Chapter6/7-Logarithmicand

ExponentialFunctions

LessonPackage

MHF4U

Chapter6/7OutlineUnitGoal:Bytheendofthisunit,youwillbeabletodemonstrateanunderstandingoftherelationshipbetweenexponentialandlogarithmicexpressions.Youwillalsobeabletosolveexponentialandlogarithmicequations.

Section Subject LearningGoals CurriculumExpectations

L1 LogasInverse

-recognizetheoperationoffindingthelogarithmtobetheinverseoperationofexponentiation-evaluatesimplelogarithmicexpressions-understandthatthelogarithmofanumbertoagivenbaseistheexponenttowhichthebasemustberaisedtogetthenumber

A1.1,1.2,1.3,2.1,2.2

L2 PowerLawofLogarithms

-uselawsoflogarithmstosimplifyexpressions-understandchangeofbaseformula A1.4

L3 ProductandQuotientLawsofLogarithms

-uselawsoflogarithmstosimplifyexpressions A1.4

L4 SolvingExponentialEquations

-recognizeequivalentalgebraicexpressions-solveexponentialequations A3.1,3.2

L5 SolvingLogarithmicEquations

-solvelogarithmicequations A3.3

L5 ApplicationsofLogarithms

-Solveproblemsarisingfromrealworldapplicationsinvolvingexponentialandlogarithmicequations

A3.4

Assessments F/A/O MinistryCode P/O/C KTACNoteCompletion A P PracticeWorksheetCompletion F/A P

Quiz–LogRules F P PreTestReview F/A P Test–LogandExponentialFuncitons O

A1.1,1.2,1.3,1.4A2.1,2.2

A3.1,3.2,3.3,3.4P K(21%),T(34%),A(10%),

C(34%)

L1–6.1/6.2–IntrotoLogarithmsandReviewofExponentialsMHF4UJensenInthissectionyouwilllearnabouthowalogarithmicfunctionistheinverseofanexponentialfunction.Youwillalsolearnhowtoexpressexponentialequationsinlogarithmicform.Part1:ReviewofExponentialFunctionsEquation:𝒚 = 𝒂(𝒃)𝒙𝑎 =initialamount𝑏 =growth(𝑏 > 1)ordecay(0 < 𝑏 < 1)factor𝑦 =futureamount𝑥 =numberoftimes𝑎hasincreasedordecreasedTocalculate𝑥,usetheequation:𝑥 = 010230567

05675002879:1;1<7=;1>0?1;@7A2BC7;51@

Example1:Aninsectcolonyhasacurrentpopulationof50insects.Itspopulationdoublesevery3days.

a)Whatisthepopulationafter12days?

𝑦 = 50 2FGH

𝑦 = 50 2 I𝑦 = 800b)Howlonguntilthepopulationreaches25600?

25600 = 50 20H

512 = 20H

log 512 = log 20H

log 512 =𝑡3 log 2

log 512log 2 =

𝑡3

9 =𝑡3

𝑡 = 27days

Part2:ReviewofInverseFunctionsInverse of a function: · The inverse of a function f is denoted as 𝑓TF

· The function and its inverse have the property that if f(a) = b, then 𝑓TF(b) = a · So if 𝑓 5 = 13, then 𝑓TF 13 = 5 · More simply put: The inverse of a function has all the same points as the original function, except that the x's and y's have been reversed. The graph of 𝑓TF(𝑥) is the graph of 𝑓(𝑥) reflected in the line 𝑦 = 𝑥. This is true for all functions and their inverses. Example 2: Determine the equation of the inverse of the function 𝑓 𝑥 = 3(𝑥 − 5)G + 1 𝑦 = 3(𝑥 − 5)G + 1 𝑥 = 3(𝑦 − 5)G + 1 𝑥 − 13 = (𝑦 − 5)G

±𝑥 − 13 = 𝑦 − 5

5 ±𝑥 − 13 = 𝑦

Part3:ReviewofExponentLaws

Name Rule

ProductRule 𝑥2 ∙ 𝑥Y = 𝑥2ZY

QuotientRule𝑥2

𝑥Y = 𝑥2TY

PowerofaPowerRule 𝑥2 Y = 𝑥2×Y

NegativeExponentRule 𝑥T2 = F\]

