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3 Exponential Functions Logarithmic Functions Exponential Functions as Mathematical Models Exponential and Logarithmic Functions
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3.1 Exponential Functions
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Exponential Function The function defined by is called an exponential function with base b and exponent x. The domain of f is the set of all real numbers.
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Example The exponential function with base 2 is the function with domain (– , ). The values of f(x) for selected values of x follow:
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Example The exponential function with base 2 is the function with domain (– , ). The values of f(x) for selected values of x follow:
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Laws of Exponents Let a and b be positive numbers and let x and y be real numbers. Then, 1. 1. 2. 2. 3. 3. 4. 4. 5. 5.
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Examples Let f(x) = 2 2x – 1. Find the value of x for which f(x) = 16. Solution We want to solve the equation 2 2x – 1 = 16 = 2 4 But this equation holds if and only if 2x – 1 = 4 giving x =.
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Examples Sketch the graph of the exponential function f(x) = 2 x. Solution First, recall that the domain of this function is the set of real numbers. Next, putting x = 0 gives y = 2 0 = 1, which is the y-intercept. (There is no x-intercept, since there is no value of x for which y = 0)
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Examples Sketch the graph of the exponential function f(x) = 2 x. Solution Now, consider a few values for x: Note that 2 x approaches zero as x decreases without bound: ✦ There is a horizontal asymptote at y = 0. Furthermore, 2 x increases without bound when x increases without bound. Thus, the range of f is the interval (0, ). x – 5 – 4 – 3 – 2 – 1 012345 y1/321/161/81/41/212481632
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Examples Sketch the graph of the exponential function f(x) = 2 x. Solution Finally, sketch the graph: x y – 2 2 – 2 2 42 f(x) = 2 x
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Examples Sketch the graph of the exponential function f(x) = (1/2) x. Solution First, recall again that the domain of this function is the set of real numbers. Next, putting x = 0 gives y = (1/2) 0 = 1, which is the y-intercept. (There is no x-intercept, since there is no value of x for which y = 0)
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Examples Sketch the graph of the exponential function f(x) = (1/2) x. Solution Now, consider a few values for x: Note that (1/2) x increases without bound when x decreases without bound. Furthermore, (1/2) x approaches zero as x increases without bound: there is a horizontal asymptote at y = 0. As before, the range of f is the interval (0, ). x – 5 – 4 – 3 – 2 – 1 012345 y321684211/21/41/81/161/32
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Examples Sketch the graph of the exponential function f(x) = (1/2) x. Solution Finally, sketch the graph: x y – 2 2 – 2 2 42 f(x) = (1/2) x
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Examples Sketch the graph of the exponential function f(x) = (1/2) x. Solution Note the symmetry between the two functions: x y – 2 2 4242 f(x) = (1/2) x f(x) = 2 x
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Properties of Exponential Functions The exponential function y = b x (b > 0, b ≠ 1) has the following properties: 1.Its domain is (– , ). 2.Its range is (0, ). 3.Its graph passes through the point (0, 1) 4.It is continuous on (– , ). 5.It is increasing on (– , ) if b > 1 and decreasing on (– , ) if b 1 and decreasing on (– , ) if b < 1.
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The Base e Exponential functions to the base e, where e is an irrational number whose value is 2.7182818…, play an important role in both theoretical and applied problems. It can be shown that
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Examples Sketch the graph of the exponential function f(x) = e x. Solution Since e x > 0 it follows that the graph of y = e x is similar to the graph of y = 2 x. Consider a few values for x: x – 3 – 2 – 1 0123 y0.050.140.3712.727.3920.09
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531Examples Sketch the graph of the exponential function f(x) = e x. Solution Sketching the graph: x y – 3 – 11 3 – 3 – 11 3 f(x) = e x
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Examples Sketch the graph of the exponential function f(x) = e –x. Solution Since e –x > 0 it follows that 0 0 it follows that 0 < 1/e < 1 and so f(x) = e –x = 1/e x = (1/e) x is an exponential function with base less than 1. f(x) = e –x = 1/e x = (1/e) x is an exponential function with base less than 1. Therefore, it has a graph similar to that of y = (1/2) x. Consider a few values for x: x – 3 – 2 – 1 0123 y20.097.392.7210.370.140.05
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531531Examples Sketch the graph of the exponential function f(x) = e –x. Solution Sketching the graph: x y – 3 – 11 3 f(x) = e –x
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3.2 Logarithmic Functions
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Logarithms We’ve discussed exponential equations of the form y = b x (b > 0, b ≠ 1) But what about solving the same equation for y? You may recall that y is called the logarithm of x to the base b, and is denoted log b x. ✦ Logarithm of x to the base b y = log b x if and only if x = b y (x > 0)
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Examples Solve log 3 x = 4 for x: Solution By definition, log 3 x = 4 implies x = 3 4 = 81.
