathrm{log}}_{5}1=0[/latex]

1.[latex]{\mathrm{log}}_{5}1=0[/latex]  since [latex]{5}^{0}=1[/latex]

2.[latex]{\mathrm{log}}_{5}5=1[/latex] since [latex]{5}^{1}=5[/latex]

Exponential and logarithmic functions are inverses of each other and we can take advantage of this to evaluate and solve expressions and equations involving logarithms and exponentials. The inverse property of logarithms and exponentials gives us an explicit way to rewrite an exponential as a logarithm or a logarithm as an exponential.

Inverse Property of Logarithms and Exponentials

[latex]\begin{array}{c}\hfill \\ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \\ \text{ }{b}^{{\mathrm{log}}_{b}x}=x, x>0, b>0, b\ne1\hfill \end{array}[/latex]

Answer: 1.Rewrite the logarithm as [latex]{\mathrm{log}}_{10}\left({10}^{2}\right)[/latex], and then apply the inverse property [latex]{\mathrm{log}}_{b}\left({b}^{x}\right)=x[/latex] to get [latex]{\mathrm{log}}_{10}\left({10}^{2}\right)=2[/latex]. 2.Rewrite the logarithm as [latex]{e}^{{\mathrm{log}}_{e}7}[/latex], and then apply the inverse property [latex]{b}^{{\mathrm{log}}_{b}x}=x[/latex] to get [latex]{e}^{{\mathrm{log}}_{e}7}=7[/latex]

The O ne-To-One  Property of Logarithms

[latex]{\mathrm{log}}_{b}M={\mathrm{log}}_{b}N\text{ if and only if}\text{ }M=N[/latex]

Answer: In order for this equation to be true we must find a value for x such that [latex]3x=2x+5[/latex] [latex]\begin{array}{c}3x=2x+5\hfill & \text{Set the arguments equal}\text{.}\hfill \\ x=5\hfill & \text{Subtract 2}x\text{.}\hfill \end{array}[/latex] Check your answer by substituting 5 for x. [latex]\begin{array}{c}{\mathrm{log}}_{3}\left(3\cdot5\right)={\mathrm{log}}_{3}\left(2\cdot5+5\right)\\{\mathrm{log}}_{3}\left(15\right)={\mathrm{log}}_{3}\left(15\right)\end{array}[/latex] This is a true statement, so we must have found the correct value for x.

• Zero and Identity Exponent Rule: [latex]{\mathrm{log}}_{b}1=0[/latex], b>0, and [latex]{\mathrm{log}}_{b}b=1[/latex], b>0
• Inverse Property: [latex]\begin{array}{c}\hfill \\ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \\ \text{ }{b}^{{\mathrm{log}}_{b}x}=x,x>0\hfill \end{array}[/latex]
• One-To-One Property: [latex]{\mathrm{log}}_{b}M={\mathrm{log}}_{b}N\text{ if and only if}\text{ }M=N[/latex]

The Product Rule for Logarithms

Recall that we use the product rule of exponents to combine the product of exponents by adding: [latex]{x}^{a}{x}^{b}={x}^{a+b}[/latex]. We have a similar property for logarithms, called the product rule for logarithms , which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents, and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.

The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.

Repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. Consider the following example:

Answer: [latex-display]{\mathrm{log}}_{b}\left(wxyz\right)={\mathrm{log}}_{b}w+{\mathrm{log}}_{b}x+{\mathrm{log}}_{b}y+{\mathrm{log}}_{b}z[/latex-display]

We begin by factoring the argument completely, expressing 30 as a product of primes.

Next we write the equivalent equation by summing the logarithms of each factor.

Analysis of the Solution

The quotient rule for logarithms.

For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine the quotient of exponents by subtracting: [latex]{x}^{\frac{a}{b}}={x}^{a-b}[/latex]. The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.

 The Quotient Rule for Logarithms

The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.

Let [latex]m={\mathrm{log}}_{b}M[/latex] and [latex]n={\mathrm{log}}_{b}N[/latex]. In exponential form, these equations are [latex]{b}^{m}=M[/latex] and [latex]{b}^{n}=N[/latex]. It follows that

Factoring and canceling we get,

Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.

First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.

Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule, noting that the prime factors of the factor 15 are 3 and 5.

 Analysis of the Solution

There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for [latex]x=-\frac{4}{3}[/latex] and x  = 2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm, that x  > 0, x  > 1, [latex]x>-\frac{4}{3}[/latex], and x  < 2. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.

Using the Power Rule for Logarithms

We’ve explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as [latex]{x}^{2}[/latex]? One method is as follows:

Notice that we used the product rule for logarithms to simplify the example above. By doing so, we have derived the power rule for logarithms , which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,

 The Power Rule for Logarithms

The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.

The argument is already written as a power, so we identify the exponent, 5, and the base, x , and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.

Expressing the argument as a power, we get [latex]{\mathrm{log}}_{3}\left(25\right)={\mathrm{log}}_{3}\left({5}^{2}\right)[/latex].

Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.

Because the logarithm of a power is the product of the exponent times the logarithm of the base, it follows that the product of a number and a logarithm can be written as a power. For the expression [latex]4\mathrm{ln}\left(x\right)[/latex], we identify the factor, 4, as the exponent and the argument, x , as the base, and rewrite the product as a logarithm of a power:

Expand and Condense Logarithms

Taken together, the product rule, quotient rule, and power rule are often called "laws of logs." Sometimes we apply more than one rule in order to simplify an expression. For example:

We can also use the power rule to expand logarithmic expressions involving negative and fractional exponents. Here is an alternate proof of the quotient rule for logarithms using the fact that a reciprocal (fraction) has a negative power:

We can also apply the product rule to express a sum or difference of logarithms as the logarithm of a product.

