6.6: Logarithmic Properties (2024)

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Learning Objectives

In this section, you will:

Use the product rule for logarithms.
Use the quotient rule for logarithms.
Use the power rule for logarithms.
Expand logarithmic expressions.
Condense logarithmic expressions.
Use the change-of-base formula for logarithms.

Figure1The pH ofhydrochloric acid is tested with litmus paper. (credit: DavidBerardan)

In chemistry,pHis used as a measure of the acidityor alkalinity of a substance. The pH scale runs from 0 to 14.Substances with a pH less than 7 are considered acidic, andsubstances with a pH greater than 7 are said to be basic. Ourbodies, for instance, must maintain a pH close to 7.35 in order forenzymes to work properly. To get a feel for what is acidic and whatis basic, consider the following pH levels of some commonsubstances:

Battery acid: 0.8
Stomach acid: 2.7
Orange juice: 3.3
Pure water: 7 (at 25° C)
Human blood: 7.35
Fresh coconut: 7.8
Sodium hydroxide (lye): 14

To determine whether a solution is acidic or basic, we find itspH, which is a measure of the number of active positive hydrogenions in the solution. The pH is defined by the following formula,whereH+H+is the concentration of hydrogen ion in thesolution

pH=−log([H+])=log(1[H+])pH=−log([H+])=log(1[H+])

The equivalenceof−log([H+])−log([H+])andlog(1[H+])log(1[H+])isone of the logarithm properties we will examine in thissection.

Using the Product Rule forLogarithms

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

logb1=0logbb=1logb1=0logbb=1

Forexample,log51=0log51=0since50=1.50=1.Andlog55=1log55=1since51=5.51=5.

Next, we have the inverse property.

logb(bx)=xblogbx=x,x>0logb(bx)=xblogbx=x,x>0

For example, to evaluatelog(100),log(100),we canrewrite the logarithm aslog10(102),log10(102),and thenapply the inverse propertylogb(bx)=xlogb(bx)=xtogetlog10(102)=2.log10(102)=2.

To evaluateeln(7),eln(7),we can rewrite thelogarithm aseloge7,eloge7,and then apply the inversepropertyblogbx=xblogbx=xtogeteloge7=7.eloge7=7.

Finally, we have theone-to-oneproperty.

logbM=logbNifandonlyifM=NlogbM=logbNifandonlyifM=N

We can use the one-to-one property to solve theequationlog3(3x)=log3(2x+5)log3(3x)=log3(2x+5)forx.x.Sincethe bases are the same, we can apply the one-to-one property bysetting the arguments equal and solving forx:x:

3x=2x+5x=5Settheargumentsequal.Subtract2x.3x=2x+5Settheargumentsequal.x=5Subtract2x.

But what about theequationlog3(3x)+log3(2x+5)=2?log3(3x)+log3(2x+5)=2?Theone-to-one property does not help us in this instance. Before wecan solve an equation like this, we need a method for combiningterms on the left side of the equation.

Recall that we use theproductrule of exponentsto combine the product of powers byadding exponents:xaxb=xa+b.xaxb=xa+b.We have a similarproperty for logarithms, called theproduct rule forlogarithms, which says that the logarithm of a product isequal to a sum of logarithms. Because logs are exponents, and wemultiply like bases, we can add the exponents. We will use theinverse property to derive the product rule below.

Given any real numberxxand positive realnumbersM,N,M,N,andb,b,whereb≠1,b≠1,wewill show

logb(MN)=logb(M)+logb(N).logb(MN)=logb(M)+logb(N).

Letm=logbMm=logbMandn=logbN.n=logbN.Inexponential form, these equationsarebm=Mbm=Mandbn=N.bn=N.It follows that

logb(MN)=logb(bmbn)=logb(bm+n)=m+n=logb(M)+logb(N)SubstituteforMandN.Applytheproductruleforexponents.Applytheinversepropertyoflogs.Substituteformandn.logb(MN)=logb(bmbn)SubstituteforMandN.=logb(bm+n)Applytheproductruleforexponents.=m+nApplytheinversepropertyoflogs.=logb(M)+logb(N)Substituteformandn.

