Math 111 Name:__KEY___________________________________________________________________Grp#______.GrpAct – #18!! ! Exp!&!Logarithmic!Functions,!Equations!and!Models! ! !Sect.5.6!Change!of!Base!Formula!Let.x,.a.and.b.be.positive.real.numbers.with. ..Then.!Complete.the.following.with.the.natural.log...Complete.with.the.common.log...To.use.a.calculator,.we.convert.logarithms.to.ln.or.log.using.the.change.of.base.formula.above..!...1. Warm!Up..Evalute.the.follo w in g.lo g arit hms.witho u t.us in g.a .calc u lat or ..If.the .lo ga rith m .is.und efin ed.w rite.“U n de fined .”..a. !log3(1) = 0....b. !log218⎛⎝⎜⎞⎠⎟= −3....c. !ln(−1)= ??..Undefined....d. !!ln(e15)= 15....e. !log7(0)= ??..Undefined....2. Your.calculator.has.a.dedicated.button.for.the.natural.logarithm.and.the.common.logarithm.(ln.and.lo g )..T o .c ompute.lo g ar ith m s .o f.a .d iffe re nt .b as e .us in g .a.c a lcu la to r,.we.can.first.use.the.change.of.base.formula.to.convert.to.log.or.ln..Use.the.change.of.base.formula.to.compute.each.of.the.following.logarithms.using.a.calculator...a. ! log7(0.91)=ln(0.91)ln(7)! −0.0485..b. ! log13(45) =log(45)log(13)! 1.4841...We.have.been.talking.about.exponential.and.logarithmic.functions.for.the.past.two.weeks..This.activity.combines.those.topics..Exponential.and.logarithmic.functions.are.incredibly.powerful.in.modeling.phenomena,.and.you’ll.get.to.see.some.of.those.examples.here..Need!a!hint!for!how!to!get!started!on!these?..You.can.either.think.about.applying.the.inverse.of.the.exponential.function,.or.you.can.think.about.rewriting.the.exponential.equation.as.a.log.equation...3. Solve.each.of.the.following.equations..Give.exact.and.approximate.solutions..If.no.solution.exists.write.“No.Solution.”..a. 9.5e0.05x.=.19..!! x =10.05ln(2) = 20ln(2) ! 13.8629........b. 3(2)x-2.=.99..!! x = 2+ln(33)ln(2)! 7.0444.......c. log3(1-x).=.1..!!x = −2.......d. 2x.=.-4..!!x = ??...No.Solution......e. 2x.=.1..!!x = 0.... .4. The.half-life.of.carbon-14.is.5700.years..A.sample.of.carbon-14.originally.contained.75.grams...Let.A(t).be.th e.amount.of.carb on-14.left.in.the.sample.after.t.years...a. Give.two.ordered.pairs.( ),()tAt.for.this.situation...Exp lain.what.they.represent.in.context..!0, 75( )& 5700, 37.5( )...b. The.amount.of.carbon-14.in.the.sample.decays.exponentially.over.time..Give.an.exponential.decay.model.for.this.in.the .fo rm.A(t ).=.Cekt...!!A(t ) = 75e−0.0001216t...c. What.are.a.reasonable.domain.and.range.for.your.model.in.the.context.of.this.problem?.Indicate.what.units.are.associated.with.elements.of.the.domain.and.the.elements.of.the.range..!!D = 0,∞⎡⎣)yearsR = 0,75(⎤⎦grams...d. Use.your.model.to.predict.the.amoun t.of.ca rbo n-14.left.in.the.sample.after.2000.years...!!A(2000) = 58.81...grams..e. What.percentage.of.carbon-14.will.be.remaining.after.8000.years?..!!A(8000) = 28.35...grams..f. How.long.will.it.take.for.the.sample.to.contain.only.10.grams.of.carbon-14?..!!A(t ) = 10 ⇒ t = 16596.9..years..g. How.long.will.it.take.for.8 0 % .o f .th e.c a rb o n-14.in.the.sample.to.decay?..!!A(t ) = 0.2 ⇒ t = 13235.5.years...h. Using.your.model.from.part.(b),.solve.for.t.to.give.a.formula.tha t.com p u tes.the .age .of.the .sam p le,.if.yo u .a re .given.the.amount.A.of.carbon-14.remaining.in.the.sample..!!t = t(A) =1−0.0001216lnA75⎛⎝⎜⎞⎠⎟...i. Give.a.formula.that.computes.the.age.of.the.sample.if.you.know.the.percentage.p.of.carbon-14.remaining.in.the.sample...!!t = t(P) =1−0.0001216lnP100⎛⎝⎜⎞⎠⎟.. .5. Solve.each.of.the.following.equations..Check.your.solutions..If.no.solution.exists.write,.“No.Solution.”..a. ln(x).+.ln(2 x ).= .2.. . .!!x =e2...b. ln(x).+.ln (3 x - 1).=.ln(10).. . .!!x = 2...c. 24377xx−=... . .!!x = 1,3...d. ( ) ( )35 27xx=... . .!!x =ln(2)−ln(3)ln(5)−ln(7)=ln23⎛⎝⎜⎞⎠⎟ln57⎛⎝⎜⎞⎠⎟... . . .e. log2(2x).=.4.–.lo g2(x.+.2)... .!!x = 2......6. The.salinity.of.the.oceans.changes.with.latitude.and.with.depth...In.the.tropics,.the.salinity.increase s.o n .th e .su rf ac e .of .th e .oc e an .d u e .to .ra p id .ev ap o ra tio n ...In .th e .h ig he r.la tit u de s ,.there.is.less.evapora tion .and .rainfa ll.cause s.the.sa linity.to.be .less.on.the.surface.than.at.lower.de p th s ...T h e.fu n c tio n .S(x) = 31.5 + 1.1 log(x + 1)( ).models.salinity.to.depths.of.1000.meters.at.a.latitude.of.57.5°N.(we.are.ap pro xim a tely.at.45°N)...The.input.x.is.the.depth.in.m ete rs.an d.the .ou tpu t.S(x).is.in.grams.per.kilo gr am.of.seawater...a. Evaluate.S(500).and.interpr e t.your.results.in.a.written.sentence..!!S(500) = 34.47..grams/kilogram.At.a.depth.of.500.meters.the.salinity.of.the.ocean.water.is.34.47..grams/kilogram..b. Find.the.depth.where.the.salinity.equals.33..!!S(x) = 31.5+1.1 log(x +1)( )= 33 ⇒ x = 22.10.meters..c. What.is.the.average.rate.of.change.of.the.salinity.between.depths.ofx = 5and.x = 25?.Include.units..!!Save 5,25[ ]= 0.035gramskilogrammeter..d. What.is.the.effect.on.the.salinity.if.the.depth.is.multip lied .b y .1 0?