and Logarithmic Functions Review of Exponential and Logarithmic Functions Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Exponential Functions Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

An exponential function is a function of the form where a is a positive real number (a > 0) and . The domain of f is the set of all real numbers. Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

(1, 6) (1, 3) (-1, 1/6) (-1, 1/3) (0, 1) Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

a >1 Summary of the characteristics of the graph of The domain is all real numbers. Range is set of positive numbers. No x-intercepts; y-intercept is 1. The x-axis (y=0) is a horizontal asymptote as With a>1, is an increasing function and is one-to-one. The graph contains the points (0,1); (1,a), and (-1, 1/a). The graph is smooth continuous with no corners or gaps. Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

(-1, 6) (-1, 3) (0, 1) (1, 1/3) (1, 1/6) Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

0 <a <1 Summary of the characteristics of the graph of The domain is all real numbers. Range is set of positive numbers. No x-intercepts; y-intercept is 1. The x-axis (y=0) is a horizontal asymptote as With 0<a<1, is a decreasing function and is one-to-one. The graph contains the points (0,1); (1,a), and (-1, 1/a). The graph is smooth continuous with no corners or gaps. Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

(1, 3) (0, 1) Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

(-1, 3) (0, 1) Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Domain: All real numbers Range: { y | y >2 } or (-1, 5) (0, 3) y = 2 Domain: All real numbers Range: { y | y >2 } or Horizontal Asymptote: y = 2 Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

More Exponential Functions (Shifts) : An equation in the form f(x) = ax. Recall that if 0 < a < 1 , the graph represents exponential decay and that if a > 1, the graph represents exponential growth Examples: f(x) = (1/2)x f(x) = 2x Exponential Decay Exponential Growth We will take a look at how these graphs “shift” according to changes in their equation... Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

f(x) = (1/2)x f(x) = (1/2)x + 1 f(x) = (1/2)x – 3 Take a look at how the following graphs compare to the original graph of f(x) = (1/2)x : f(x) = (1/2)x f(x) = (1/2)x + 1 f(x) = (1/2)x – 3 Vertical Shift: The graphs of f(x) = ax + k are shifted vertically by k units. Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

f(x) = (2)x f(x) = (2)x – 3 f(x) = (2)x + 2 – 3 Take a look at how the following graphs compare to the original graph of f(x) = (2)x : f(x) = (2)x f(x) = (2)x – 3 f(x) = (2)x + 2 – 3 (3,1) (0,1) (-2,-2) Notice that f(0) = 1 Notice that this graph is shifted 3 units to the right. Notice that this graph is shifted 2 units to the left and 3 units down. Horizontal Shift: The graphs of f(x) = ax – h are shifted horizontally by h units. Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

f(x) = (2)x f(x) = –(2)x f(x) = –(2)x + 2 – 3 Take a look at how the following graphs compare to the original graph of f(x) = (2)x : f(x) = (2)x f(x) = –(2)x f(x) = –(2)x + 2 – 3 (0,1) (0,-1) (-2,-4) Notice that f(0) = 1 This graph is a reflection of f(x) = (2)x . The graph is reflected over the x-axis. Shift the graph of f(x) = (2)x ,2 units to the left. Reflect the graph over the x-axis. Then, shift the graph 3 units down Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Logarithmic Functions Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Logarithmic Functions A logarithmic function is the inverse of an exponential function. For the function y = 2x, the inverse is x = 2y. In order to solve this inverse equation for y, we write it in logarithmic form. x = 2y is written as y = log2x and is read as “y = the logarithm of x to base 2”. y = 2x 1 2 4 8 16 1 2 4 8 16 y = log2x (x = 2y) Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Graphing the Logarithmic Function y = x y = 2x y = log2x Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Comparing Exponential and Logarithmic Function Graphs y = 2x y = log2x The y-intercept is 1. There is no y-intercept. There is no x-intercept. The x-intercept is 1. The domain is All Reals The domain is x > 0 The range is y > 0. The range is All Reals There is a horizontal asymptote at y = 0. There is a vertical asymptote at x = 0. The graph of y = 2x has been reflected in the line of y = x, to give the graph of y = log2x. Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

2 is the exponent of the power, to which 7 is raised, to equal 49. Logarithms Consider 72 = 49. 2 is the exponent of the power, to which 7 is raised, to equal 49. The logarithm of 49 to the base 7 is equal to 2 (log749 = 2). Exponential notation Logarithmic form log749 = 2 72 = 49 In general: If bx = N, then logbN = x. State in logarithmic form: State in exponential form: a) 63 = 216 log6216 = 3 a) log5125 = 3 53 = 125 b) log2128= 7 27 = 128 b) 42 = 16 log416 = 2 Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