[latex]\begin{array}{c}\mathrm{log}\left(10+100\right)\overset{?}{=}\end{array}\mathrm{log}\left(10\right)+\mathrm{log}\left(100\right)\\\mathrm{log}\left(110\right)\overset{?}{=}1+2\\2.04\ne3[/latex]

First, because we have a quotient of two expressions, we can use the quotient rule:

Then seeing the product in the first term, we use the product rule:

Finally, we use the power rule on the first term:

Answer: [latex]\begin{array}{c}\mathrm{log}\left(\sqrt{x}\right)\hfill & =\mathrm{log}{x}^{\left(\frac{1}{2}\right)}\hfill \\ \hfill & =\frac{1}{2}\mathrm{log}x\hfill \end{array}[/latex]

Think About it

Answer: No. There is no way to expand the logarithm of a sum or difference inside the argument of the logarithm. Rewrite the expression as an equation and express it as an exponential to give yourself some proof. [latex-display]m=\mathrm{ln}\left({x}^{2}+{y}^{2}\right)[/latex-display] If you rewrite this as an exponential you get: [latex-display]e^m={x}^{2}+{y}^{2}[/latex-display] From here, there's not much more you can do to make this expression more simple.

We can expand by applying the Product and Quotient Rules.

Condense Logarithms

We can use the rules of logarithms we just learned to condense sums, differences, and products with the same base as a single logarithm. It is important to remember that the logarithms must have the same base to be combined. We will learn later how to change the base of any logarithm before condensing.

Using the product and quotient rules

This reduces our original expression to

Then, using the quotient rule

We apply the power rule first:

Next we apply the product rule to the sum:

Finally, we apply the quotient rule to the difference:

Change of Base

Most calculators can evaluate only common and natural logs. In order to evaluate logarithms with a base other than 10 or [latex]e[/latex], we use the change-of-base formula to rewrite the logarithm as the quotient of logarithms of any other base; when using a calculator, we would change them to common or natural logs.

To derive the change-of-base formula, we use the one-to-one property and power rule for logarithms .

Given any positive real numbers M , b , and n , where [latex]n\ne 1 [/latex] and [latex]b\ne 1[/latex], we show

Let [latex]y={\mathrm{log}}_{b}M[/latex]. By taking the log base [latex]n[/latex] of both sides of the equation, we arrive at an exponential form, namely [latex]{b}^{y}=M[/latex]. It follows that

For example, to evaluate [latex]{\mathrm{log}}_{5}36[/latex] using a calculator, we must first rewrite the expression as a quotient of common or natural logs. We will use the common log.

Because we will be expressing [latex]{\mathrm{log}}_{5}3[/latex] as a quotient of natural logarithms, the new base, n  = e .

We rewrite the log as a quotient using the change-of-base formula. The numerator of the quotient will be the natural log with argument 3. The denominator of the quotient will be the natural log with argument 5.

The Change-of-Base Formula

The change-of-base formula can be used to evaluate a logarithm with any base.

For any positive real numbers M , b , and n , where [latex]n\ne 1 [/latex] and [latex]b\ne 1[/latex],

It follows that the change-of-base formula can be used to rewrite a logarithm with any base as the quotient of common or natural logs.

According to the change-of-base formula, we can rewrite the log base 2 as a logarithm of any other base. Since our calculators can evaluate the natural log, we might choose to use the natural logarithm, which is the log base e .

Think About It

Answer: Yes. Remember that [latex]\mathrm{log}9[/latex] means [latex]{\text{log}}_{\text{10}}\text{9}[/latex]. So, [latex]\mathrm{log}9=\frac{\mathrm{ln}9}{\mathrm{ln}10}[/latex] .

• Factor the argument completely, expressing each whole number factor as a product of primes.
• Write the equivalent expression by summing the logarithms of each factor.

You can use the quotient rule of logarithms to write an equivalent difference of logarithms in the following way:

• Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
• Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
• Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.

To use the power rule of logarithms to write an equivalent product of a factor and a logarithm consider the following:

• Express the argument as a power, if needed.
• Write the equivalent expression by multiplying the exponent times the logarithm of the base.

Given a sum, difference, or product of logarithms with the same base, write an equivalent expression as a single logarithm.

• Apply the power property first. Identify terms that are products of factors and a logarithm, and rewrite each as the logarithm of a power.
• Next apply the product property. Rewrite sums of logarithms as the logarithm of a product.
• Apply the quotient property last. Rewrite differences of logarithms as the logarithm of a quotient.

Given a logarithm with the form [latex]{\mathrm{log}}_{b}M[/latex]

• Determine the new base n , remembering that the common log, [latex]\mathrm{log}\left(x\right)[/latex], has base 10, and the natural log, [latex]\mathrm{ln}\left(x\right)[/latex], has base e .
• The numerator of the quotient will be a logarithm with base n  and argument M .
• The denominator of the quotient will be a logarithm with base n  and argument b .

Why learn about exponential and logarithmic equations?

[latex]R=\mathrm{log}\left(\frac{A}{A_{0}}\right)[/latex]

The question provided her with the quantity [latex]\frac{A}{A_{0}}[/latex] and asked her to solve for R.  [latex]{A_{0}}[/latex] is a baseline measure of ground movement as detected by a seismometer, seen in the image above. When there is an earthquake, wave amplitudes recorded by seismometers are expressed relative to this baseline.  For example, Joan's book asked the following question:

Joan hoped to give her grandfather the Richter scale magnitude for the Alaska quake, 8.5, and see if he could find how much greater the wave amplitude of that quake was than the baseline, [latex]{A_{0}}[/latex].