Note that repeated applications of the product rule forlogarithms allow us to simplify the logarithm of the product of anynumber of factors. For example,considerlogb(wxyz).logb(wxyz).Using the product rulefor logarithms, we can rewrite this logarithm of a product as thesum of logarithms of its factors:

logb(wxyz)=logbw+logbx+logby+logbzlogb(wxyz)=logbw+logbx+logby+logbz

THEPRODUCT RULE FOR LOGARITHMS

Theproduct rule for logarithmscanbe used to simplify a logarithm of a product by rewriting it as asum of individual logarithms.

logb(MN)=logb(M)+logb(N)forb>0logb(MN)=logb(M)+logb(N)forb>0

HOWTO

Given the logarithm of a product, use the product ruleof logarithms to write an equivalent sum oflogarithms.

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

EXAMPLE1

Using the Product Rule forLogarithms

Expandlog3(30x(3x+4)).log3(30x(3x+4)).

Answer

TRYIT#1

Expandlogb(8k).logb(8k).

Using the Quotient Rule forLogarithms

For quotients, we have a similar rule for logarithms. Recallthat we use thequotient rule ofexponentsto combine the quotient of exponents bysubtracting:xaxb=xa−b.xaxb=xa−b.Thequotientrule for logarithmssays that the logarithm of aquotient is equal to a difference of logarithms. Just as with theproduct rule, we can use the inverse property to derive thequotient rule.

Given any real numberxxand positive realnumbersM,M,N,N,andb,b,whereb≠1,b≠1,wewill show

logb(MN)=logb(M)−logb(N).logb(MN)=logb(M)−logb(N).

Letm=logbMm=logbMandn=logbN.n=logbN.Inexponential form, these equationsarebm=Mbm=Mandbn=N.bn=N.It follows that

logb(MN)=logb(bmbn)=logb(bm−n)=m−n=logb(M)−logb(N)SubstituteforMandN.Applythequotientruleforexponents.Applytheinversepropertyoflogs.Substituteformandn.logb(MN)=logb(bmbn)SubstituteforMandN.=logb(bm−n)Applythequotientruleforexponents.=m−nApplytheinversepropertyoflogs.=logb(M)−logb(N)Substituteformandn.

For example, toexpandlog(2x2+6x3x+9),log(2x2+6x3x+9),we must firstexpress the quotient in lowest terms. Factoring and canceling weget,

log(2x2+6x3x+9)=log(2x(x+3)3(x+3))=log(2x3)Factorthenumeratoranddenominator.Cancelthecommonfactors.log(2x2+6x3x+9)=log(2x(x+3)3(x+3))Factorthenumeratoranddenominator.=log(2x3)Cancelthecommonfactors.

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

log(2x3)=log(2x)−log(3)=log(2)+log(x)−log(3)log(2x3)=log(2x)−log(3)=log(2)+log(x)−log(3)

THEQUOTIENT RULE FOR LOGARITHMS

Thequotient rule for logarithmscanbe used to simplify a logarithm or a quotient by rewriting it asthe difference of individual logarithms.

logb(MN)=logbM−logbNlogb(MN)=logbM−logbN

HOWTO

Given the logarithm of a quotient, use the quotient ruleof logarithms to write an equivalent difference oflogarithms.

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

EXAMPLE2

Using the Quotient Rule forLogarithms

Expandlog2(15x(x−1)(3x+4)(2−x)).log2(15x(x−1)(3x+4)(2−x)).

Answer

Analysis

There are exceptions to consider in this and later examples.First, because denominators must never be zero, this expression isnot defined forx=−43x=−43andx=2.x=2.Also,since the argument of a logarithm must be positive, we note as weobserve the expanded logarithm,thatx>0,x>0,x>1,x>1,x>−43,x>−43,andx<2.x<2.Combiningthese conditions is beyond the scope of this section, and we willnot consider them here or in subsequent exercises.

TRYIT#2

Expandlog3(7x2+21x7x(x−1)(x−2)).log3(7x2+21x7x(x−1)(x−2)).

