Skill 6 Exponential and Logarithmic Functions

Transcription

1 Skill 6 Exponential and Logarithmic Functions Skill 6a: Graphs of Exponential Functions Skill 6b: Solving Exponential Equations (not requiring logarithms) Skill 6c: Definition of Logarithms Skill 6d: Graphs of Logarithms Skill 6e: Properties of logarithms (product rule, quotient rule, power rule, change of base) Skill 6f: Logarithmic Equations (not requiring exponentials) Skill 6g: Exponential and Logarithmic Equations requiring inverse operations

2 Skill 6a: Graphs of Exponential Functions An exponential function is defined as an expression with a constant base with a variable exponent. The following are examples of exponential functions: f(x) = x g(x) = π 3x h(x) = 3 x + In general an exponential function is of the form f(x) = a x, where a > 0 and a. Why is it necessary for a > 0? What can't a =? Say, a = then f ( ) = ( ) = i, but we want real numbers. Say, a =, then the range would be {}.. Complete the table below for the function f(x) = x. Then graph the function at the right. x f(x) What is the domain of the basic exponential function? D = (, ) What is the range of the basic exponential function? R = (0, ) What is the equation of the horizontal asymptote of the basic exponential function? y = 0. Complete the table for the function f(x) = x. 3. Complete the table for f(x) = ( )x. Then graph the function above Then graph the function above. x f(x) x f(x)

3 . If < a < b. Sketch a graph that illustrates the difference between f(x) = a x and g(x) = b x 5. If 0 < a < b <. Sketch a graph that illustrates the difference between f(x) = a x and g(x) = b x 6. If a >, how does the graph of f(x) = a x compare to the graphs of g(x) = ( a )x and h(x) =? a x They all have the same graph (they re equivalent). Match the function below with the correct graph. A 7. y = 3 x C 8. y = 3 x D 9. y = 3 x + 5 C E B F E C 0. y = ( 3 )x. y = ( 3 )x. y = ( 3 )x 3. y = 3 x. y = ( 3 ) x 5. y = 3 x A 6. y = 3 x

4 7. What is the domain, range, y-intercept, and the equation of the horizontal asymptote for f(x) = x+ 3. Domain = (, ) Range = ( 3, ) The number e is defined as the value of ( + n )n as n approaches infinity. e is an irrational number, but to ten decimal places it can be approximated as When e is the base of an exponential function, it is called the natural exponential function. 8. Sketch the graph of y = 3e 0.5x x y = Skill 6b: Solving Exponential Equations (not requiring logarithms) Some exponential equations can be solved by rewriting constants values in terms of the base. Solve for x:. 3 x = x 7 + = 0 3 x = 3 x = 6 3x 7 = x 7 = 6 3x 7 = 3x = 9 x = = ( x ). 6 x = 6 6 = x = x x = ( 3 ) x = 3x = 3x = x = 3

5 5. 5 x = x+3 = 6 3x 5 (5 3 ) x = 5 ( 5 ) (x+3) = ( ) (3x 5) 5 6x = 5 6x = x = 3 5x+5 = x 0 5x + 5 = x 0 35 = 7x x = 5 Skill 6c: Definition of Logarithms A logarithm is defined as the inverse of an exponential function.. f(x) = x, A) What is f(3)? B) What is f (8)? f(3) = 3 = 8 f (x) = log x f (8) = log 8 = 3 The exponential equation 3 = 8 can be written as the logarithmic (or log) equation log 8 = 3. Rewrite the following exponential equations as logarithmic equations.. 5 = = = e log 5 65 = log 3 3 = 5 log 000 = 3 ln Note that log 0 x is usually written log x, so instead of writing log 0 00 =, write log 00 =. Also log e x is written ln x. Rewrite the following logarithmic equations as exponential equations. 5. log 7 3 = 3 6. log = 7. log = 8. ln = = 3 0 = = 5 e 0 = Rewrite the following logarithmic equations as exponential equations and determine the value of x. 9. log x = 0. log 6 = x. log x = 5 = x x = 6 x = 6 x = 3 x = 3 5 = x x = 3

6 . log 5 5 = x 3. log x 8 =. log,000,000 = x 5 x = 5 5 x = 5 x = x = x = 8 x = 9 Note: The base must be greater than zero and not equal to one. 0 x =,000,000 0 x = 0 6 x = 6 Skill 6d: Graphs of Logarithms Since a logarithm is the inverse of an exponential function, the graph of a y = log x is the reflection of the graph of y = x across the line y = x. x x x log x 0 0 For a basic logarithm: Domain: (0, ) Range: (, ) Vertical Asymptote: x = 0 X - Intercept: x = State the domain, range, x-intercept, and give the equation of the vertical asymptote for each function below:. f(x) = 5log (x). f(x) = log 5 (x ) Domain: (0, ) Range: (, ) Domain: (, ) Range: (, ) Vertical Asymptote: x = 0 Vertical Asymptote: x = X - Intercept: x = 5 X - Intercept: x = 5