ExponentofZero 𝑥^ = 1

Equationofinverse:

𝑓TF(𝑥) = 5 ± _𝑥 − 13

Part4:InverseofanExponentialFunctionExample3:a)Findtheequationoftheinverseof𝑓 𝑥 = 2\.𝑦 = 2\𝑥 = 2Blog 𝑥 = log 2Blog 𝑥 = 𝑦 log 2

𝑦 =log 𝑥log 2

𝑦 = logG 𝑥𝑓TF 𝑥 = logG 𝑥b)Graphtheboth𝑓 𝑥 and𝑓TF(𝑥).

Note:justswap𝑥and𝑦coordinatestogetkeypointsfortheinverseofafunction.Thegraphshouldappeartobeareflectionacrosstheline𝑦 = 𝑥.c)Completethechartofkeypropertiesforbothfunctions

𝒚 = 𝟐𝒙 𝒚 = 𝐥𝐨𝐠𝟐 𝒙𝑥-int:none 𝑥-int:(1,0)𝑦-int:(0,1) 𝑦-int:noneDomain:{𝑋 ∈ ℝ} Domain: 𝑋 ∈ ℝ 𝑥 > 0}Range: 𝑌 ∈ ℝ 𝑦 > 0} Range:{𝑌 ∈ ℝ}Asymptote:horizontalasymptoteat𝑦 = 0 Asymptote:verticalasymptoteat𝑥 = 0

𝒇 𝒙 = 𝟐𝒙𝒙 𝒚-2 0.25-1 0.50 11 22 4

𝒇T𝟏 𝒙 = 𝒍𝒐𝒈𝟐𝒙𝒙 𝒚

0.25 -20.5 -11 02 14 2

Thisstepusesthe‘changeofbase’formulathatwewillcoverlaterintheunit.

logY 𝑚 = pqr6pqr Y

Part5:WhatisaLogarithmicFunction?Thelogarithmicfunctionistheinverseoftheexponentialfunctionwiththesamebase.

Inotherwords,thesolutiontoalogarithmisalwaysanEXPONENT.ThelogarithmicfunctionismostusefulforsolvingforunknownexponentsCommonlogarithmsarelogarithmswithabaseof10.Itisnotnecessarytowritethebaseforcommonlogarithms:log 𝑥meansthesameaslogF^ 𝑥Part6:WritingEquivalentExponentialandLogarithmicExpressionsExponentialequationscanbewritteninlogarithmicform,andviceversa𝑦 = 𝑏\àx = logY 𝑦𝑦 = logY 𝑥à𝑥 = 𝑏BExample4:Rewriteeachequationinlogarithmicform

Example5:Writeeachlogarithmicequationinexponentialform

Thelogarithmicfunctionisdefinedas𝑦 = logY 𝑥,or𝑦equalsthelogarithmof𝑥tothebase𝑏.Thefunctionisdefinedonlyfor𝒃 > 𝟎, 𝒃 ≠ 𝟏Inthisnotation,𝒚istheexponenttowhichthebase,𝒃,mustberaisedtogivethevalueof𝒙.

a)16 = 2IlogG 16 = 4

b)𝑚 = 𝑛Hlog<𝑚 = 3

c)3TG = Fy

logH z19{ = −2

a)logI 64 = 34H = 64

b)𝑦 = log𝑥10B = 𝑥Note:becausethereisnobasewritten,thisisunderstoodtobethecommonlogarithmof𝑥.

Part7:EvaluateaLogarithmExample6:Evaluateeachlogarithmwithoutacalculator

Note:eitheroftherulespresentedaboveareappropriatetouseforevaluatinglogarithmicexpressions

