Report

Section 6.3 – Exponential Functions Laws of Exponents If s, t, a, and b are real numbers where a > 0 and b > 0, then: ∙ = + 1 =1 ( ) = 0 =1 () = − 1 1 = = Definition: An Exponential Function is in the form, = “a” is a positive real number and does not equal 1 “a” is the base and is the Growth Factor “C” is a real number and does not equal 0 “C” is the Initial Value because 0 = 0 = The domain of f(x) is the set of all real numbers ( + 1) +1 1 = → = = () Section 6.3 – Exponential Functions Examples () ( + 1) = () −1 2 3 1 3 = 2/3 2 0 1 3/2 3 = 1 2 1 3 2 9/4 3 = 3/2 2 2 9 4 27/8 3 = 9/4 2 3 27 8 = = 0 = 1, ℎ = 1 3 =1 2 3 2 3 = 2 Section 6.3 – Exponential Functions Examples () ( + 1) = () −1 1 2 1/4 1 = 1/2 2 0 1 4 1/8 1 = 1/4 2 1 1 8 1/16 1 = 1/8 2 2 1 16 1/32 1 = 1/16 2 3 1 32 = 0 = 1/4, ℎ = 1/4 1 3 = 4 2 1 2 = Section 6.3 – Exponential Functions Properties of the Exponential Function, = , > 1 The domain is the set of all real numbers. The range is the set of all positive real numbers. The y-intercept is 1; x-intercepts do not exist. The x-axis (y = 0) is a horizontal asymptote, as x. If a > 1, the f(x) is increasing function. The graph contains the points (0, 1), (1, a), and (-1, 1/a). The graph is smooth and continuous. Section 6.3 – Exponential Functions = , > 1 The graph of the exponential function is shown below. = = a Section 6.3 – Exponential Functions Properties of the Exponential Function, = , 0 < < 1 The domain is the set of all real numbers. The range is the set of all positive real numbers. The y-intercept is 1; x-intercepts do not exist. The x-axis (y = 0) is a horizontal asymptote, as x. If 0 < a < 1, then f(x) is a decreasing function. The graph contains the points (0, 1), (1, a), and (-1, 1/a). The graph is smooth and continuous. Section 6.3 – Exponential Functions = , 0 < < 1 The graph of the exponential function is shown below. = , 0 < < 1 Section 6.3 – Exponential Functions Euler’s Constant – e The value of the following expression approaches e, 1 1+ as n approaches . Using calculus notation, 1 = lim 1 + →∞ Applications of e Growth and decay Compound interest Differential and Integral calculus with exponential functions Infinite series Section 6.3 – Exponential Functions Theorem If = , then = . Solving Exponential Equations 1) 32−1 = 3 2 − 1 = x=1 2) 22 = 32 22 = 4 ∙ 8 3) 4+2 = 8 22 +2 = 23 22 = 22 ∙ 23 22(+2) = 23 22 = 25 2 = 5 5 = 2 22+4 = 23 2 + 4 = 3 1 =− 2 Section 6.3 – Exponential Functions Solving Exponential Equations 4) 1 +2 3 3−1 93 = +2 = 32 3−−2 = 36 − − 2 = 6 =− 2 7 3 5) 81 ∙ 9−2−2 = 27 9 ∙ 9 ∙ 9−2−2 = 27 32 ∙ 32 32 −2−2 = 33 34 ∙ 3−4−4 = 33 3−4 = 33 −4 = 3 3 =− 4 Section 6.4 – Logarithmic Functions The exponential and logarithmic functions are inverses of each other. The logarithmic function is defined by = = The domain is the set of all positive real numbers 0, ∞ . The range is the set of all real numbers −∞, ∞ . The x-intercept is 1 and the y-intercept does not exist. The y-axis (x = 0) is a vertical asymptote. If 0 < a < 1, then the logarithmic function is a decreasing function. If a > 1, then the logarithmic function is an increasing function. The graph contains the points (1, 0), (a, 1), and (1/a, –1). The graph is smooth and continuous. Section 6.4 – Logarithmic Functions The graph of the logarithmic function is shown below. , 1 = The natural logarithmic function = = ln = = 10 The common logarithmic function = 10 = Section 6.4 – Logarithmic Functions Graphs of = = , 1 = = = a Section 6.4 – Logarithmic Functions Graphs of = = Inverse Functions: = = = = a , 1 = Section 6.4 – Logarithmic Functions Graphs of = = ln Inverse Functions: = = ln Section 6.4 – Logarithmic Functions Graph = + 2 = = a , 1 = Section 6.4 – Logarithmic Functions Graph = +1 + 2 = = a , 1 = Section 6.4 – Logarithmic Functions Graph = − 1 = = a , 1 = Section 6.4 – Logarithmic Functions Graph = ( − 2) − 1 = = a , 1 = Section 6.4 – Logarithmic Functions Change the exponential statements to logarithmic statements 5 = 6.7 8 = 9.2 3 = = 4 5 = 6.7 = 8 6.7 3 = ln = ln 4 Change the logarithmic statements to exponential statements 3 5 = 7 = 4 ln = 6 5 = 3 7 = 4 = 6 Solve the following equations 3 (2) = 1 2 (5 + 1) = 4 2 32 = −3 + 9 2 = 31 3 = 2 5 + 1 = 24 32 = 2−3+9 5 + 1 = 16 25 = 2−3+9 5 = 15 =3 5 = −3 + 9 −4 = −3 4 = 3 Section 6.4 – Logarithmic Functions Solve the following equations = 10 7 = 15 8 + 2 = 12 = ln 10 7 = ln 15 ln 15 = 7 2 = 4 = 2 = ln 2