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4.3 Using Exponential Equations Given a y , if I am looking for “y”….no big deal, just tell me what “a” x equals and what “x” equals. 23 8 Given a y , if I am looking for “x” and I know what “a” and “y” it’s not x 2 15 always so easy. The answer is going to be somewhere between x = 3 and x = 4 (closer to 4). We need a function related to exponents. If we are looking for the exponent and know everything else, we can find it. That mathematical function is called a logarithm. x log 2 15 2 2 x 15 The log, base 2, of 15 equals “x”. This whole concept is similar to when you first learned about radicals (square root, cubic roots, etc.) They were basically the inverse of Where is that exponents. We could use them to “undo” an exponent. exponent? Logarithms are used when you are LOOKING FOR THE EXPONENT Introduction to logarithms https://www.khanacademy.org/math/algebra2/exponential-andlogarithmic-functions/introduction-to-logarithms/v/logarithms There are lots of properties/rules for using exponents. We will talk about one…The Power Rule Let x log a M is equivalent to ax M ax M p p Now raise both side to the power of “p” a xp M p Now change back to a log Substitute log a M in for x. Change the order of the multiplication on the right side and we get log a M p xp ` equivalent log a M p log a Mp log a M p p log a M When you take the log of any value raised to a power it equals that power times the log of the number. Your book give you a short cut, but I wanted you to know why. 2x 8 log 8 x log 2 x3 Example 4.3.1: Solving an Exponential Equation Suppose we want to model the decrease in a river’s flow in the summer to predict when it will be 300 CFS. We observe the level to be 1860 CFS (cubic feet per second) on Monday, 1702 CFS on Tuesday, and 1557 CFS on Wednesday. Recall from section 4.1 that exponential equations take the form y = abx; where “a” represents the initial population, “b” represents the rate of growth, and “x” is the time. Solution: 1702 1557 Notice that ≈ 0.915 and ≈ 0.915 so you could say the level is dropping around 8.5% per day, or 1860 1702 each day you have about 91.5% of what you had the day before. The exponential equation F = 1860•0.915d is a good model since the level is dropping at a consistent rate; where F = flow and d = day. We want to know what day the flow will drop to 300 CFS. F = 1860•0.915d 300 = 1860•0.915d replace F with 300 d .1613 ≈ 0.915 divide both sides by 1860 d ≈ Log 0.915 0.1613 d ≈ (Log 0.1613)/(log0.915) introduce logarithms to solve for d d ≈ 20.5 enter Log(0.1613)/Log(0.915) in your calculator Final Answer: If the pattern continues, the level will hit 300 CFS about half way through day 20. This would be on Sunday about 3 weeks away. Note: It is important to recognize that growth or decline is always measured relative to 100%. If a population of 490 elk is increasing at 13% the model would be y = 490(100%+13%)x or y = 490•1.13x. If the population was decreasing at 13% the model would be y = 490(100%-13%)x or y = 490•0.87x. We could have also entered our data into the calculator and found our model using exponential regression. Remember that regression is better when the percent decrease is less consistent since regression will take an average. 0 1860 1 1702 2 1557 Sample Problem 4.3.2 The Southern White Rhino, considered extinct in the 1800’s, has enjoyed a steady increase in population in the last hundred years. a) Consider 1929 to be year 0 and use regression to find an exponential equation to model the data. Round the numbers in your equation to 3 decimal places. b) It is clear in the table that the population reached 1000 between 1948 and 1968. Use your equation to find the year, accurate to the nearest tenth place. c) Use your equation to predict the year the population will reach Years 30,000, accurate to the nearest tenth place. Yea r Since Population 1929 0 19 39 58 68 80 81 Solutions: a) Enter into the calculator the data from the table, then select the 1929 1948 1968 1987 1997 2009 2010 150 550 1800 4665 8440 11640 20170 ExpReg option to generate the regression equation: y 164.466 1.061 x b) Using the equation from a), c) Using the equation from a), substitute 1,000 for y and solve for x. substitute 30,000 for y and solve for x. 1000 164.466 1.061x 1000 1.061x 164.466 6.0802 1.061x 30000 164.466 1.061x 30000 1.061x 164.466 182.4085 1.061x log 6.0802 log1.061x log 6.0802 x log1.061 log182.4085 log1.061x log182.4085 x log1.061 log 6.0802 log1.061 x 30.5 x year 1959.5 (1929 30.5) x x log182.4085 log1.061 x 87.9 x year 2016.9 (1929 87.9) Sample Problem 4.3.4 Temperature and moisture content in wood are related, in that as the temperature increases the percent moisture in the wood decreases. a) Use regression to find an exponential equation to model the data. Round the numbers in your equation to 3 decimal places. b) Use your equation to accurately find the temperature, accurate to the nearest tenth place, needed to get the moisture content to 5%. c) Use your equation to predict the temperature, accurate to the nearest tenth place that would be necessary to get the moisture content to 2%. Solutions: a) Enter into the calculator the data from the table, then select the ExpReg option to generate the regression equation: y 9.834 0.972 b) x Using the equation from a), substitute 5 for y and solve for x. c) Using the equation from a), substitute 2 for y and solve for x. 5 9.834 0.972 x 5 0.972 x 9.834 0.5084 0.972 x 2 9.834 0.972 x 2 0.972 x 9.834 0.2033 0.972 x log 0.5084 log 0.972 x log 0.2033 log 0.972 x log 0.2033 x log 0.972 log 0.2033 x log 0.972 x 56.1° log 0.5084 x log 0.972 x log 0.5084 x log 0.972 x 23.82°