Rule:if𝑥2 = 𝑥Y,then𝑎 = 𝑏a)𝑦 = logH 813B = 813B = 3I𝑦 = 4

b)𝑦 = log| FF^^}

10B = F

F^^

10B = | FF^}G

10B = 10TG𝑦 = −2

c)𝑦 = logG |F~}

𝑦 = logG |FG}H

𝑦 = logG 2TH𝑦 = −3

Rule:log2(𝑎Y) = 𝑏a)𝑦 = logI 64𝑦 = logI(4H)𝑦 = 3

Writeinexponentialform

Raisebothsidestotheexponentof𝑛

Applypowerlawofexponents

Writeasalogarithmicexpression

Substitute𝑤 = log' 𝑥

L2–6.4–PowerLawofLogarithmsMHF4UJensenPart1:SolvingforanUnknownExponentExample1:Supposeyouinvest$100inanaccountthatpays5%interest,compoundedannually.Theamount,𝐴,indollars,intheaccountafteranygiventime,𝑡,inyears,isgivenby𝐴 = 100 1.05 /.Howlongwillittakefortheamountinthisaccounttodouble?200 = 100 1.05 /2 = 1.05 /log 2 = log 1.05/log 2 = 𝑡 log 1.05

𝑡 =log 2log 1.05

𝑡 ≅ 14.2yearsInthisexample,weusedthepowerlawoflogarithmstohelpsolveforanunknownexponent.

ProofofPowerLawofLogarithms:Let𝑤 = log' 𝑥

𝑤 = log' 𝑥

𝑥 = 𝑏4

𝑥5 = 𝑏4 5

𝑥5 = 𝑏45

log' 𝑥5 = 𝑤𝑛

log' 𝑥5 = 𝑛 log' 𝑥

PowerLawofLogarithms:

𝐥𝐨𝐠𝒃 𝒙𝒏 = 𝒏 𝐥𝐨𝐠𝒃 𝒙,𝑏 > 0, 𝑏 ≠ 1, 𝑥 > 0

Part2:PracticethePowerLawofLogarithmsExample2:Evaluateeachofthefollowinga)log@ 9B

Method1:SimplifyandEvaluateusingrulesfromlastlessonRule:logC(𝑎') = 𝑏log@ 9B = log@(3H)B = log@ 3I = 8

Method2:UsePowerLawofLogarithmsRule:log' 𝑥5 = 𝑛 log' 𝑥log@ 9B = 4 log@ 9 = 4 log@ 3H = 4(2) = 8

b)logH 8KlogH 8K = 5 logH(2@) = 5(3) = 15

c)logK √125

logK √125 =12 logK

(5@) = M

H(3)

= @

H

Part3:ChangeofBaseFormulaThinkingbacktoexample1,wehadtheequation:2 = 1.05/WecouldhavewrittenthisinlogarithmicformaslogM.NK 2 = 𝑡,butunfortunately,thereisnoeasywaytochange2toapowerwithbase1.05andyoucan’tjusttypeonyourcalculatortoevaluatebecausemostscientificcalculatorscanonlyevaluatelogarithmsinbase10.Soweusedthepowerlawoflogarithmsinstead.Anytimeyouwanttoevaluatealogarithmthatisnotbase10,suchaslogM.NK 2,youcanusetheCHANGEOFBASEFORMULA:

Usingthisformula,wecoulddeterminethatlogM.NK 2 =

OPQ HOPQ M.NK

,whichisexactlywhatweendedupwithby

usingthepowerlawoflogarithms.Part4:EvaluateLogarithmswithVariousBasesExample3:Evaluate,correcttothreedecimalplaces

Example4:Solvefor𝑦intheequation100 = 2S

OR

Tocalculatealogarithmwithanybase,expressintermsofcommonlogarithmsusethechangeofbaseformula:

𝐥𝐨𝐠𝒃 𝒎 = 𝐥𝐨𝐠𝒎𝐥𝐨𝐠 𝒃

,𝑚 > 0, 𝑏 > 0, 𝑏 ≠ 1

a)logK 17

=log17log5

≅ 1.760

b)logXY10

= OPQ MN

OPQZXY[

≅ −3.322

𝑦 = logH 100

𝑦 =log100log2

𝑦 ≅ 6.644

log100 = log2S log100 = 𝑦 log2

𝑦 =log100log2

𝑦 ≅ 6.644

L3–7.3–ProductandQuotientLawsofLogarithmsMHF4UJensenPart1:ProofofProductLawofLogarithmsLet𝑥 = log& 𝑚and𝑦 = log& 𝑛Writteninexponentialform:𝑏+ = 𝑚and𝑏, = 𝑛𝑚𝑛 = 𝑏+𝑏,𝑚𝑛 = 𝑏+-,log& 𝑚𝑛 = 𝑥 + 𝑦log& 𝑚𝑛 = log& 𝑚 + log& 𝑛Part2:SummaryofLogRules