In this module you will learn how to solve problems such as the one Joan is planning to try to stump her grandfather with. We will come back to Joan and her grandfather at the end of this module to see if she was able to ask him a question he didn't know how to answer.

In this module, you will learn about the properties of exponential and logarithmic functions in the same way that you learned about the properties of exponents.  You will use the properties of logarithms and exponentials to solve equations that involve them.

The learning outcomes for this module include:

• Define and use the properties of logarithms to expand, condense, and change the base of a logarithmic expression
• Use the properties of logarithms and exponentials to solve equations

Licenses & Attributions

Cc licensed content, original.

• Revision and Adaptation. Provided by: Lumen Learning License: CC BY: Attribution .

CC licensed content, Shared previously

• Precalculus. Provided by: OpenStax Authored by: Jay Abramson, et al.. Located at: https://openstax.org/books/precalculus/pages/1-introduction-to-functions. License: CC BY: Attribution . License terms: Download For Free at : http://cnx.org/contents/ [email protected] ..

Please add a message.

Message received. Thanks for the feedback.

Module 12: Exponential and Logarithmic Equations and Models

Properties of logarithms, learning outcomes.

• Rewrite a logarithmic expression using the power rule, product rule, or quotient rule.
• Expand logarithmic expressions using a combination of logarithm rules.
• Condense logarithmic expressions using logarithm rules.

Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties are easy to prove.

[latex]\begin{array}{l}{\mathrm{log}}_{b}1=0\\{\mathrm{log}}_{b}b=1\end{array}[/latex]

For example, [latex]{\mathrm{log}}_{5}1=0[/latex] since [latex]{5}^{0}=1[/latex] and [latex]{\mathrm{log}}_{5}5=1[/latex] since [latex]{5}^{1}=5[/latex].

Next, we have the inverse property.

[latex]\begin{array}{l}\hfill \\ {\mathrm{log}}_{b}\left({b}^{x}\right)=x\hfill \\ \text{ }{b}^{{\mathrm{log}}_{b}x}=x,x>0\hfill \end{array}[/latex]

For example, to evaluate [latex]\mathrm{log}\left(100\right)[/latex], we can rewrite the logarithm as [latex]{\mathrm{log}}_{10}\left({10}^{2}\right)[/latex] and then apply the inverse property [latex]{\mathrm{log}}_{b}\left({b}^{x}\right)=x[/latex] to get [latex]{\mathrm{log}}_{10}\left({10}^{2}\right)=2[/latex].

To evaluate [latex]{e}^{\mathrm{ln}\left(7\right)}[/latex], we can rewrite the logarithm as [latex]{e}^{{\mathrm{log}}_{e}7}[/latex] and then apply the inverse property [latex]{b}^{{\mathrm{log}}_{b}x}=x[/latex] to get [latex]{e}^{{\mathrm{log}}_{e}7}=7[/latex].

Finally, we have the one-to-one property.

[latex]{\mathrm{log}}_{b}M={\mathrm{log}}_{b}N\text{ if and only if}\text{ }M=N[/latex]

We can use the one-to-one property to solve the equation [latex]{\mathrm{log}}_{3}\left(3x\right)={\mathrm{log}}_{3}\left(2x+5\right)[/latex] for x . Since the bases are the same, we can apply the one-to-one property by setting the arguments equal and solving for x :

[latex]\begin{array}{l}3x=2x+5\hfill & \text{Set the arguments equal}\text{.}\hfill \\ x=5\hfill & \text{Subtract 2}x\text{.}\hfill \end{array}[/latex]

But what about the equation [latex]{\mathrm{log}}_{3}\left(3x\right)+{\mathrm{log}}_{3}\left(2x+5\right)=2[/latex]? The one-to-one property does not help us in this instance. Before we can solve an equation like this, we need a method for combining logarithms on the left side of the equation.

Using the Product Rule for Logarithms

Recall that we use the product rule of exponents to combine the product of exponents by adding: [latex]{x}^{a}{x}^{b}={x}^{a+b}[/latex]. We have a similar property for logarithms, called the product rule for logarithms , which says that the logarithm of a product is equal to a sum of logarithms. Because logs are exponents and we multiply like bases, we can add the exponents. We will use the inverse property to derive the product rule below.

Given any real number x  and positive real numbers M , N , and b , where [latex]b\ne 1[/latex], we will show

[latex]{\mathrm{log}}_{b}\left(MN\right)\text{= }{\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)[/latex].

Let [latex]m={\mathrm{log}}_{b}M[/latex] and [latex]n={\mathrm{log}}_{b}N[/latex]. In exponential form, these equations are [latex]{b}^{m}=M[/latex] and [latex]{b}^{n}=N[/latex]. It follows that

[latex]\begin{array}{lllllllll}{\mathrm{log}}_{b}\left(MN\right)\hfill & ={\mathrm{log}}_{b}\left({b}^{m}{b}^{n}\right)\hfill & \text{Substitute for }M\text{ and }N.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left({b}^{m+n}\right)\hfill & \text{Apply the product rule for exponents}.\hfill \\ \hfill & =m+n\hfill & \text{Apply the inverse property of logs}.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)\hfill & \text{Substitute for }m\text{ and }n.\hfill \end{array}[/latex]

Note that repeated applications of the product rule for logarithms allow us to simplify the logarithm of the product of any number of factors. For example, consider [latex]\mathrm{log}_{b}(wxyz)[/latex]. Using the product rule for logarithms, we can rewrite this logarithm of a product as the sum of logarithms of its factors:

A General Note: The Product Rule for Logarithms

The product rule for logarithms can be used to simplify a logarithm of a product by rewriting it as a sum of individual logarithms.

