3 Exponential and Logarithmic Functions u Exponential Functions

3 Exponential and Logarithmic Functions u Exponential Functions u Logarithmic Functions u Exponential Functions as Mathematical Models

3. 1 Exponential Functions

Exponential Function u The function defined by is called an exponential function with base b and exponent x. u The domain of f is the set of all real numbers.

Example u The exponential function with base 2 is the function with domain (– , ). u The values of f(x) for selected values of x follow:

Example u The exponential function with base 2 is the function with domain (– , ). u The values of f(x) for selected values of x follow:

Laws of Exponents u Let a and b be positive numbers and let x and y be real numbers. Then, 1. 2. 3. 4. 5.

Examples 2 x – 1 u Let f(x) = 2 . Find the value of x for which f(x) = 16. Solution u We want to solve the equation 22 x – 1 = 16 = 24 u But this equation holds if and only if 2 x – 1 = 4 giving x = .

Examples u Sketch the graph of the exponential function f(x) = 2 x. Solution u First, recall that the domain of this function is the set of real numbers. u Next, putting x = 0 gives y = 20 = 1, which is the y-intercept. (There is no x-intercept, since there is no value of x for which y = 0)

Examples u Sketch the graph of the exponential function f(x) = 2 x. Solution u Now, consider a few values for x: x – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 y 1/32 1/16 1/8 1/4 1/2 1 2 4 8 16 32 u Note that 2 x approaches zero as x decreases without bound: ✦ There is a horizontal asymptote at y = 0. u Furthermore, 2 x increases without bound when x increases without bound. u Thus, the range of f is the interval (0, ).

Examples u Sketch the graph of the exponential function f(x) = 2 x. Solution u Finally, sketch the graph: y 4 f (x ) = 2 x 2 – 2 2 x

Examples u Sketch the graph of the exponential function f(x) = (1/2)x. Solution u First, recall again that the domain of this function is the set of real numbers. u 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)

Examples u Sketch the graph of the exponential function f(x) = (1/2)x. Solution u Now, consider a few values for x: x – 5 – 4 – 3 – 2 – 1 0 1 2 3 y 32 16 8 4 2 1 1/2 1/4 1/8 4 5 1/16 1/32 u Note that (1/2)x increases without bound when x decreases without bound. u Furthermore, (1/2)x approaches zero as x increases without bound: there is a horizontal asymptote at y = 0. u As before, the range of f is the interval (0, ).

Examples u Sketch the graph of the exponential function f(x) = (1/2)x. Solution u Finally, sketch the graph: y 4 2 f(x) = (1/2)x – 2 2 x

Examples u Sketch the graph of the exponential function f(x) = (1/2)x. Solution u Note the symmetry between the two functions: y 4 f (x ) = 2 x 2 f(x) = (1/2)x – 2 2 x

Properties of Exponential Functions u The exponential function y = bx (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.

The Base e u 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. u It can be shown that

Examples u Sketch the graph of the exponential function f(x) = ex. Solution u Since ex > 0 it follows that the graph of y = ex is similar to the graph of y = 2 x. u Consider a few values for x: x – 3 – 2 – 1 0 1 2 3 y 0. 05 0. 14 0. 37 1 2. 72 7. 39 20. 09

Examples u Sketch the graph of the exponential function f(x) = ex. Solution u Sketching the graph: 5 y f (x ) = e x 3 1 – 3 – 1 1 3 x

Examples u Sketch the graph of the exponential function f(x) = e–x. Solution u Since e–x > 0 it follows that 0 < 1/e < 1 and so f(x) = e–x = 1/ex = (1/e)x is an exponential function with base less than 1. u Therefore, it has a graph similar to that of y = (1/2)x. u Consider a few values for x: x y – 3 – 2 – 1 0 1 2 3 20. 09 7. 39 2. 72 1 0. 37 0. 14 0. 05

Examples u Sketch the graph of the exponential function f(x) = e–x. Solution u Sketching the graph: 5 y 3 1 – 3 – 1 f(x) = e–x 1 3 x

3. 2 Logarithmic Functions

Logarithms u We’ve discussed exponential equations of the form y = bx (b > 0, b ≠ 1) u But what about solving the same equation for y? u You may recall that y is called the logarithm of x to the base b, and is denoted logbx. ✦ Logarithm of x to the base b y = logbx if and only if x = by (x > 0)

Examples u Solve log 3 x = 4 for x: Solution u By definition, log 3 x = 4 implies x = 34 = 81.

