MATH 234, Calculus for BusinessLecture 4, Textbook Section 2.6Logarithms, Growth and Decay;Mathematics of FinanceUniversity of Illinois, Urbana-ChampaignJanuary 30th, 2017Logarithms, Growth and Decay; Mathematics of Finance (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business January 30th, 2017 1 / 18AnnouncementsQuiz 2 is tomorrow in section. The coverage is textbook Sections 2.1,2.2, 2.3, and 2.4. In particular, Sections 2.5 (logarithms) and 2.6(applications) will not be on the quiz.HW2 is due Thursday 2/2, 11:59pm; late HWs are not accepted.Logarithms, Growth and Decay; Mathematics of Finance (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business January 30th, 2017 2 / 18Review of Sections 2.4, 2.5 Comparison of Exponential and Logarithmic FunctionsRemarkThe value of y = loga(x) is the number that solves ay= x.Theorem (Exponentials and logarithms)Fix a, b. For variables x and u, the rules are listed as counter-parts(1) a1= a(2) a0= 1(3) (ax)u= ax∗u(4) ax∗ au= ax+u(5)axau= ax−u(6) axbx= (ab)x(7) aloga(x)= x(8) ax= ex∗ln(a)(1) loga(a) = 1(2) loga(1) = 0(3) loga(xu) = u ∗ loga(x)(4) loga(x ∗ u) = loga(x) + loga(u)(5) logaxu= loga(x) − loga(u)(6) logab(x) =loga(x)1 + loga(b)(7) loga(ax) = x(8) loga(x) =ln(x)ln(a)Logarithms, Growth and Decay; Mathematics of Finance (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business January 30th, 2017 3 / 18Review of Sections 2.4, 2.5 Learn through mistakesExample (Learn through mistakes)Find as many mistakes per line as possible, and explain what went wrong.ln(x)2−ln(y)2ln(xy)=ln(x2) − 2 ln(y)ln(x) ln(y)=ln(x2) − 2ln(x)=ln(x2− 2)ln(x)= logx(x2− 2)Line 1: First,ln(x)26= ln(x2). Second,ln(y)26= 2 ln(y).Third, ln(xy) 6= ln(x) ln(y), the rule is ln(xy ) = ln(x) + ln(y)Line 2: Incorrect cancelling of fractions.Line 3: ln(x2) − 2 6= ln(x2− 2).Line 4: The rule loga(x2− 2) =ln(x2−2)ln(a)only works if a is a constant.The base of a logarithm cannot be a variable, x.Logarithms, Growth and Decay; Mathematics of Finance (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business January 30th, 2017 4 / 18Review of Sections 2.4, 2.5 Learn through mistakesExample (The Correct Version)ln(x)2−ln(y)2ln(xy)=ln(x) − ln(y)∗ln(x) + ln(y)ln(xy)Factor=ln(x) − ln(y)∗ln(x) + ln(y)ln(x) + ln(y)Rule 4= ln(x) − ln(y ) CancelLogarithms, Growth and Decay; Mathematics of Finance (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business January 30th, 2017 5 / 18Section 2.6, Applications Growth and DecayDefinitionLet y0be the initial quantity at time t = 0. A quantity grows or decaysexponentially at rate k if the quantity at time t isy = y0ektRemarkIf k > 0, then quantity grows exponentially, and k is the growth constant.If k < 0, then quantity decays exponentially, and k is the decay constant.RemarkThe value of e is the most commonly used base. Other bases also work:y = y0∗ (6kt) or y = y0∗ (0.5kt)are also exponential growth/decay functions.Logarithms, Growth and Decay; Mathematics of Finance (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business January 30th, 2017 6 / 18Section 2.6, Applications Growth and DecayDefinition (... from previous slide)Let y0be the initial quantity at time t = 0. A quantity grows or decaysexponentially at rate k if the quantity at time t is y = y0ekt.ExampleInitially, there are 2 grams of yeast in a sugar solution. The amount ofyeast grows exponentially to 7 grams after 12 hours.Determine the growth function, in terms of t hours.Solution (Step 1: Determine values of y0and k.)The initial quantity is y0= 2. Note that time t = 12 gives quantity y = 7,so the equation y = y0ektbecomes7 = 2 ∗ ek∗12ln(7/2) = 12k ∗ ln(e) = 12k ∗ 1ln72= ln(e12k)ln(7/2)12= kLogarithms, Growth and Decay; Mathematics of Finance (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business January 30th, 2017 7 / 18Section 2.6, Applications Growth and DecayDefinition (... from previous slide)Let y0be the initial quantity at time t = 0. A quantity grows or decaysexponentially at rate k if the quantity at time t is y = y0ektExampleInitially, there are 2 grams of yeast in a sugar solution. The amount ofyeast grows exponentially to 7 grams after 12 hours.Determine the growth function, in terms of t hours.Solution (Step 2: Determine growth function using approximate values.)Recall y0= 2 and k =ln(7/2)12. Suppose question wants k to 3 decimals...k =ln(7/2)12≈ 0.104and so the growth function isy = y0ekt≈ 2 ∗ e0.104∗tLogarithms, Growth and Decay; Mathematics of Finance (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business January 30th, 2017 8 / 18Section 2.6, Applications Growth and DecayDefinition (... from previous slide)Let y0be the initial quantity at time t = 0. A quantity grows or decaysexponentially at rate k if the quantity at time t is y = y0ektExampleInitially, there are 2 grams of yeast in a sugar solution. The amount ofyeast grows exponentially to 7 grams after 12 hours.Determine the growth function, in terms of t hours.Solution (Alternate Step 2: Determine growth function using exact values.)Recall y0= 2 and k =ln(7/2)12. The growth function isy = y0ekt= 2 ∗ eln(7/2)12∗t= 2 ∗ eln(7/2)∗t12y = 2 ∗eln(7/2)t/12= 2 ∗72t/12Logarithms, Growth and Decay; Mathematics of Finance (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business January 30th, 2017 9 / 18Section 2.6, Applications Growth and DecayExample (... with exact values)Initially, there are 2 grams of yeast in a sugar solution. The amount ofyeast grows exponentially to 7 grams after 12 hours.The exact growth function is y = 2 ∗72t/12RemarkThe exact solution is useful for verifying the calculations: at time t = 0,y = 2 ∗720/12= 2 ∗720= 2 ∗ 1 = 2and at time t = 12 hours,y = 2 ∗7212/12= 2 ∗721= 2 ∗72= 7.Logarithms, Growth and Decay; Mathematics of Finance (University of Illinois, Urbana-Champaign)MATH 234, Calculus for Business January 30th, 2017 10 / 18Section 2.6, Applications Mathematics of FinanceRemarkPrincipal is P dollars, interest is r percent, m is times compounded/year.In terms of growth rate:simple interest < compound interest < continuous compound interestP + Prt < P ∗1 +rmmt< P ∗ (ert)DefinitionNominal rate is r . The effective rate is the actual