PowerLawofLogarithms 𝐥𝐨𝐠𝒃 𝒙𝒏 = 𝒏 𝐥𝐨𝐠𝒃 𝒙for𝑏 > 0, 𝑏 ≠ 1, 𝑥 > 0

ProductLawofLogarithms 𝐥𝐨𝐠𝒃 𝒎𝒏 = 𝐥𝐨𝐠𝒃 𝒎 + 𝐥𝐨𝐠𝒃 𝒏for𝑏 > 0, 𝑏 ≠ 1,𝑚 > 0, 𝑛 > 0

QuotientLawofLogarithms 𝐥𝐨𝐠𝒃𝒎𝒏

= 𝐥𝐨𝐠𝒃 𝒎 − 𝐥𝐨𝐠𝒃 𝒏for𝑏 > 0, 𝑏 ≠ 1,𝑚 > 0, 𝑛 > 0

ChangeofBaseFormula 𝐥𝐨𝐠𝒃 𝒎 = 𝐥𝐨𝐠𝒎𝐥𝐨𝐠 𝒃

,𝑚 > 0, 𝑏 > 0, 𝑏 ≠ 1

ExponentialtoLogarithmic 𝑦 = 𝑏+àx = log& 𝑦

LogarithmictoExponential 𝑦 = log& 𝑥à𝑥 = 𝑏,

Otherusefultips log> 𝑎& = 𝑏log 𝑎 = log@A 𝑎log& 𝑏 = 1

Part3:PracticeUsingLogRulesExample1:Writeasasinglelogarithma)logB 6 + logB 8 − logB 16

= logB6×816

= logB 3

b)log 𝑥 + log 𝑦 + log(3𝑥) − log 𝑦= log 𝑥 + log 3𝑥 = log 𝑥 3𝑥 = log 3𝑥I c)JKLM N

JKLM B

= logB 7d)log 12 − 3 log 2 + 2 log 3= log 12 − log 2Q + log 3I= log 12 − log 8 + log 9

= log12×98

= log272

Example2:Writeasasinglelogarithmandthenevaluate

Can’tusepowerlawbecausetheexponent2appliesonlyto𝑥,notto3𝑥.

Startedbycollectingliketerms.Musthavesamebaseandargument.

Usedchangeofbaseformula.

a)logS 4 + logS 16= logS(4×16)= logS 64

=log64log8

= 2

b)logQ 405 − logQ 5

= logQ V4055 W

= logQ 81

=log81log3

= 4

c)2 log5 + @Ilog16

= log5I + log√16= log25 + log4= log(25×4)= log100= 2

Example3:WritetheLogarithmasaSumorDifferenceofLogarithms

Example4:Simplifythefollowingalgebraicexpressions

a)logQ(𝑥𝑦)= logQ 𝑥 + logQ 𝑦

b)log20= log4 + log5

c)log(𝑎𝑏I𝑐)= log𝑎 + log𝑏I + log 𝑐= log𝑎 + 2 log𝑏 + log 𝑐

a)log Z√++M[

= log\𝑥@I

𝑥]I^

= log𝑥_QI

= −32 log𝑥

b)log`√𝑥aQ+ log𝑥I − log√𝑥

= log𝑥QI + log𝑥I − log𝑥

@I

=32 log𝑥 + 2 log𝑥 −

12 log𝑥

=32 log𝑥 +

42 log𝑥 −

12 log𝑥

= 3 log𝑥

c)log(2𝑥 − 2) − log(𝑥I − 1)

= logV2𝑥 − 2𝑥I − 1W

= log b2(𝑥 − 1)

(𝑥 − 1)(𝑥 + 1)c

= log2

𝑥 + 1

L4–7.1/7.2–SolvingExponentialEquationsMHF4UJensenPart1:ChangingtheBaseofPowersExponentialfunctionscanbewritteninmanydifferentways.Itisoftenusefultoexpressanexponentialexpressionusingadifferentbasethantheonethatisgiven.Example1:Expresseachofthefollowingintermsofapowerwithabaseof2.