[latex]{\mathrm{log}}_{b}\left(MN\right)={\mathrm{log}}_{b}\left(M\right)+{\mathrm{log}}_{b}\left(N\right)\text{ for }b>0[/latex]

Example: Using the Product Rule for Logarithms

Expand [latex]{\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right)[/latex].

We begin by writing an equal equation by summing the logarithms of each factor.

[latex]{\mathrm{log}}_{3}\left(30x\left(3x+4\right)\right)={\mathrm{log}}_{3}\left(30x\right)+{\mathrm{log}}_{3}\left(3x+4\right)={\mathrm{log}}_{3}\left(30\right)+{\mathrm{log}}_{3}\left(x\right)+{\mathrm{log}}_{3}\left(3x+4\right)[/latex]

The final expansion looks like this. Note how the factor [latex]30x[/latex] can be expanded into the sum of two logarithms:

[latex]{\mathrm{log}}_{3}\left(30\right)+{\mathrm{log}}_{3}\left(x\right)+{\mathrm{log}}_{3}\left(3x+4\right)[/latex]

Expand [latex]{\mathrm{log}}_{b}\left(8k\right)[/latex].

[latex]{\mathrm{log}}_{b}8+{\mathrm{log}}_{b}k[/latex]

Using the Quotient Rule for Logarithms

For quotients, we have a similar rule for logarithms. Recall that we use the quotient rule of exponents to combine the quotient of exponents by subtracting: [latex]{x}^{\frac{a}{b}}={x}^{a-b}[/latex]. The quotient rule for logarithms says that the logarithm of a quotient is equal to a difference of logarithms. Just as with the product rule, we can use the inverse property to derive the quotient rule.

Given any real number x  and positive real numbers M , N , and b , where [latex]b\ne 1[/latex], we will show

[latex]{\mathrm{log}}_{b}\left(\frac{M}{N}\right)\text{= }{\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)[/latex].

[latex]\begin{array}{l}{\mathrm{log}}_{b}\left(\frac{M}{N}\right)\hfill & ={\mathrm{log}}_{b}\left(\frac{{b}^{m}}{{b}^{n}}\right)\hfill & \text{Substitute for }M\text{ and }N.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left({b}^{m-n}\right)\hfill & \text{Apply the quotient rule for exponents}.\hfill \\ \hfill & =m-n\hfill & \text{Apply the inverse property of logs}.\hfill \\ \hfill & ={\mathrm{log}}_{b}\left(M\right)-{\mathrm{log}}_{b}\left(N\right)\hfill & \text{Substitute for }m\text{ and }n.\hfill \end{array}[/latex]

For example, to expand [latex]\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right)[/latex], we must first express the quotient in lowest terms. Factoring and canceling, we get

[latex]\begin{array}{lllll}\mathrm{log}\left(\frac{2{x}^{2}+6x}{3x+9}\right) & =\mathrm{log}\left(\frac{2x\left(x+3\right)}{3\left(x+3\right)}\right)\hfill & \text{Factor the numerator and denominator}.\hfill \\ & \text{}=\mathrm{log}\left(\frac{2x}{3}\right)\hfill & \text{Cancel the common factors}.\hfill \end{array}[/latex]

Next we apply the quotient rule by subtracting the logarithm of the denominator from the logarithm of the numerator. Then we apply the product rule.

[latex]\begin{array}{lll}\mathrm{log}\left(\frac{2x}{3}\right) & =\mathrm{log}\left(2x\right)-\mathrm{log}\left(3\right)\hfill \\ \text{} & =\mathrm{log}\left(2\right)+\mathrm{log}\left(x\right)-\mathrm{log}\left(3\right)\hfill \end{array}[/latex]

A General Note: The Quotient Rule for Logarithms

The quotient rule for logarithms can be used to simplify a logarithm or a quotient by rewriting it as the difference of individual logarithms.

[latex]{\mathrm{log}}_{b}\left(\frac{M}{N}\right)={\mathrm{log}}_{b}M-{\mathrm{log}}_{b}N[/latex]

How To: Given the logarithm of a quotient, use the quotient rule of logarithms to write an equivalent difference of logarithms

• Express the argument in lowest terms by factoring the numerator and denominator and canceling common terms.
• Write the equivalent expression by subtracting the logarithm of the denominator from the logarithm of the numerator.
• Check to see that each term is fully expanded. If not, apply the product rule for logarithms to expand completely.

Example: Using the Quotient Rule for Logarithms

Expand [latex]{\mathrm{log}}_{2}\left(\frac{15x\left(x - 1\right)}{\left(3x+4\right)\left(2-x\right)}\right)[/latex].

First we note that the quotient is factored and in lowest terms, so we apply the quotient rule.

[latex]{\mathrm{log}}_{2}\left(\frac{15x\left(x - 1\right)}{\left(3x+4\right)\left(2-x\right)}\right)={\mathrm{log}}_{2}\left(15x\left(x - 1\right)\right)-{\mathrm{log}}_{2}\left(\left(3x+4\right)\left(2-x\right)\right)[/latex]

Notice that the resulting terms are logarithms of products. To expand completely, we apply the product rule.