State in logarithmic form: Logarithms State in logarithmic form: a) b) log2 32 = 3x + 2 Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Evaluating Logarithms Think – What power must you raise 2 to, to get 128? 2. log327 Note: log2128 = log227 = 7 log327 = log333 = 3 log2128 = x 2x = 128 2x = 27 x = 7 log327 = x 3x = 27 3x = 33 x = 3 logaam = m because logarithmic and exponential functions are inverses. 3. log556 = 6 4. log816 5. log81 log816 = x 8x = 16 23x = 24 3x = 4 log81 = x 8x = 1 8x = 80 x = 0 loga1 = 0 Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Evaluating Logarithms 6. log4(log338) 7. = x log48 = x 4x = 8 22x = 23 2x = 3 2x = 1 9. Given log165 = x, and log84 = y, express log220 in terms of x and y. 8. log165 = x log84 = y = 23 = 8 16x = 5 24x = 5 8y = 4 23y = 4 log220 = log2(4 x 5) = log2(23y x 24x) = log2(23y + 4x) = 3y + 4x Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Logarithmic Functions x = 2y is an exponential equation. Its inverse (solving for y) is called a logarithmic equation. Let’s look at the parts of each type of equation: Exponential Equation x = ay Logarithmic Equation y = loga x exponent /logarithm base number It is helpful to remember: “The logarithm of a number is the power to which the base must be raised to get the given number.” Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Example: Rewrite in exponential form and solve loga64 = 2 base number exponent a2 = 64 a = 8 Example: Solve log5 x = 3 Rewrite in exponential form: 53 = x x = 125 Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Example: Solve 7y = 1 49 y = –2 An equation in the form y = logb x where b > 0 and b ≠ 1 is called a logarithmic function. Logarithmic and exponential functions are inverses of each other logb bx = x blogb x = x Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Examples. Evaluate each: a. log8 84 b. 6[log6 (3y – 1)] logb bx = x log8 84 = 4 blogb x = x 6[log6 (3y – 1)] = 3y – 1 Here are some special logarithm values: 1. loga 1 = 0 because a0 = 1 2. loga a = 1 because a1 = a 3. loga ax = x because ax = ax Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Laws of Logarithms Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Laws of Logarithms Consider the following two problems: Simplify log3 (9 • 27) = log3 (32• 33) = log3 (32 + 3) = 2 + 3 Simplify log3 9 + log3 27 = log3 32 + log3 33 = 2 + 3 These examples suggest the Law: Product Law of Logarithms: For all positive numbers m, n and b where b ≠ 1, logb mn = logb m + logb n Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Consider the following: a. b. = log3 34 33 = log3 34 – 3 = 4 – 3 = log3 34 – log3 33 = 4 – 3 These examples suggest the following Law: Quotient Law of Logarithms: For all positive numbers m, n and b where b ≠ 1, logb m = logb m – logb n n Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

The logarithm of a product equals the sum of the logarithms. The Product and Quotient Laws Product Law: logb(mn) = logbm + logbn The logarithm of a product equals the sum of the logarithms. Quotient Law: The logarithm of a quotient equals the difference of the logarithms. Express as a sum and difference of logarithms: = log3A + log3B - log3C Evaluate: log210 + log212.8 = log2(10 x 12.8) = log2(128) = log2(27) = 7 Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Simplifying Logarithms Solve: x = log550 - log510 x = log55 = 1 Given that log79 = 1.129, find the value of log763: log763 = log7(9 x 7) = log79 + log77 = 1.129 + 1 = 2.129 Evaluate: x = log45a + log48a3 - log410a4 x = 1 x = log44 Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Consider the following: Evaluate a. log3 94 b. 4 log3 9 = 2 • 4 = (log3 32) • 4 = 2 • 4 These examples suggest the following Law: Power Law of Logarithms: For all positive numbers m, n and b where b ≠ 1, logb mp = p • logb m Consider the following: Evaluate a. log3 94 b. 4 log3 9 Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

The logarithm of a number to a power equals the The Power Law Power Law: logbmn = n logbm The logarithm of a number to a power equals the power times the logarithm of the number. Express as a single log: = log5216 Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Applying the Power Laws Evaluate: = 2(1) + 4(1) Given that log62 = 0.387 and log65 = 0.898 solve Thursday, November 29, 2018Thursday, November 29, 2018 = 0.418 Mr M Kennedy

Applying the Power Laws Evaluate: If log28 = x, express each in terms of x: log28 = x log223 = x 3log22 = x a) log2512 b) log22 = log283 = 3log28 = 3x Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

= log12 9 12 = log12 9 – log12 12 = 0.884 – 1 = –0.116 = log12 18 9 = log12 18 – log12 9 = 1.163 – 0.884 = 0.279 More examples: Given log12 9 = 0.884 and log12 18 = 1.163, find each: a. b. log12 2 Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Example: Solve 2 log6 4 – 1 log6 8 = log6 x 3 Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Natural Exponential Functions The most commonly used base for exponential and logarithmic functions is the irrational number e. Exponential functions to base e are called natural exponential functions and are written y = ex. Natural exponential function follows the same rules as other exponential functions. Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Exponential Function y > 0 for all x passes through (0,1) positive slope increasing Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

eln x = x [Let’s y = ln x and x = ey  x = elnx] Natural Logarithms logarithms to base e ( 2.71828) loge x or ln x (Note: These mean the same thing!) ln x is simply the exponent or power to which e must be raised to get x. y = ln x  x = ey Since natural exponential functions and natural logarithmic functions are inverses of each other, one is generally helpful in solving the other. Mindful that ln x signifies the power to which e must be raised to get x, for a > 0, eln x = x [Let’s y = ln x and x = ey  x = elnx] ln ex = x [Let’s y = ln ex  ey = ex  y = x] eln x = ln ex = x Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

ln e = ln 1 = ln 2 = ln 40 = ln 0.1 = Ex) the natural logarithm of x Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

ln e = 1 since e1 = e ln 1 = 0 since e0 = 1 Ex) the natural logarithm of x ln e = 1 since e1 = e ln 1 = 0 since e0 = 1 ln 2 = 0.6931... since e0.6931... = 2 ln 40 = 3.688... since e3.688.. = 40 ln 0.1 = -2.3025 since e-2.3025. = -1 Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy

Natural Logarithmic Function y > 0 for x > 1 y < 0 for 0 < x < 1 passes through (1,0) positive slope (increasing) Thursday, November 29, 2018Thursday, November 29, 2018 Mr M Kennedy