Partd)showsthatanypositivenumbercanbeexpressedasapowerofanyotherpositivenumber.Example2:SolveeachequationbygettingacommonbaseRemember:if𝑥" = 𝑥$,then𝑎 = 𝑏

a)8= 2)

b)4)= (2,))= 2.

c)√16×3√325 6)

= 1678×32

95

= (2:);,×(2<)

)<

= 2,×2)= 2<

d)122= = 12log2= = log12𝑥 log 2 = log 12

𝑥 =log12log2

∴ 12 = 2BCD ;,BCD ,

b)4,= = 8=E)(2,),= = (2))=E)2:= = 2)=EF4𝑥 = 3𝑥 − 9𝑥 = −9

a)4=I< = 64=4=I< = (4))=

4=I< = 4)=𝑥 + 5 = 3𝑥5 = 2𝑥𝑥 = <

,

Part2:SolvingExponentialEquationsWhenyouhavepowersinyourequationwithdifferentbasesanditisdifficulttowritewiththesamebase,itmaybeeasiertosolvebytakingthelogarithmofbothsidesandapplyingthepowerlawoflogarithmstoremovethevariablefromtheexponent.Example3:Solveeachequationa)4,=E; = 3=I,log 4,=E; = log 3=I,2𝑥 − 1 log 4 = 𝑥 + 2 log 32𝑥 log 4 − log 4 = 𝑥 log 3 + 2 log 32𝑥 log 4 − 𝑥 log 3 = 2 log 3 + log 4𝑥 2 log 4 − log 3 = 2 log 3 + log 4

𝑥 =2 log 3 + log 42 log 4 − log 3

𝑥 ≅ 2.14b)2=I; = 3=E;log 2=I; = log 3=E;𝑥 + 1 log 2 = 𝑥 − 1 log 3𝑥 log 2 + log 2 = 𝑥 log 3 − log 3𝑥 log 2 − 𝑥 log 3 = − log 3 − log 2𝑥 log 2 − log 3 = − log 3 − log 2

𝑥 =− log 3 − log 2log 2 − log 3

𝑥 ≅ 4.419

Takelogofbothsides

Usepowerlawoflogarithms

Usedistributivepropertytoexpand

Movevariabletermstooneside

Commonfactor

Isolatethevariable

Part3:ApplyingtheQuadraticFormulaSometimesthereisnoobviousmethodofsolvinganexponentialequation.Ifyounoticetwopowerswiththesamebaseandanexponentof𝑥,theremaybeahiddenquadratic.Example4:Solvethefollowingequation2= − 2E= = 42= 2= − 2E= = 2= 4 2,= − 2N = 4 2= 2,= − 4 2= − 1 = 02= , − 4 2= − 1 = 0Let𝑘 = 2=toseethequadratic𝑘, − 4𝑘 − 1 = 0

𝑘 = E$± $8E:"R,"

𝑘 =4 ± −4 , − 4(1)(−1)

2(1)

𝑘 =4 ± 20

2

𝑘 =4 ± 2 5

2

𝑘 =2 2 ± 5

2

𝑘 = 2 ± 5Nowsubstitute2=backinfor𝑘andsolve

Case12= = 2 + √5log2= = log(2 + √5)

𝑥 =log(2 + √5)

log2

𝑥 ≅ 2.08

Case22= = 2 − √5log2= = log(2 − √5)Can’ttakethelogofanegativenumber,thereforethisisanextraneousroot(Nosolution).

Multiplybothsidesby2=

Distribute

Rearrangeintostandardform𝑎𝑥, + 𝑏𝑥 + 𝑐 = 0

Solveusingquadraticformula

Don’tforgettosimplifytheradicalexpression

Part4:ApplicationQuestionRemember:Equation:𝒚 = 𝒂(𝒃)𝒙

𝑎 =initialamount

𝑏 =growth(𝑏 > 1)ordecay(0 < 𝑏 < 1)factor

𝑦 =futureamount

𝑥 =numberoftimes𝑎hasincreasedordecreased

Tocalculate𝑥,usetheequation:𝑥 = [\["][_`a[_`a_[["bacd\e\fage\h[i\ejaR"klae_\j

Example5:Abacteriaculturedoublesevery15minutes.Howlongwillittakeforacultureof20bacteriatogrowtoapopulationof163840?