[latex]\begin{array}{l}{\mathrm{log}}_{2}\left(15x\left(x - 1\right)\right)-{\mathrm{log}}_{2}\left(\left(3x+4\right)\left(2-x\right)\right) \\\text{}= \left[{\mathrm{log}}_{2}\left(15\right)+{\mathrm{log}}_{2}\left(x\right)+{\mathrm{log}}_{2}\left(x - 1\right)\right]-\left[{\mathrm{log}}_{2}\left(3x+4\right)+{\mathrm{log}}_{2}\left(2-x\right)\right]\hfill \\ \text{}={\mathrm{log}}_{2}\left(15\right)+{\mathrm{log}}_{2}\left(x\right)+{\mathrm{log}}_{2}\left(x - 1\right)-{\mathrm{log}}_{2}\left(3x+4\right)-{\mathrm{log}}_{2}\left(2-x\right)\hfill \end{array}[/latex]

Analysis of the Solution

There are exceptions to consider in this and later examples. First, because denominators must never be zero, this expression is not defined for [latex]x=-\frac{4}{3}[/latex] and x  = 2. Also, since the argument of a logarithm must be positive, we note as we observe the expanded logarithm that x  > 0, x  > 1, [latex]x>-\frac{4}{3}[/latex], and x  < 2. Combining these conditions is beyond the scope of this section, and we will not consider them here or in subsequent exercises.

Expand [latex]{\mathrm{log}}_{3}\left(\frac{7{x}^{2}+21x}{7x\left(x - 1\right)\left(x - 2\right)}\right)[/latex].

[latex]{\mathrm{log}}_{3}\left(x+3\right)-{\mathrm{log}}_{3}\left(x - 1\right)-{\mathrm{log}}_{3}\left(x - 2\right)[/latex]

Using the Power Rule for Logarithms

We have explored the product rule and the quotient rule, but how can we take the logarithm of a power, such as [latex]{x}^{2}[/latex]? One method is as follows:

[latex]\begin{array}{l}{\mathrm{log}}_{b}\left({x}^{2}\right)\hfill & ={\mathrm{log}}_{b}\left(x\cdot x\right)\hfill \\ \hfill & ={\mathrm{log}}_{b}x+{\mathrm{log}}_{b}x\hfill \\ \hfill & =2{\mathrm{log}}_{b}x\hfill \end{array}[/latex]

Notice that we used the product rule for logarithms to find a solution for the example above. By doing so, we have derived the power rule for logarithms , which says that the log of a power is equal to the exponent times the log of the base. Keep in mind that although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example,

[latex]\begin{array}{lll}100={10}^{2}, \hfill & \sqrt{3}={3}^{\frac{1}{2}}, \hfill & \frac{1}{e}={e}^{-1}\hfill \end{array}[/latex]

A General Note: The Power Rule for Logarithms

The power rule for logarithms can be used to simplify the logarithm of a power by rewriting it as the product of the exponent times the logarithm of the base.

[latex]{\mathrm{log}}_{b}\left({M}^{n}\right)=n{\mathrm{log}}_{b}M[/latex]

How To: Given the logarithm of a power, use the power rule of logarithms to write an equivalent product of a factor and a logarithm

• Express the argument as a power, if needed.
• Write the equivalent expression by multiplying the exponent times the logarithm of the base.

Example: Expanding a Logarithm with Powers

Rewrite [latex]{\mathrm{log}}_{2}{x}^{5}[/latex].

The argument is already written as a power, so we identify the exponent, 5, and the base, x , and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.

[latex]{\mathrm{log}}_{2}\left({x}^{5}\right)=5{\mathrm{log}}_{2}x[/latex]

Rewrite [latex]\mathrm{ln}{x}^{2}[/latex].

[latex]2\mathrm{ln}x[/latex]

Example: Rewriting an Expression as a Power before Using the Power Rule

Rewrite [latex]{\mathrm{log}}_{3}\left(25\right)[/latex] using the power rule for logs.

Expressing the argument as a power, we get [latex]{\mathrm{log}}_{3}\left(25\right)={\mathrm{log}}_{3}\left({5}^{2}\right)[/latex].

Next we identify the exponent, 2, and the base, 5, and rewrite the equivalent expression by multiplying the exponent times the logarithm of the base.

[latex]{\mathrm{log}}_{3}\left({5}^{2}\right)=2{\mathrm{log}}_{3}\left(5\right)[/latex]

Rewrite [latex]\mathrm{ln}\left(\frac{1}{{x}^{2}}\right)[/latex].

[latex]-2\mathrm{ln}\left(x\right)[/latex]

Contribute!

Improve this page Learn More

• Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
• Question ID 63350. Authored by : Brin,Leon. License : CC BY: Attribution . License Terms : IMathAS Community License CC-BY + GPL
• College Algebra. Authored by : Abramson, Jay et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

• Math Article
• Properties Of Logarithms

Properties of Logarithms

In Mathematics, properties of logarithms functions are used to solve logarithm problems. We have learned many properties in basic maths such as commutative, associative and distributive, which are applicable for algebra. In the case of logarithmic functions , there are basically five properties.

The logarithmic number is associated with exponent and power, such that if x n = m, then it is equal to log x m=n. Hence, it is necessary that we should also learn exponent law . For example, the logarithm of 10000 to base 10 is 4, because 4 is the power to which ten must be raised to produce 10000: 10 4 = 10000, so log 10 10000 = 4.

With the help of these properties, we can express the logarithm of a product as a sum of logarithms, the log of the quotient as a difference of log and log of power as a product.

Only positive real numbers have real number logarithms, negative and complex numbers have complex logarithms.

Logarithm Base Properties

Before we proceed ahead for logarithm properties, we need to revise the law of exponents, so that we can compare the properties.

For exponents, the laws are:

• Product rule: a m .a n =a m+n
• Quotient rule: a m /a n  = a m-n
• Power of a Power: (a m ) n  = a mn

Now let us learn the  properties of logarithmic functions .

Product Property

If a, m and n are positive integers and a ≠ 1, then;

log a (mn) = log a m + log a n

Thus, the log of two numbers m and n, with base ‘a’  is equal to the sum of log m and log n with the same base ‘a’.