7 3. f(x) = log 3 (9x 7). f(x) = ln( x + ) Domain: ( 7, ) Range: (, ) Domain: (, ) Range: (, ) 9 Vertical Asymptote: x = 7 9 Vertical Asymptote: x = X - Intercept: x = X - Intercept: x = e Match the function below with the correct graph. D 5. y = log 3 x A B C B 6. y = log 3 x C 7. y = log 3 ( x) B 8. y = log x 3 F 9. y = log 3 (x ) D E F E 0. y = log3(x) A. y = log 3 ( x 6 ) B. y = log 3 ( x ) Note:

8 Match the function below with the correct graph. B 3. y = ln x A B C C. y = log 5 x A 5. y = log x Skill 6e: Properties of Logarithms Derivation of the Product Rule log(ab) = y, a = 0 m, and b = 0 n 0 y = ab 0 y = 0 m+n so, y = m + n since a = 0 m and b = 0 n, m = log a and n = log b So, log(ab) = y log(ab) = m + n log(ab) = log a + log b Product Rule of Logarithms log(ab) = log a + log b Also since log(a n ) = log(a a a) = log a + log a + log a = n log a Power Rule of Logarithms log(a n ) = n log a And recall log = log b b Quotient Rule of Logarithms log ( a ) = log a log b b

9 Rewrite the following using the properties of logarithms:. log 3x. log x log x 0 = log 3 + log x = 5 + log x = log x log 00 = log x = + log x = 0 log x. log 3 x 3 y = log 3 x 3 log 3 y = 3 log 3 x log 3 y 5. log 5 ab = log 5 (ab) = log 5 (ab) = [log 5 a + log 5 b] = log 5 a log 5 b 3 x 6. log = log 7 x log 7 7 = log 7 (x) 3 = 3 log 7 x Combine the following using the properties of logarithms into a single logarithm: 7. log(x) + log(y) log (z) 8. + log x 9. log 3 x + log 3 y = log(x) + log(y) log (z) = log x + log y log z = log (x y ) log z = log ( x y z ) = log 6 + log x = log (6x) = log 3 x log log 3 y = log 3 ( x 3 ) + log 3 y = log 3 ( yx 3 )

10 If log and log 8 3 = 0. 58, determine the following: 0. log 8 5. log 8 5. log 8 30 = log 8 5 = log 8 5 (0.77) = log = log 8 5 log = 0.6 = log 8 (5 9) = log log 8 9 = log log 8 3 = log log (0.58) =.83. log = log 8 5 log 8 8 = log 8 (5 3 ) = 3 log 8 (5) 3(0.77) =.3 = log 8 (5 6) = log log 8 6 = log log 8 6 = log log 8 8 = log log = log 8 0 = log 8 ( 5) = log 8 + log 8 5 = log log 8 5 = 3 log log Changing Bases: log a b = c can be rewritten as a c = b so, or, log a c = log b c log a = log b so, c = log b log a log a b = c log a b = log b log a

11 So with just a 'log' or 'ln' button on a calculator, any logarthin can be found. Change of Base Rule for Logarithms log a b = log b log a or log a b = ln b ln a Determine the following to four decimal places: 6. log log 3 8. log 7 ( ) = log 60 log.953 = ln ln = undefined Skill 6f: Logarithmic Equations (not requiring inverse operations) Solve for x:. log(5) + log(x) = log(3) + log (0). log 3 + log x = log 5 + log (x ) log(5x) = log (30) 5x = 30 x = 6 log (3x) = log (5(x )) log (3x) = log (5x 0) 3x = 5x 0 x = 0 x = 5 3. log (x ) = log (5) log (x). log x = log + log (3x ) log (x ) = log ( 5 x ) x = 5 x x x 5 = 0 (x 5)(x + ) = 0 x = 5, x = 5 x = is extraneous log x = log((3x )) log x = log(6x 8) x = 6x 8 x 6x + 8 = 0 (x )(x ) = 0 x =, x =

12 5. log 3 (5 x) = 3 6. log (x + ) + log (x) = = 5 x x = x = log (x(x + )) = 3 log (x + x) = 3 3 = x + x x + x 8 = 0 (x + )(x ) = 0 x = (is extraneous) x = x = Skill 6g: Logarithmic and Exponential Equations Exponential Functions and Logarithmic Functions are inverses of each other; f(x) = x g(x) = ln x f (x) = log x g (x) = e x Simplify the following expressions:. 3 log 3(x+3). log 6 6 x 3. e ln(x 5). ln e (9 x) = x + 3 = x = x 5 = 9 x Solve each equation using inverse functions. Approximate solutions to 3 decimal places when needed x = log(x) 6 = 7. e 5 x = log 0 x = log 50 x log 0 = log 50 x = log 50 x.699 log(x) 6 = log( x ) = = x x = 08 x = ± 08 ln e 5 x = ln (5 x) ln e = ln 5 x = ln x = 5 ln x 3.6 x = ± 0 x = 50 (is extraneous) x = 50

13 8. 5ln(x ) = x+ = x x = 5 x ln(x ) = 3 e 3 = x x = + e 3 x.086 log 5 (x+) = log (x+) (x + ) log 5 = (x + ) log x log 5 + log 5 = x log + log x log 5 x log = log log 5 x(log 5 log ) = log log 5 x (log 5 log ) = log log 5 x log 5 6 = log 5 x = log( 5 ) log( 5 6 ) log 0 ( x) = log 5 x ( x) log 0 = x log 5 x = x log 5 = x + x log 5 = x( + log 5) x = (+log 5) x.589 x.9