Using the Power Rule forLogarithms

We’ve explored the product rule and the quotient rule, but howcan we take the logarithm of a power, such asx2?x2?Onemethod is as follows:

logb(x2)=logb(x⋅x)=logbx+logbx=2logbxlogb(x2)=logb(x⋅x)=logbx+logbx=2logbx

Notice that we used theproduct rule for logarithmstofind a solution for the example above. By doing so, we have derivedthepower rule for logarithms, which saysthat the log of a power is equal to the exponent times the log ofthe base. Keep in mind that, although the input to a logarithm maynot be written as a power, we may be able to change it to a power.For example,

100=1023–√=3121e=e−1100=1023=3121e=e−1

THEPOWER RULE FOR LOGARITHMS

Thepower rule for logarithmscan beused to simplify the logarithm of a power by rewriting it as theproduct of the exponent times the logarithm of the base.

logb(Mn)=nlogbMlogb(Mn)=nlogbM

HOWTO

Given the logarithm of a power, use the power rule oflogarithms to write an equivalent product of a factor and alogarithm.

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

EXAMPLE3

Expanding a Logarithm withPowers

Expandlog2x5.log2x5.

Answer

TRYIT#3

Expandlnx2.lnx2.

EXAMPLE4

Rewriting an Expression as a Powerbefore Using the Power Rule

Expandlog3(25)log3(25)using the power rule forlogs.

Answer

TRYIT#4

Expandln(1x2).ln(1x2).

EXAMPLE5

Using the Power Rule inReverse

Rewrite4ln(x)4ln(x)using the power rule for logs toa single logarithm with a leading coefficient of 1.

Answer

TRYIT#5

Rewrite2log342log34using the power rule for logs toa single logarithm with a leading coefficient of 1.

Expanding LogarithmicExpressions

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

logb(6xy)=logb(6x)−logby=logb6+logbx−logbylogb(6xy)=logb(6x)−logby=logb6+logbx−logby

We can use the power rule to expand logarithmic expressionsinvolving negative and fractional exponents. Here is an alternateproof of the quotient rule for logarithms using the fact that areciprocal is a negative power:

logb(AC)=logb(AC−1)=logb(A)+logb(C−1)=logbA+(−1)logbC=logbA−logbClogb(AC)=logb(AC−1)=logb(A)+logb(C−1)=logbA+(−1)logbC=logbA−logbC

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

With practice, we can look at a logarithmic expression andexpand it mentally, writing the final answer. Remember, however,that we can only do this with products, quotients, powers, androots—never with addition or subtraction inside the argument of thelogarithm.

EXAMPLE6

Expanding Logarithms Using Product,Quotient, and Power Rules

Rewriteln(x4y7)ln(x4y7)as a sum or difference oflogs.

Answer

TRYIT#6

Expandlog(x2y3z4).log(x2y3z4).

EXAMPLE7

Using the Power Rule for Logarithmsto Simplify the Logarithm of a Radical Expression

Expandlog(x−−√).log(x).

Answer

TRYIT#7

Expandln(x2−−√3).ln(x23).

Q&A

Can weexpandln(x2+y2)?ln(x2+y2)?

No. There is no way to expand thelogarithm of a sum or difference inside the argument of thelogarithm.

EXAMPLE8

Expanding Complex LogarithmicExpressions

Expandlog6(64x3(4x+1)(2x−1)).log6(64x3(4x+1)(2x−1)).

Answer

TRYIT#8

Expandln((x−1)(2x+1)2√(x2−9)).ln((x−1)(2x+1)2(x2−9)).

Condensing LogarithmicExpressions

We can use the rules of logarithms we just learned to condensesums, differences, and products with the same base as a singlelogarithm. It is important to remember that the logarithms musthave the same base to be combined. We will learn later how tochange the base of any logarithm before condensing.

HOWTO

Given a sum, difference, or product of logarithms withthe same base, write an equivalent expression as a singlelogarithm.

Apply the power property first. Identify terms that areproducts of factors and a logarithm, and rewrite each as thelogarithm of a power.
Next apply the product property. Rewrite sums of logarithms asthe logarithm of a product.
Apply the quotient property last. Rewrite differences oflogarithms as the logarithm of a quotient.

EXAMPLE9

Using the Product and Quotient Rulesto Combine Logarithms

Writelog3(5)+log3(8)−log3(2)log3(5)+log3(8)−log3(2)as asingle logarithm.

Answer

TRYIT#9

Condenselog3−log4+log5−log6.log3−log4+log5−log6.