Example: log 3 (9.25)

= log 3 (9) + log 3 (27)

= log 3 (3 2 ) + log 3 (3 3 )

= 2 + 3 (By property: log b b x = x)

Quotient Property

If m, n and a are positive integers and a ≠ 1, then;

log a (m/n) = log a m – log a n

In the above expression, the logarithm of a quotient of two positive numbers m and n results in a diffe rence of log of m and log n with the same base ‘a’.

Example:  log 2 (21/8)

log 2 (21/8) = log 2  21 – log 2  8

If a and m are positive numbers, a ≠ 1 and n is a real number, then;

log a m n = n log a m

The above property defines that logarithm of a positive number m to the power n is equal to the product of n and log of m.

log 2 10 3  = 3 log 2 10

The above three properties are the important ones for logarithms. Some other properties are given below along with suitable examples.

Change of Base rule

If m, n and p are positive numbers and n ≠ 1, p ≠ 1, then;

Log n m = log p m/log p n

log 2  10 = log p  10/log p  2

Reciprocal rule

If m and n are the positive numbers other than 1, then;

log n m = 1/log m n

log 2  10 = 1/log 10  2

Also, read:

Comparison of Exponent law and Logarithm law

As you can see these log properties are very much similar to laws of exponents. Let us compare here both the properties using a table:

Natural Logarithm Properties

The natural log (ln) follows the same properties as the base logarithms do.

• ln(pq) = ln p + ln q
• ln(p/q) = ln p – ln q
• ln p q = q log p

Applications of Logarithms

The application of logarithms is enormous inside as well as outside the mathematics subject. Let us discuss brief description of common applications of logarithms in our real life :

• They are used for the calculation of the magnitude of the earthquake.
• Logarithms are being utilized in finding the level of noise in terms of decibels, such as a sound made by a bell.
• In chemistry, the logarithms are applied in order to find acidity or pH level.
• They are used in finding money growth on a certain rate of interest.
• Logarithms are widely used for measuring the time taken by something to decay or grow exponentially, such as bacteria growth, radioactive decay, etc.
• They can also be used in the calculations where multiplication has to be turned into addition or vice versa.

Frequently Asked Questions – FAQs

What are the properties of logarithms, what are the 4 properties of logarithms, what is the purpose of logarithms, how do you use the properties of logarithms, can the base of a log be negative, what are the properties of natural logarithms.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

Visit BYJU’S for all Maths related queries and study materials

Your result is as below

Request OTP on Voice Call

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Post My Comment

• Share Share

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

Properties of Logarithms

When pre-calculus students learn about logarithmic functions, one of the most important lessons they come across is the properties of logarithms. This is because students can simplify and evaluate logarithms with the help of these properties.

Even though log lessons may be challenging, math teachers can help make them more engaging and accessible by using various teaching strategies. We share a few such strategies in this article. Read on and learn more!

Strategies for Teaching Properties of Logarithms

Review logarithms.

Start your lesson on the properties of logarithms by briefly reviewing what logarithms are. Remind students that a logarithm is an exponent. That is, log a x (“log base a of x”) is the exponent to which a must be raised to get x . You can present this in the following manner:

Where a > 0, a ≠ 1, and x > 0.

So logarithms are the opposite of exponentials, as they basically “undo” exponentials. You can also play this video in your class. The video introduces what logarithms represent, by using examples.

Then, check if there are any gaps in what students have learned so far on logarithms. For example, write a simple log on the whiteboard, such as log 2 16 = x, and ask students to transform it into an exponent. Can most students easily say that the equivalent exponent is 2 x = 16 ?

What about evaluating logarithms? Have students acquired proficiency in evaluating a given log? For instance, write log 3 81 = x on the whiteboard. Can students easily determine what this evaluates to, that is, log 3 81 = 4? Practice a bit more and address potential gaps.

For more advanced practice examples on evaluating logarithms, use this brief online activity by Khan Academy. If you require detailed guidelines on teaching logarithms, as well as fun activities to practice logarithms, feel free to check out this article.

Now that you’ve briefly reviewed them, you can proceed with explaining the properties of logarithms. For starters, highlight that we use the properties of logarithms for simplifying and evaluating logarithms.

Add that with the help of these properties, we can rewrite logarithmic expressions, that is, we can expand or condense them. Point out that you will look into three such properties in this class, including:

• multiplication property of logarithms
• division property of logarithms
• power rule of logarithms

Multiplication Property of Logarithms

Explain to students that the multiplication property of logarithms states that the logarithm of the product of two numbers is equal to the sum of individual logarithms of each number. Present this property on the whiteboard in the following way:

Division Property of Logarithms

Point out that according to the division property of logarithms, the logarithm of the quotient of two numbers is equal to the difference of the individual logarithm of each number. Present this property on the whiteboard in the following way:

Power Rule of Logarithms

Finally, explain that the power rule of logarithms states that the logarithm of a number raised to a certain power is equal to the product of power and logarithm of the number. Present this property on the whiteboard in the following way:

log 2 8 + log 2 32 = log 2 (8 × 32)

log 2 8 + log 2 32 = log 2 256

To check if this is correct, we can evaluate the logarithms, that is:

log 2 8 = 3, because 2 3 = 8

log 2 32 = 5, because 2 5 = 32

log 2 256 = 8, because 2 8 = 256

If we simply replace these values above in the statement log 2 8 + log 2 32 = log 2 256, we’ll get the following:

So there you have it! We see that the multiplication property is true.

Additional Resources:

If you have the technical possibilities, you can also complement your lesson with multimedia material, such as videos. For example, use this video by Khan Academy to introduce the multiplication, as well as the division property of logarithms.

Afterward, play this video by Khan Academy to illustrate the power rule of logarithms. By rewriting and simplifying log5(x3) as 3log5(x), the video demonstrates how the logarithmic power rule applies.

