Exponential and Logarithmic Equations Chapter 4 Section 4.6 Exponential and Logarithmic Equations Logarithmic Functions This module explores more detailed solutions of exponential and logarithmic equations. Section 4.6 v5.0.1 10/26/2012

Exponential and Logarithmic Equations Chapter 4 Section 4.4 Exponential and Logarithmic Equations Logarithmic Functions This module explores more detailed solutions of exponential and logarithmic equations. Section 4.6 v5.0.1 10/26/2012

Solving Equations: Examples Exponential and Logarithmic Equations 1. World population 1950 1960 1970 1980 1990 2000 2010 2020 2030 2040 2050 15 14 13 12 11 10 9 8 7 6 5 4 3 2 t P(t) (x109) t P( t) P3  P(t) = 3(1.018)t–1960 1950 2.5098 1960 3.0000 1970 3.5859 1980 4.2862 1990 5.1233 2000 6.1239 World Population Growth Here we tabulate population on decade intervals and show the associated growth curve represented by the graph of the population function. Note that the vertical scale is measured in billions of people (that is, 109) and the horizontal scale is marked off in decades. The graph is a typical exponential function graph. Note that the change in population from 2000 to 2050 is based on the extrapolation of the given population model to 2050. The population at that point is rounded to four decimal places to conform to the rest of the tabular data. However, there is room in the table for only three decimal places once the population exceeds 10 billion. Note that the population in 1950 is to the left of the benchmark point (t = 1960) on the graph. When t = 1960, the exponent shown for P(t) is zero, leaving 3 multiplied by 1 as the value of P(t). This is our benchmark point. Because of the nature of the exponential function, valid population points occur on the graph well to the left of 1960 and continue to decrease as t is chosen further and further to the left of 1960.  2010 7.3199 P2 2020 8.7495 2030 10.458 P1   2040 12.500 2050 14.942 10/26/2012 Section 4.6 v5.0.1 Section 4.6 v5.0.1 10/26/2012

Solving Equations: Examples Exponential and Logarithmic Equations 1. World population P(t) (billions) in year t is given by P(t) = 3(1.018)t–1960 What was the population in 1950 ? In 2000 ? P(1950) = 3(1.018)1950–1960 P(2000) = 3(1.018)2000–1960 What is the growth factor ? What is the average annual increase from 1950 to 2000? From 2000 to 2050 ? = 3(1.018)–10 = 2.50982 billion = 3(1.018)40 = 6.12396 billion 1.018 World Population Growth After years of study demographers have created and verified an exponential model model of world population growth. The benchmark year is 1960, when the exponent is zero and the population is an even 3 billion people. We can establish population levels at various times over the past century or two, and assuming the model is correct, we can make predictions about future growth of the population. The growth factor, in this case defined to be the base of the exponential function, is 1.0118. This means that each year the current population is multiplied by 1.018 to get the following year’s population. In 1950 the exponent is negative, so the population in 1950 is to the left of the benchmark point (t = 1960) on the graph. Because of the nature of the exponential function, valid population points occur on the graph well to the left of 1960 and less than the 3 billion benchmark value. The average annual increase is just the slope of the secant line from the point on the graph where t = 1950 to the point on the graph where t = 2000. So, over the half century from 1950 to 2000 the average annual increase is 72,282,704 people per year. We note that this half-century average over the next half century is more than twice as high at 176,369,423 people per year. Typical of all exponential growth functions, the rate of growth increases as the population increases. That is, the rate of growth depends on the current size of the population, so the graph gets steeper and steeper as we move toward larger values of t. P t = 6.12396 – 2.50982 2000 – 1950 = 3.61414 50 = 72,282,704 people/yr P(2050) – P(2000) 2050 – 2000 = 176,369,423 people/yr 10/26/2012 Section 4.6 v5.0.1 Section 4.6 v5.0.1 10/26/2012