EXAMPLE10

Condensing Complex LogarithmicExpressions

Condenselog2(x2)+12log2(x−1)−3log2((x+3)2).log2(x2)+12log2(x−1)−3log2((x+3)2).

Answer

TRYIT#10

Rewritelog(5)+0.5log(x)−log(7x−1)+3log(x−1)log(5)+0.5log(x)−log(7x−1)+3log(x−1)asa single logarithm.

EXAMPLE11

Rewriting as a SingleLogarithm

Rewrite2logx−4log(x+5)+1xlog(3x+5)2logx−4log(x+5)+1xlog(3x+5)asa single logarithm.

Answer

TRYIT#11

Condense4(3log(x)+log(x+5)−log(2x+3)).4(3log(x)+log(x+5)−log(2x+3)).

EXAMPLE12

Applying of the Laws ofLogs

Recall that, in chemistry,pH=−log[H+].pH=−log[H+].Ifthe concentration of hydrogen ions in a liquid is doubled, what isthe effect on pH?

Answer

TRYIT#12

How does the pH change when the concentration of positivehydrogen ions is decreased by half?

Using the Change-of-Base Formula forLogarithms

Most calculators can evaluate only common and natural logs. Inorder to evaluate logarithms with a base other than 10ore,e,we use thechange-of-baseformulato rewrite the logarithm as the quotient oflogarithms of any other base; when using a calculator, we wouldchange them to common or natural logs.

To derive the change-of-base formula, we usetheone-to-oneproperty andpower rule forlogarithms.

Given any positive realnumbersM,b,M,b,andn,n,wheren≠1n≠1andb≠1,b≠1,weshow

logbM=lognMlognblogbM=lognMlognb

Lety=logbM.y=logbM.By exponentiating both sides withbasebb, we arrive at an exponential form,namelyby=M.by=M.It follows that

logn(by)ylognbylogbM=lognM=lognM=lognMlognb=lognMlognbApplytheone-to-oneproperty.Applythepowerruleforlogarithms.Isolatey.Substitutefory.logn(by)=lognMApplytheone-to-oneproperty.ylognb=lognMApplythepowerruleforlogarithms.y=lognMlognbIsolatey.logbM=lognMlognbSubstitutefory.

For example, to evaluatelog536log536using acalculator, we must first rewrite the expression as a quotient ofcommon or natural logs. We will use the common log.

log536=log(36)log(5)≈2.2266Applythechangeofbaseformulausingbase10.Useacalculatortoevaluateto4decimalplaces.log536=log(36)log(5)Applythechangeofbaseformulausingbase10.≈2.2266Useacalculatortoevaluateto4decimalplaces.

THECHANGE-OF-BASE FORMULA

Thechange-of-base formulacan beused to evaluate a logarithm with any base.

For any positive realnumbersM,b,M,b,andn,n,wheren≠1n≠1andb≠1,b≠1,

logbM=lognMlognb.logbM=lognMlognb.

It follows that the change-of-base formula can be used torewrite a logarithm with any base as the quotient of common ornatural logs.

logbM=lnMlnblogbM=lnMlnb

and

logbM=logMlogblogbM=logMlogb

HOWTO

Given a logarithm with theformlogbM,logbM,use the change-of-base formula torewrite it as a quotient of logs with any positivebasen,n,wheren≠1.n≠1.

Determine the new basen,n,remembering that thecommon log,log(x),log(x),has base 10, and the naturallog,ln(x),ln(x),has basee.e.
Rewrite the log as a quotient using the change-of-base formula
1. The numerator of the quotient will be a logarithm withbasennand argumentM.M.
2. The denominator of the quotient will be a logarithm withbasennand argumentb.b.

EXAMPLE13

Changing Logarithmic Expressions toExpressions Involving Only Natural Logs

Changelog53log53to a quotient of naturallogarithms.

Answer

TRYIT#13

Changelog0.58log0.58to a quotient of naturallogarithms.

Q&A

Can we change common logarithms to naturallogarithms?

Yes. Rememberthatlog9log9meanslog109.log109.So,log9=ln9ln10.log9=ln9ln10.

EXAMPLE14

Using the Change-of-Base Formula witha Calculator

Evaluatelog2(10)log2(10)using the change-of-baseformula with a calculator.