Examples u Solve log 164 = x for x: Solution u log 164 = x is equivalent to 4 = 16 x = (42)x = 42 x, or 41 = 42 x, from which we deduce that

Examples u Solve logx 8 = 3 for x: Solution u By definition, we see that logx 8 = 3 is equivalent to

Logarithmic Notation log x = log 10 x Common logarithm ln x Natural logarithm = loge x

Laws of Logarithms u If m and n are positive numbers, then 1. 2. 3. 4. 5.

Examples u Given that log 2 ≈ 0. 3010, log 3 ≈ 0. 4771, and log 5 ≈ 0. 6990, use the laws of logarithms to find

Examples u Given that log 2 ≈ 0. 3010, log 3 ≈ 0. 4771, and log 5 ≈ 0. 6990, use the laws of logarithms to find

Examples u Given that log 2 ≈ 0. 3010, log 3 ≈ 0. 4771, and log 5 ≈ 0. 6990, use the laws of logarithms to find

Examples u Given that log 2 ≈ 0. 3010, log 3 ≈ 0. 4771, and log 5 ≈ 0. 6990, use the laws of logarithms to find

Examples u Expand simplify the expression:

Examples u Expand simplify the expression:

Examples u Expand simplify the expression:

Examples u Use the properties of logarithms to solve the equation for x: Law 2 Definition of logarithms

Examples u Use the properties of logarithms to solve the equation for x: Laws 1 and 2 Definition of logarithms

Logarithmic Function u The function defined by is called the logarithmic function with base b. u The domain of f is the set of all positive numbers.

Properties of Logarithmic Functions u The logarithmic function y = logbx (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.

Example u Sketch the graph of the function y = ln x. Solution u We first sketch the graph of y = ex. u The required graph is the mirror image of the y x graph of y = e with respect to the line y = x: y = ex y=x y = ln x 1 1 x

Properties Relating Exponential and Logarithmic Functions u Properties relating ex and ln x: eln x =x (x > 0) ln ex =x (for any real number x)

Examples u Solve the equation 2 ex + 2 = 5. Solution u Divide both sides of the equation by 2 to obtain: u Take the natural logarithm of each side of the equation and solve:

Examples u Solve the equation 5 ln x + 3 = 0. Solution u Add – 3 to both sides of the equation and then divide both sides of the equation by 5 to obtain: and so:

3. 3 Exponential Functions as Mathematical Models 1. Growth of bacteria 2. Radioactive decay 3. Assembly time

Applied Example: Growth of Bacteria u 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. u Suppose 10, 000 bacteria are present initially in the culture and 60, 000 present two hours later. u How many bacteria will there be in the culture at the end of four hours?

Applied Example: Growth of Bacteria Solution u We are given that Q(0) = Q 0 = 10, 000, so Q(t) = 10, 000 ekt. u At t = 2 there are 60, 000 bacteria, so Q(2) = 60, 000, thus: u Taking the natural logarithm on both sides we get: u So, the number of bacteria present at any time t is given by:

Applied Example: Growth of Bacteria Solution u At the end of four hours (t = 4), there will be or 360, 029 bacteria.

Applied Example: Radioactive Decay u Radioactive substances decay exponentially. u 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. u The half-life of a radioactive substance is the time required for a given amount to be reduced by one-half. u The half-life of radium is approximately 1600 years. u 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?

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) = 200 e–kt The half-life of radium is 1600 years, so Q(1600) = 100, thus

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:

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.

Applied Example: Assembly Time u The Camera Division of Eastman Optical produces a single lens reflex camera. u 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?

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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