163840 = 20 2[;<

8192 = 2[;<

log 8192 = log 2[;<

log 8192 =𝑡15 log 2

log 8192log 2 =

𝑡15

13 =𝑡15

𝑡 = 195minutesExample6:Oneminuteaftera100-mgsampleofPolonium-218isplacedintoanuclearchamber,only80-mgremains.Whatisthehalf-lifeofpolonium-218?

80 = 10012

;i

0.8 = 0.5;i

log 0.8 = log 0.5;i

log 0.8 =1ℎ log 0.5

log 0.8log 0.5 =

1ℎ

ℎ =log 0.5log 0.8

ℎ ≅ 3.1minutes

L5–7.4–SolvingLogarithmicEquationsMHF4UJensenPart1:TryandSolveaLogarithmicEquationSolvetheequationlog 𝑥 + 5 = 2 log(𝑥 − 1)Hint:applythepowerlawoflogarithmstotherightsideoftheequationlog(𝑥 + 5) = log(𝑥 − 1)-𝑥 + 5 = (𝑥 − 1)-𝑥 + 5 = 𝑥- − 2𝑥 + 10 = 𝑥- − 3𝑥 − 40 = (𝑥 − 4)(𝑥 + 1)𝑥 = 4or𝑥 = −1Reject𝑥 = −1becauselog(𝑥 − 1)isundefinedforthisvalueof𝑥.Therefore,theonlysolutionis𝑥 = 4Part1:SolveSimpleLogarithmicEquationsExample2:Solveeachofthefollowingequationsa)log(𝑥 + 4) = 1

Note:Iflog1 𝑎 = log1 𝑏,then𝑎 = 𝑏.

Method1:re-writeinexponentialform𝑥 + 4 = 104𝑥 + 4 = 10𝑥 = 6

Method1:expressbothsidesasalogarithmofthesamebaselog(𝑥 + 4) = log(10)𝑥 + 4 = 10𝑥 = 6

Tocompletethislesson,youwillneedtorememberhowtochangefromlogarithmictoexponential:

𝑦 = log7 𝑥à𝑥 = 𝑏8

b)log9(2𝑥 − 3) = 25- = 2𝑥 − 325 = 2𝑥 − 328 = 2𝑥14 = 𝑥Part2:ApplyFactoringStrategiestoSolveEquationsExample3:Solveeachequationandrejectanyextraneousrootsa)log 𝑥 − 1 − 1 = − log 𝑥 + 2 log 𝑥 − 1 + log(𝑥 + 2) = 1log (𝑥 − 1)(𝑥 + 2) = 1log(𝑥- + 𝑥 − 2) = 1𝑥- + 𝑥 − 2 = 104𝑥- + 𝑥 − 12 = 0𝑥 + 4 𝑥 − 3 = 0𝑥 = −4or𝑥 = 3Reject𝑥 = −4becausebothoftheoriginalexpressionsareundefinedforthisvalue.Theonlysolutionis𝑥 = 3

Example4:Iflog; 𝑏 = 3,thenuselogrulestofindthevalueof…a)log; 𝑎𝑏-= log; 𝑎 + log; 𝑏-= log; 𝑎 + 2log; 𝑏= 1 + 2 3 = 7b)log7 𝑎

=log; 𝑎log; 𝑏

=13

Hint:needtochangethebase

𝐥𝐨𝐠𝒃 𝒎 =𝐥𝐨𝐠𝒎𝐥𝐨𝐠 𝒃

b)log √𝑥- + 48𝑥C = -D

log(𝑥- + 48𝑥)4D =

23

13 log(𝑥

- + 48𝑥) =23

3 E13 log(𝑥

- + 48𝑥)F = 3 G23H

log(𝑥- + 48𝑥) = 2𝑥- + 48𝑥 = 10-𝑥- + 48𝑥 − 100 = 0(𝑥 + 50)(𝑥 − 2) = 0𝑥 = −50or𝑥 = 2Botharevalidsolutionsbecausetheybothmaketheargumentofthelogarithmpositive.