Answer

TRYIT#14

Evaluatelog5(100)log5(100)using the change-of-baseformula.

MEDIA

Access these online resources for additional instruction andpractice with laws of logarithms.

6.5 SectionExercises

Verbal

How does the power rule for logarithms help when solvinglogarithms with the formlogb(x−−√n)?logb(xn)?

What does the change-of-base formula do? Why is it useful whenusing a calculator?

Algebraic

For the following exercises, expand each logarithm as much aspossible. Rewrite each expression as a sum, difference, or productof logs.

logb(7x⋅2y)logb(7x⋅2y)

ln(3ab⋅5c)ln(3ab⋅5c)

logb(1317)logb(1317)

log4(xzw)log4(xzw)

ln(14k)ln(14k)

log2(yx)log2(yx)

For the following exercises, condense to a single logarithm ifpossible.

ln(7)+ln(x)+ln(y)ln(7)+ln(x)+ln(y)

10.

log3(2)+log3(a)+log3(11)+log3(b)log3(2)+log3(a)+log3(11)+log3(b)

11.

logb(28)−logb(7)logb(28)−logb(7)

12.

ln(a)−ln(d)−ln(c)ln(a)−ln(d)−ln(c)

13.

−logb(17)−logb(17)

14.

13ln(8)13ln(8)

For the following exercises, use the properties of logarithms toexpand each logarithm as much as possible. Rewrite each expressionas a sum, difference, or product of logs.

15.

log(x15y13z19)log(x15y13z19)

16.

ln(a−2b−4c5)ln(a−2b−4c5)

17.

log(x3y−4−−−−−√)log(x3y−4)

18.

ln(yy1−y−−−√)ln(yy1−y)

19.

log(x2y3x2y5−−−−√3)log(x2y3x2y53)

For the following exercises, condense each expression to asingle logarithm using the properties of logarithms.

20.

log(2x4)+log(3x5)log(2x4)+log(3x5)

21.

ln(6x9)−ln(3x2)ln(6x9)−ln(3x2)

22.

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

23.

log(x)−12log(y)+3log(z)log(x)−12log(y)+3log(z)

24.

4log7(c)+log7(a)3+log7(b)34log7(c)+log7(a)3+log7(b)3

For the following exercises, rewrite each expression as anequivalent ratio of logs using the indicated base.

25.

log7(15)log7(15)to baseee

26.

log14(55.875)log14(55.875)to base1010

For the following exercises,supposelog5(6)=alog5(6)=aandlog5(11)=b.log5(11)=b.Usethe change-of-base formula along with properties of logarithms torewrite each expression in termsofaaandb.b.Show the steps for solving.

27.

log11(5)log11(5)

28.

log6(55)log6(55)

29.

log11(611)log11(611)

Numeric

For the following exercises, use properties of logarithms toevaluate without using a calculator.

30.

log3(19)−3log3(3)log3(19)−3log3(3)

31.

6log8(2)+log8(64)3log8(4)6log8(2)+log8(64)3log8(4)

32.

2log9(3)−4log9(3)+log9(1729)2log9(3)−4log9(3)+log9(1729)

For the following exercises, use the change-of-base formula toevaluate each expression as a quotient of natural logs. Use acalculator to approximate each to five decimal places.

33.

log3(22)log3(22)

34.

log8(65)log8(65)

35.

log6(5.38)log6(5.38)

36.

log4(152)log4(152)

37.

log12(4.7)log12(4.7)

Extensions

38.

Use the product rule for logarithms to findallxxvalues suchthatlog12(2x+6)+log12(x+2)=2.log12(2x+6)+log12(x+2)=2.Showthe steps for solving.

39.

Use the quotient rule for logarithms to findallxxvalues suchthatlog6(x+2)−log6(x−3)=1.log6(x+2)−log6(x−3)=1.Showthe steps for solving.

40.

Can the power property of logarithms be derived from the powerproperty of exponents using the equationbx=m?bx=m?Ifnot, explain why. If so, show the derivation.

41.

Prove thatlogb(n)=1logn(b)logb(n)=1logn(b)for anypositiveintegersb>1b>1andn>1.n>1.

42.

Doeslog81(2401)=log3(7)?log81(2401)=log3(7)?Verifythe claim algebraically.