Solving Equations: Examples Exponential and Logarithmic Equations 1. World population (continued) What is the per cent change in population from 1950 to 2000 ? What is the per cent change in population from 2000 to 2050 ? Exponential Fact: On equal intervals exponential growth functions always grow at a rate proportional to the initial value on the interval NOTE: The same is true for exponential decay functions P(2000) – P(1950) P(1950) (100) 6.1239 – 2.5098 2.5098 (100) = = 143.999 % P(2050) – P(2000) P(2000) (100) 14.9424 – 6.1239 6.1239 (100) = = 143.999 % World Population Growth (continued) We proceed to find the percent growth over equal time intervals. To find the percent change in population over a half-century, we start by finding the absolute change, and then compute what percentage this is of the initial value at the beginning of the time interval. Note that the percent change in population is the same over successive time intervals, even though the rate of growth increases with time. This reflects the fact that as the population grows, the rate of growth has to increase to maintain a constant percent change. This fact is summarized at the bottom of the illustration showing that the rate of growth is always proportional to the initial value on each time interval. The proportionality constant, represented by the percent change, is the same over equal intervals. To illustrate this fact, one might try computing the population over each decade, instead of each half-century. 10/26/2012 Section 4.6 v5.0.1 Section 4.6 v5.0.1 10/26/2012

Solving Equations: Examples Exponential and Logarithmic Equations 1. World population (continued) 1950 1960 1970 1980 1990 2000 2010 2020 2030 2040 2050 15 14 13 12 11 10 9 8 7 6 5 4 3 2 t P(t) (x109) t P( t) t = 50 P3  1950 2.5098 1960 3.0000 P = 8.8185 Average annual rate of change 1970 3.5859 P P2 (100) = 143.99 % 1980 4.2862 1990 5.1233 P t = 3.6141 50 billion/yr P t = 8.8184 50 2000 6.1239 billion/yr World Population Growth (continued) Here we tabulate population on decade intervals and show the associated growth curve represented by the graph of the population function. Note that the vertical scale is measured in billions of people (that is, 109) and the horizontal scale is marked off in decades. The graph is a typical exponential function graph. The dashed lines clearly show the rate of increase of population increasing over time. However, we also see the percent change holding steady over equal intervals. We should note that the change in population from 2000 to 2050 is based on the extrapolation of the given population model to 2050. The population at that point is rounded to four decimal places to conform to the rest of the tabular data. However, there is room in the table for only three decimal places once the exceeds 10 billion. Consequently, the actual change is 14.94243154 – 6.123960415 = 8.818471125 ≈ 8.8185 which differs slightly from the difference in the truncated tabular values shown. What will be the actual population in 2100 and the rate of change from 2050 to 2100 ? billion people. Thus = 0.4303404732 So, P(2100) = 36,459,455,200 people and rate is 430,340,473,100 people per yr. = 72,282,704 people/yr  = 176,369,423 people/yr 2010 7.3199 P2 Average annual rate of change 2020 8.7495 P P1 (100) = 143.99 % P = 3.6141 2030 10.458 P1  2040 12.500 t = 50 2050 14.942 10/26/2012 Section 4.6 v5.0.1 Section 4.6 v5.0.1 10/26/2012

Solving Equations: Examples Exponential and Logarithmic Equations 1. World population (continued) When will the population reach 20 billion ? World population P(t) in billions in year t is given by P(t) = 3(1.018)t–1960 20 = 3(1.018)t–1960 Recall that logb xt = t logb x log 20 = log (3 (1.018)t–1960 ) = log 3 + (t – 1960)log (1.018) t = 1960 + log 20 – log 3 log 1.018 World Population Growth (continued) Here we fall back on the symbolic form of the exponential growth function to find the point in time when the population reaches 20 billion. As before, to move the variable out of the exponent we apply the inverse function, that is, the common log (or base-10 log) to both sides of the exponential equation. This gives us a simple linear equation in t, requiring only that we be able to compute the logs of 20, 3, and 1.018. The result is that we expect to reach a population of 20 billion people in the year 2066. Since world population was about 2.5 billion in 1950, this means that in a little over a century (116 years) world population will be eight times its value in 1950. The population will be 40 billion at The time from 20 billion to 40 billion, that is a doubling, takes 2105 – 2066 = 39 years In yet another 39 years, the population will double again to 80 billion. = 1960 + 106.34 ≈ 2066 Question: How long till P(t) is 40 billion ? … about 39 years What is the doubling time for P(t) ? … about 39 years 10/26/2012 Section 4.6 v5.0.1 ≈ 2105 Section 4.6 v5.0.1 10/26/2012