c)logD 𝑥 − logD(𝑥 − 4) = 2

logD I𝑥

𝑥 − 4J = 2𝑥

𝑥 − 4 = 3-𝑥

𝑥 − 4 = 9𝑥 = 9(𝑥 − 4)𝑥 = 9𝑥 − 3636 = 8𝑥92 = 𝑥

L6–6.5–ApplicationsofLogarithmsinPhysicalSciencesMHF4UJensenPart1:ReviewofSolvingLogarithmicEquationsExample1:Solvefor𝑥inthefollowingequationlog% 𝑥 − 6 =4 − log% 𝑥log% 𝑥 − 6 + log% 𝑥 =4log% 𝑥 − 6 (𝑥) = 42. = (𝑥 − 6)(𝑥)16 = 𝑥% − 6𝑥0 = 𝑥% − 6𝑥 − 160 = (𝑥 − 8)(𝑥 + 2)𝑥 = 8Reject𝑥 = −2asbotoriginallogarithmicexpressionsareundefinedforthisvaluePart2:pHScaleThepHscaleisusedtomeasuretheacidityoralkalinityofachemicalsolution.Itisdefinedas:

𝒑𝑯 = − 𝐥𝐨𝐠 𝑯7 where 𝐻7 istheconcentrationofhydroniumions,measuredinmolesperliter.

Example2:AnswerthefollowingpHscalequestionsa)Tomatojuicehasahydroniumionconcentrationofapproximately0.0001mol/L.WhatisitspH?𝑝𝐻 = − log 0.0001𝑝𝐻 = −(−4)𝑝𝐻 = 4b)Bloodhasahydroniumionconcentrationofapproximately4×10<=mol/L.Isbloodacidicoralkaline?𝑝𝐻 = − log(4×10<=)𝑝𝐻 ≅ 6.4Sincethisisbelowtheneutralvalueof7,bloodisacidic.c)OrangejuicehasapHofapproximately3.Whatistheconcentrationofhydroniumionsinorangejuice?3 = − log 𝐻7 −3 = log 𝐻7 10<A = 𝐻7 𝐻7 = 0.001mol/LPart3:DecibelScaleSomecommonsoundlevelsareindicatedonthedecibelscaleshown.Thedifferenceinsoundlevels,indecibels,canbefoundusingtheequation:

𝜷𝟐 − 𝜷𝟏 = 𝟏𝟎 𝐥𝐨𝐠𝑰𝟐𝑰𝟏

where,𝛽% − 𝛽Histhedifferenceinsoundlevels,indecibels,andIJIKistheratiooftheirsoundintensities,where𝐼ismeasuredin

wattspersquaremeter 𝑊/𝑚%

Example3:Answerthefollowingquestionsaboutdecibelsa)Howmanytimesasintenseasawhisperisthesoundofanormalconversation

60 − 30 = 10 log𝐼%𝐼H

30 = 10 log𝐼%𝐼H

3 = log𝐼%𝐼H

10A =𝐼%𝐼H

𝐼%𝐼H= 1000

Aconversationsounds1000timesasintenseasawhisper.b)Thesoundlevelinnormalcitytrafficisapproximately85dB.Thesoundlevelwhileridingasnowmobileisabout32timesasintense.Whatisthesoundlevelwhileridingasnowmobile,indecibels?𝛽% − 85 = 10 log 32 𝛽% = 10 log 32 + 85𝛽% ≅ 100dBPart4:RichterScaleThemagnitude,𝑀,ofanearthquakeismeasuredusingtheRichterscale,whichisdefinedas:

𝑴 = 𝐥𝐨𝐠𝑰𝑰𝟎

where𝐼istheintensityoftheearthquakebeingmeasuredand𝐼Sistheintensityofastandard,low-levelearthquake.

Example4:AnswerthefollowingquestionsabouttheRichterScalea)Howmanytimesasintenseasastandardearthquakeisanearthquakemeasuring2.4ontheRichterscale?

2.4 = log𝐼𝐼S

10%.. =𝐼𝐼S

𝐼𝐼S=≅ 251.19

Itisabout251timesasintenseasastandardearthquake.b)Whatisthemagnitudeofanearthquake1000timesasintenseasastandardearthquake?𝑀 = log 1000 𝑀 = 3