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Examples Solve log 16 4 = x for x: Solution log 16 4 = x is equivalent to 4 = 16 x = (4 2 ) x = 4 2x, or 4 1 = 4 2x, from which we deduce that
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Examples Solve log x 8 = 3 for x: Solution By definition, we see that log x 8 = 3 is equivalent to
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Logarithmic Notation log x= log 10 x Common logarithm ln x= log e x Natural logarithm
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Laws of Logarithms If m and n are positive numbers, then 1. 1. 2. 2. 3. 3. 4. 4. 5. 5.
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Examples Given that log 2 ≈ 0.3010, log 3 ≈ 0.4771, and log 5 ≈ 0.6990, use the laws of logarithms to find
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Examples
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Examples
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Examples
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Examples Expand and simplify the expression:
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Examples
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Examples
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Examples Use the properties of logarithms to solve the equation for x: Law 2 Definition of logarithms
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Examples Use the properties of logarithms to solve the equation for x: Laws 1 and 2 Definition of logarithms
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Logarithmic Function The function defined by is called the logarithmic function with base b. The domain of f is the set of all positive numbers.
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Properties of Logarithmic Functions The logarithmic function y = log b x(b > 0, b ≠ 1) has the following properties: 1.Its domain is (0, ). 2.Its range is (– , ). 3.Its graph passes through the point (1, 0). 4.It is continuous on (0, ). 5.It is increasing on (0, ) if b > 1 and decreasing on (0, ) if b 1 and decreasing on (0, ) if b < 1.
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Example Sketch the graph of the function y = ln x. Solution We first sketch the graph of y = e x. 1 x y 1 y = e x y = ln x y = x The required graph is the mirror image of the graph of y = e x with respect to the line y = x:
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Properties Relating Exponential and Logarithmic Functions Properties relating e x and ln x: e ln x = x(x > 0) ln e x = x (for any real number x)
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Examples Solve the equation 2e x + 2 = 5. Solution Divide both sides of the equation by 2 to obtain: Take the natural logarithm of each side of the equation and solve:
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Examples Solve the equation 5 ln x + 3 = 0. Solution Add – 3 to both sides of the equation and then divide both sides of the equation by 5 to obtain: and so:
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3.3 Exponential Functions as Mathematical Models 1.Growth of bacteria 2.Radioactive decay 3.Assembly time
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Applied Example: Growth of Bacteria In a laboratory, the number of bacteria in a culture grows according to where Q 0 denotes the number of bacteria initially present in the culture, k is a constant determined by the strain of bacteria under consideration, and t is the elapsed time measured in hours. Suppose 10,000 bacteria are present initially in the culture and 60,000 present two hours later. How many bacteria will there be in the culture at the end of four hours?
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Applied Example: Growth of Bacteria Solution We are given that Q(0) = Q 0 = 10,000, so Q(t) = 10,000e kt. At t = 2 there are 60,000 bacteria, so Q(2) = 60,000, thus: Taking the natural logarithm on both sides we get: So, the number of bacteria present at any time t is given by:
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Applied Example: Growth of Bacteria Solution At the end of four hours (t = 4), there will be or 360,029 bacteria.
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Applied Example: Radioactive Decay Radioactive substances decay exponentially. For example, the amount of radium present at any time t obeys the law where Q 0 is the initial amount present and k is a suitable positive constant. The half-life of a radioactive substance is the time required for a given amount to be reduced by one-half. The half-life of radium is approximately 1600 years. Suppose initially there are 200 milligrams of pure radium. a.Find the amount left after t years. b.What is the amount after 800 years?
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Applied Example: Radioactive Decay Solution a.Find the amount left after t years. The initial amount is 200 milligrams, so Q(0) = Q 0 = 200, so Q(t) = 200e –kt The half-life of radium is 1600 years, so Q(1600) = 100, thus
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Applied Example: Radioactive Decay Solution a.Find the amount left after t years. Taking the natural logarithm on both sides yields: Therefore, the amount of radium left after t years is:
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Applied Example: Radioactive Decay Solution b.What is the amount after 800 years? In particular, the amount of radium left after 800 years is: or approximately 141 milligrams. or approximately 141 milligrams.
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Applied Example: Assembly Time The Camera Division of Eastman Optical produces a single lens reflex camera. Eastman’s training department determines that after completing the basic training program, a new, previously inexperienced employee will be able to assemble model F cameras per day, t months after the employee starts work on the assembly line. a.How many model F cameras can a new employee assemble per day after basic training? b.How many model F cameras can an employee with one month of experience assemble per day? c.How many model F cameras can the average experienced employee assemble per day?
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Applied Example: Assembly Time Solution a.The number of model F cameras a new employee can assemble is given by b.The number of model F cameras that an employee with 1, 2, and 6 months of experience can assemble per day is given by or about 32 cameras per day. c.As t increases without bound, Q(t) approaches 50. Hence, the average experienced employee can be expected to assemble 50 model F cameras per day.
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End of Chapter
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