Download Exponential and Logarithmic Functions: Modeling Population Growth and more Study notes Mathematics in PDF only on Docsity! Haberman / Kling MTH 111c Section II: Exponential and Logarithmic Functions Module 2: Exponential Models EXAMPLE: Suppose that the function models the population of a certain species of monkeys in South America t years after 1998. Describe this monkey population. (( ) 50 0.945 tm t = ) SOLUTION: Based on our study in Module 1, we know that an exponential function of the form ( ) (1 )xf x a r= ⋅ + has initial value a and growth rate r per unit of x. Thus, we can determine immediately that the monkey population modeled by has initial value 50 and, since (( ) 50 0.945 tm t = ) 0.945 1 ( 0.055)= + − , the growth rate is per year (since the growth rate is negative, the population is decreasing). 5.5%− CONCLUSION: There were 50 monkeys in South America in 1998, and the population is decreasing at the rate of per year. 5.5% EXAMPLE: Suppose that the population of a certain species of snakes in South America was 136 in 1998. If this snake population is increasing at the rate of 2.7% per year, find a function that models this snake population. SOLUTION: Based on our study in Module 1, we know that an exponential function of the form ( ) (1 )xf x a r= ⋅ + has initial value a and growth rate r per unit of x. Thus, we can determine immediately that this snake population is modeled by the exponential function where t is years after 1998. (( ) 136 1.027 ts t = ) 2 EXAMPLE: Suppose that the population of a certain species of rodent in South America was 6996 in 1998. Suppose further that in 2002 the population of this same species of rodent in South America was 8077. If this rodent population is increasing exponentially, find a function that models this rodent population t years after 1998. SOLUTION: Since we are told that the rodent population is increasing exponentially, we know that an exponential function will model the population. So we know that the function we are looking for has form . Since ( ) tr t a b= ⋅ 0t = represents 1998, we know that the “initial value” is the rodent population in 1998. So we can immediately see that , and that . To find b we can use the fact that in 2002 the population of this same species of rodent in South America was 8077. Since 2002 corresponds with 6996a = ( ) 6996 tr t b= ⋅ 4t = we see that: ( ) ( ) ( ) 4 4 4 1 4 1 4 1 4 8077 6996 8077 6996 8077 6996 (4) 8077 6996 8077 r b b b b = ⇒ ⋅ = ⇒ = ⇒ = ⇒ = Thus, ( ) ( ) 1 4 4 8077 6996 8077 6996 ( ) 6996 6996 t t r t ⎛ ⎞= ⋅⎜ ⎟ ⎝ ⎠ = ⋅ (Note that we cannot cancel the two 6996’s in the algebraic rule for r since one of them is under the exponent and the other one isn’t!) 5 So it looks like ( ) 5 2xf x = ⋅ . Check: 2(2) 5 2 5 4 20 f = = ⋅ = ⋅ and 3 3 1 2 ( 3) 5 2 5 5 8 f − = − = ⋅ = ⋅ Both function values check. Therefore, ( ) 5 2xf x = ⋅ . Try this one yourself and check your answer. Find the equation of the exponential function, that passes through and . )(xg (2, 5) (5, 63) SOLUTION: Since both of these points should satisfy the equation ( ) xg x ab= , we can use them to obtain a system of equations: 2 5 (2, 5) (2) 5 5 (5, 63) (5) 63 263 1g ab g ab ⇒ = ⇒ = ⇒ = ⇒ = Solving the second equation 2 for a, we obtain 2 5a b = and we can use this equation to substitute for a in the first equation 1 : 5 5 2 63 5 63 ab b b = ⇒ = We can now solve this equation for b: 5 2 3 3 1/3 5 63 5 63 63 5 63 5 2.326967 b b b b b b = = = ⎛ ⎞= ⎜ ⎟ ⎝ ⎠ ≈ 6 So 2.326967 can be substituted for the constant b in the equation ( ) xg x ab= : ( ) (2.326967) .xg x a≈ We can now find the value of the constant b by using the fact that the ordered pair satisfies the function (2, 5) ( ) (2.326967) .xg x a≈ 25 (2.326967) 5 5.414775 5 5.414775 0.923399. a a a a ≈ ≈ ≈ ≈ So an approximation for the exponential function that passes through and is (2, 5) (5, 63) ( ) 0.923399(2.326867) .xg x ≈ Click here some practice problems for this module
