Mathematics Functions: Logarithms

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1 a place of mind F A C U L T Y O F E D U C A T I O N Department of Curriculum and Pedagogy Mathematics Functions: Logarithms Science and Mathematics Education Research Group Supported by UBC Teaching and Learning Enhancement Fund

2 Log Functions

3 Review of Inverse Functions This question set epects students to be comfortable with transformations and finding the inverse of functions. Students should also know the basic properties of logarithms and eponents. Review: Reflect a graph in the line y = to find the graph of the inverse. y The inverse of an equation in the form y f () can be found by interchanging and y, then solving the equation for y.

4 Log Functions I We are given the function y = a. What is the equation of its inverse? A. y a B. y a C. D. E. y log y log a a The function does not have an inverse Press for hint Solve for y: = a y

5 Solution Answer: D Justification: We find the inverse of the function by interchanging and y values, then solving for y: y a y a y log a interchange and y Recall the definition of logarithms: = a y log a = y

6 Log Functions II What is the graph of the function y = a if a > 1. A. B. C. D.

7 Solution Answer: A Justification: This is one of the basic graphs you should know how to quickly sketch. If a > 1, all functions in the form y = a cross the y-ais at the point (0, 1). As goes to positive infinity, the graph grows eponentially. As goes to negative infinity, the graph is still positive but decays eponentially to y = 0.

8 Log Functions III The function y = log a is the inverse of the function y = a. What is the graph of the function y = log a when a > 1? A. B. C. D.

9 Solution Answer: C Justification: To find the graph of y = log a, reflect the graph y = a across the line y = because the two are inverses of each other. y a y log a Graph B is incorrect because negative eponents are never returned by the log function. Graph D is incorrect because the log function should be able to return values when is between 0 and 1. Graph A is the graph of y = log a when 0 < a < 1. What does y = a look like when 0 < a < 1?

10 Log Functions IV What are the domain and range of f () = log 2 ()? f ( ) log 2( ) Domain Range A. B. C. D. E. > 0 y 0 > 0 All real numbers 0 y 0 0 All real numbers All real numbers All real numbers Press for hint

11 Solution Answer: B Justification: The domain of the log function is the range of an eponential function since these two functions are inverses. Notice that we do not include 0 in the domain because eponential functions do not return the value 0. The range of the log function is the domain of the eponential function. There are no restrictions on the eponent on a number, so the log function can return any number. f ( ) log 2( ) Domain: > 0 Range: All real numbers

12 Log Functions V What point eists on all graphs in the form y = log a, where a > 0? A. B. C. D. E. (0,1) (1, 0) (1,1) (1, a) Nosuch point eists

13 Solution Answer: B Justification: The graph of y = log a must pass through the point (1, 0). This is because log a 1 = 0 for all a > 0. (1, 0) This is similar to the rule that the graph y = a always passes through the point (0, 1), because a 0 = 1. Notice how the point (0, 1) on the eponential function is the inverse of the point on the logarithmic function, (1, 0).

14 Log Functions VI The graph shown to the right is in the form y = log a. What is the value of a. A. B. C. D. E Cannot be determined

15 Solution Answer: C Justification: Use the property that log a a = 1. This means that all graphs in the form y = log a pass through the point (a, 1). y log 1. 5 (1.5,1) Look at the graph and find where it crosses the line y = 1. The -value of this point is the base of the log function. In this question, the graph crosses the line y = 1 at (1.5, 1). The equation of the graph is therefore: y log 1. 5

16 Log Functions VII The graph shown to the right is in the form y = log a. What is the value of a. A. B. C. D. E Cannot be determined

17 Solution Answer: C Justification: Use the same technique eplained in the previous question to find the value of a. (0.5,1) y log 0. 5 Look at the graph and find where it crosses the line y = 1. The -value of this point is the base of the log function. In this question, the graph crosses the line y = 1 at (0.5, 1). The equation of the graph is therefore: y log 0. 5

18 Log Functions VIII What is the correct graph of the following function? f ( ) 2log 3( ) A. B. C. D.

19 Solution Answer: A Justification: We know the shape of the graph of y = log 3. The graph of f() = 2log 3 vertically stretches this graph by 2: y log 3 f ( ) 2log 3 (3,1) (3, 2) (1, 0) (1, 0)

20 Log Functions IX What is the correct graph of the following function? f ( ) log 3( 2 ) A. B. C. D.

21 Solution Answer: C Justification: The value of f() = log 3 2 is equivalent to y = 2log 3 (review the previous question to see its graph). Recall that y = log 3 () is only defined when is positive. The domain of the log function is > 0. f ( ) log 3 2 (3, 2) Since 2 is always positive, the log function f() = log 3 ( 2 ) is defined for all values of. Since 2 is symmetric across the y-ais, so is f() = log 3 2. (1, 0)

22 Log Functions X Given the graph y = log 2 ( 2 ), how many solutions does the following equation have? log 2( 2 ) A. B. C. D. E Cannot be determined y log 2( 2 )

23 Solution Answer: C Justification: The equation given is hard to solve by working only with equations. Since the graph of y = log 2 ( 2 ) is given, find the number of points where y = intersects this graph: (2,2) (4,4) There are two integer solutions at = 2 and = 4. The third solution is between 0 and -1. We do not epect the graph to cross the line y = more times when > 5 since the log function increases much more slowly than y =.

24 Log Functions XI The graph of y = log b is given below. What is the graph of y log 1 ( )? b A. B. C. D. y log b ( )

25 Solution Answer: A Justification: The first step is to rewrite logarithm in terms of a graph we are more familiar with, such as y = log b. y log 1 b log ( ) log 1 log b log log1 log b log 0 log b b Change of base property (to base 10) Since Since log log a b b log a log b 1 0 Change of base property Change of base property: log c a log b a log b c

26 Solution Continued y log 1 b ( ) log b This is a reflection of the graph y = log b across the -ais. Reflect in -ais y log 1 ( ) b y log b ( ) Recall that negative eponents takes the reciprocal of its base. This eplains the change in sign of the eponent returned by the log function. 1 b y b y