Solving Equations: Examples Exponential and Logarithmic Equations 2. Solve e2x = e5x–3 ln (e2x) = ln (e5x–3) 2x = 5x – 3 3 = 3x x = 1 3. Solve 2x = –4 Question: What are the domain and range of f(x) = 2x ? Domain = R Range = (0, ) so, 2x > 0 for all x Clearly there is no solution ... OR ... use 1-1 property Solution set: { 1 } Solving Equations Example 2 shows how to move the unknown from the exponent to a factor on each side of the equation by applying the inverse function for the natural exponential. Using the 1–1 property we extract the variable expressions from the natural logarithm to yield a simple linear equation in x. In Example 3 we note that any power of a positive base (2 in this case) will be positive. Hence, there is no value of x for which 2x = –4. The solution set is empty. Solution set: { } 10/26/2012 Section 4.6 v5.0.1 Section 4.6 v5.0.1 10/26/2012

Solving Equations: Examples Exponential and Logarithmic Equations 4. Solve Since the exponential function is 1-1 then x2 = 4x – 3 x2 – 4x + 3 = 0 (x – 1)(x – 3) = 0 x – 1 = 0 OR x – 3 = 0 x = 1 OR x = 3 NOTE: Could have applied inverse function x2 7 74x–3 = Zero product rule Solution set: { 1, 3 } Solving Equations In Example 4 we apply the 1–1 property of power functions to set the exponents equal to each other. This produces a simple quadratic equation which is easily solved by factoring and applying the zero product rule. From this we gain two solutions, 1 and 3. We could have applied the inverse function, log7 to isolate the x terms. It turns out we could also have used logarithms of other bases, for example common log and natural log. We would then have to also apply the power rule for logarithms to extract the exponents as multipliers, dividing out the log expression in common with both sides. For example, Applying the power rule, Dividing by log 7, x2 = 4x – 3 , which yields the same solutions. x2 7 74x–3 = log7 Question: Could we use log or ln instead of log7 ? 10/26/2012 Section 4.6 v5.0.1 Section 4.6 v5.0.1 10/26/2012

Solving Equations: Examples Exponential and Logarithmic Equations 5. Solve 5 ln x = 10 ln x = 2 By definition 2 is the exponent of the base that yields x Thus x = e2 ≈ (2.71828)2 ≈ 7.38906 6. Solve log3 (1 – x) = 1 By definition 31 = 1 – x Thus x = 1 – 3 = –2 Solution set: { ~ 7.38906 } Solving Equations In Example 5 we isolate ln x and then use the definition of logarithm to isolate x as the square of e. This we evaluate as approximately 7.38906. This value is only approximate, since e is irrational (in fact, transcendental) so that algebraic expressions in decimal fractions must be approximate. For Example 6 we employ the definition of logarithm to extract the expression in x from the logarithm. This linear equation is trivial to solve. Solution set: { –2 } 10/26/2012 Section 4.6 v5.0.1 Section 4.6 v5.0.1 10/26/2012

Solving Equations: Examples Exponential and Logarithmic Equations 7. Solve ln (x2 – 4) – ln (x + 2) = ln (3 – x) Using the quotient rule Since the logarithm function is 1-1 x2 – 4 = (x + 2)(3 – x) 2x2 – x – 10 = 0 2x – 5 = 0 OR x + 2 = 0 x = 5/2 OR x = –2 x2 – 4 x + 2 ln ( ) = ln (3 – x) x2 – 4 x + 2 = 3 – x = 3x – x2 + 6 – 2x = (2x – 5)(x + 2) Solving Equations Example 7 requires the use of the quotient rule for logarithms, followed by the application of the 1–1 property of logarithms. The rational expression so produced can be cleared of fractions and the resulting quadratic equations can be solved by factoring or by used of the quadratic formula. Factoring is much simpler and faster and so is used to set up the application of the zero product property. The two solutions are then –2 and 5/2. However, for x = –2 we note that ln(x2 – 4) and ln(x + 2) do not exist, since the natural logarithm is not defined at 0. So –2 is not a solution, leaving 5/2 as the only solution. This is easily seen by noting that the domain of any logarithm function is the set of all positive real numbers and the range is the set of all real numbers. Question: What are the domain and range of the ln function ? Solution set: { 5/2 } Why not –2 ? 10/26/2012 Section 4.6 v5.0.1 Section 4.6 v5.0.1 10/26/2012

Exponential and Logarithmic Equations Think about it ! 10/26/2012 Section 4.6 v5.0.1 Section 4.6 v5.0.1 10/26/2012