Activities to Practice Properties of Logarithms

Properties of logarithms game.

This is a simple online game that helps students sharpen their skills of simplifying and evaluating logarithms with the help of the properties of logarithms. To implement this game in your classroom, make sure that there is a sufficient number of devices for all students.

Students play the game individually, which makes the game suitable for parents who are homeschooling their kids, as well. Students are presented with different tasks, such as being asked to rewrite a log in a certain form by applying the properties of logarithms.

If they get stuck, students can also decide to play a video for help or use a hint. In the end, open space for discussion and reflection. Was any example particularly challenging? Why? Which properties did students use in their exercises?

This is a fun game that will help students improve their knowledge of logarithms and properties of logarithms. To play this game in your class, you’ll need to print out this free Log Race Worksheet , some dice, chips and scissors, markers, and some paper.

Print out as many copies as needed depending on the size of your class. Cut out the task cards from the worksheet and separate them into different piles, depending on what is required in them. For instance, the cards where students should expand a log go into a ‘log’ pile.

Then, draw a game board shaped like a road with designated spaces (squares), as shown on the worksheet, with a ‘start’ and ‘finish’ space. Each space (or square) on the road has a specific instruction, such as ‘expand’, ‘rewrite as exponent’, ‘evaluate’ etc.

Launch the Game!

Divide students into groups of 3, with one person as a ‘checker’. Player 1 rolls the dice and moves their chip on the game board the number of spaces that they got with the dice (ex: if they got a 3 by rolling the dice, they move their chip three spaces on the game board.

The space the student lands on with their chip indicates what kind of a card the student should take. For example, if the student landed on a space that says ‘condense’, they need to draw the top card from the ‘condense’ pile and condense the logarithmic expression written on the card.

Each student has a few minutes to solve the task on their card. If they solve it correctly, they get to roll the dice and move again. If they solve it incorrectly, they lose their turn in the next round. The designated checker in each group checks the answers and keeps track of the scores.

As the name of the game indicates, the goal is to be the first one to reach the ‘finish’ space. If no one manages to reach ‘finish’ by the end of the class, the winner is the player that was closest to ‘finish’.

Before You Leave…

If you enjoyed these tips and activities for teaching properties of logarithms, you may want to check out our lesson that’s dedicated to this topic. So if you need guidance to structure your class and teach pre-calculus, make sure to sign up for more free resources here !

Feel free to also check out our article with free resources on properties of logarithms!

You can also sign up for our membership on MathTeacherCoach or head over to our blog – you’ll find plenty of awesome resources for kids of all ages!

This article is based on:

Unit 3 – Exponential and Logarithmic Functions

• 3-1 Exponential Functions
• 3-2 Logarithmic Functions
• 3-3 Properties of Logarithms
• 3-4 Exponential and Logarithmic Equations
• 3-5 Modeling with Nonlinear Regression

• school Campus Bookshelves
• menu_book Bookshelves
• perm_media Learning Objects
• login Login
• how_to_reg Request Instructor Account
• hub Instructor Commons
• Download Page (PDF)
• Download Full Book (PDF)
• Periodic Table
• Physics Constants
• Scientific Calculator
• Reference & Cite
• Tools expand_more
• Readability

selected template will load here

This action is not available.

3.4: Solving Logarithmic Equations

• Last updated
• Save as PDF
• Page ID 40911

• Richard W. Beveridge
• Clatsop Community College

In the previous section, we took exponential equations and used the properties of logarithms to restate them as logarithmic equations. In this section, we will take logarithmic equations and use properties of logarithms to restate them as exponential equations. In the previous section, we used the property of logarithms that said \(\log _{b} M^{p}=p \log _{b} M .\) In this section, we will make use of two additional properties of logarithms: \[ \log _{b}(M * N)=\log _{b} M+\log _{b} N \] and \[ \log _{b} \frac{M}{N}=\log _{b} M-\log _{b} N \] Just as our previous property of logarithms was simply a restatement of the rules of expoenents, these two properties of logarithms depend on the rules of exponents as well. since we're interested in \(\log _{b} M\) and \(\log _{b} N,\) let's restate these in terms of exponents: If \(\log _{b} M=x\) then \(b^{x}=M\) and if \(\log _{b} N=y\) then \(b^{y}=N\) The properties of logarithms we're interested in justifying have to do with \(M * N\) and \(\frac{M}{N},\) so let's look at those expressions in terms of exponents: \[ \begin{array}{c} M * N=b^{x} * b^{y}=b^{x+y} \\ \text { and } \\ \frac{M}{N}=\frac{b^{x}}{b^{y}}=b^{x-y} \end{array} \]

If we're interested in \(\log _{b}(M * N),\) then we're asking the question "What power do we raise \(b\) to in order to get \(M * N ? "\) We can see above that raising \(b\) to the \(x+y\) power gives us \(M * N .\) since \(x=\log _{b} M\) and \(y=\log _{b} N\) then \(x+y=\) \(\log _{b} M+\log _{b} N,\) so: \[ \log _{b}(M * N)=x+y=\log _{b} M+\log _{b} N \] Likewise, if we're interested in \(\log _{b} \frac{M}{N},\) we're asking the question "What power do we raise \(b\) to in order to get \(\frac{M}{N} ? "\) since raising \(b\) to the \(x-y\) power gives us \(\frac{M}{N}\) and \(x-y=\log _{b} M-\log _{b} N,\) then: \[ \log _{b} \frac{M}{N}=x-y=\log _{b} M-\log _{b} N \]

Let's look at an example to see how we'll use this to solve equations:

Example Solve for \(x\) \[ \log _{2} x+\log _{2}(x-4)=2 \] The first thing we can do here is to combine the two logarithmic statements into one. since \(\log _{b}(M * N)=\log _{b} M+\log _{b} N,\) then \(\log _{2} x+\log _{2}(x-4)=\log _{2}[x(x-4)]\) \[ \begin{aligned} \log _{2} x+\log _{2}(x-4) &=2 \\ \log _{2}[x(x-4)] &=2 \end{aligned} \] Then we'll restate the resulting logarithmic relationship as an exponential relationship: \[ \begin{aligned} 2^{2} &=x(x-4) \\ 4 &=x^{2}-4 x \\ 0 &=x^{2}-4 x-4 \\ 4.828,-0.828 & \approx x \end{aligned} \]

Most textbooks reject answers that result in taking the logarithm of a negative number, such as would be the case for \(x \approx-0.828 .\) However, the logarithms of negative numbers result in complex valued answers, rather than an undefined quantity. For that reason, in this text, we will include all answers.

If a problem involves a difference of logarithms, we can use the other property of logarithms introduced in this section. Example Solve for \(x\) \[ \log (5 x-1)-\log (x-2)=2 \] Again, our first step is to restate the difference of logarithms using the property \(\log _{b} \frac{M}{N}=\log _{b} M-\log _{b} N:\) \[ \begin{aligned} \log (5 x-1) &-\log (x-2)=2 \\ \log \left[\frac{5 x-1}{x-2}\right] &=2 \end{aligned} \] We're working with a logarithm in base 10 in this problem, so in our next step we'll say: \[ \begin{array}{c} \log \left[\frac{5 x-1}{x-2}\right]=2 \\ \frac{5 x-1}{x-2}=10^{2} \end{array} \]

Then multiply on both sides by \(x-2\) \[ \begin{array}{c} 10^{2}=\frac{5 x-1}{x-2} \\ (x-2) * 100=\frac{5 x-1}{\cancel{(x-2)}} *\cancel{(x-2)} \end{array} \]

And, solve for \(x\) \[ \begin{array}{c} 100 x-200=5 x-1 \\ 95 x=199 \\ x=\frac{199}{95} \end{array} \] In some equations, all of the terms are stated using logartihms. These equations often come out in a form that says \(\log _{b} x=\log _{b} y .\) If this is the case, we can then conclude that \(x=y\)

It seems reasonable that if the exponent we raise \(b\) to in order to get \(x\) is the same exponent that we raise \(b\) to in order to get \(y,\) then \(x\) and \(y\) are the same thing. Assume: \[ \log _{b} x=\log _{b} y \] then \[ b^{a}=x \text { and } b^{a}=y \] if both \(x\) and \(y\) are equal to \(b^{a},\) then \(x=y\)

Example Solve for \(x\) \(\log _{5}(4-x)=\log _{5}(x+8)+\log _{5}(2 x+13)\)

First, let's use the properties of logarithms to restate the equation so that there is only one logarithm on each side. \[ \begin{array}{l} \log _{5}(4-x)=\log _{5}(x+8)+\log _{5}(2 x+13) \\ \log _{5}(4-x)=\log _{5}[(x+8)(2 x+13)] \end{array} \] Then, we'll use the property of logarithms we just discussed: \[ \text { If } \log _{b} x=\log _{b} y \] then \[ \begin{array}{c} x=y \\ \log _{5}(4-x)=\log _{5}[(x+8)(2 x+13)] \\ 4-x=(x+8)(2 x+13) \\ 0=2 x^{2}+29 x+104 \\ 0=2(x+5)(x+10) \\ -5,-10=x \end{array} \]

Exercises 3.4

Solve for the indicated variable in each equation. 1) \(\quad \log _{3} 5+\log _{3} x=2\) 2) \(\quad \log _{4} x+\log _{4} 5=1\) 3) \(\quad \log _{2} x=2+\log _{2} 3\) 4) \(\quad \log _{5} x=2+\log _{5} 3\) 5) \(\quad \log _{3} x+\log _{3}(x-8)=2\) 6) \(\quad \log _{6} x+\log _{6}(x-5)=1\) 7) \(\quad \log (3 x+2)=\log (x-4)+1\) 8) \(\quad \log (x-1)-\log x=-0.5\) 9) \(\quad \log _{2} a+\log _{2}(a+2)=3\) 10) \(\quad \log _{3} x+\log _{3}(x-2)=1\) 11) \(\quad \log _{2} y-\log _{2}(y-2)=3\) 12) \(\quad \log _{2} x-\log _{2}(x+3)=2\) 13) \(\quad \log _{3} x+\log _{3}(x+4)=2\) 14) \(\quad \log _{4} u+\log _{4}(u+1)=1\) 15) \(\quad \log 5+\log x=\log 6\) 16) \(\quad \ln x+\ln 4=\ln 2\) 17) \(\quad \log _{7} x-\log _{7} 12=\log _{7} 2\) 18) \(\quad \log 2-\log x=\log 8\) 19) \(\quad \log _{3} x-\log _{3}(x-2)=\log _{3} 4\) 20) \(\quad \log _{6} 2-\log _{6}(x-2)=\log _{6} 9\) 21) \(\quad \log _{4} x-\log _{4}(x-4)=\log _{4}(x-6)\) 22) \(\quad \log _{9}(2 x+7)-\log _{9}(x-1)=\log _{9}(x-7)\) 23) \(\quad 2 \log _{2} x=\log _{2}(2 x-1)\) 24) \(\quad 2 \log _{4} y=\log _{4}(y+2)\) 25) \(\quad 2 \log (x-3)-3 \log 2=1\) 26) \(\quad 2 \log _{5} 7-\log _{5}(x+1)=\log _{5}(2 x-5)\)