MATH 4 part 2

a = 3; b = 2 a. amplitude is = / a / =/3/ =3 b. period is P = 2π b = 2π 2 =π2. y = 1 cos x 2 Solution: a= 1; b=1 2 a. amplitude is = /a/ =/1 / 2 =1 2 b. period is P = 2π b = 2π 1 = 2π3. y = -3 sin 4x Solution: a = -3; b = 4 a. amplitude is = /a /

= /-3/ =3 b. period is P = 2π b = 2π 4 =π 24. y = cos 4x Solution: a = 1; b = 4 a. amplitude is = /a / = /1/ =1 b. period is P = 2π b = 2π 4 =π 25. y = - 2 sin x 3 Solution: a=-2; b=1 3 a. amplitude is = /a/ = /- 2 / 3

=2 3 b. period is P = 2π b = 2π 1 = 2πTry this out A. Determine the amplitude of the following functions. 1. y = 3 sin x 2. y = 2 sin 1 x 2 3. y = 3 cos 2x 2 4. y = -2 cos 2x 5. y = cos 4 x 6. y = 2 sin 2x 7. y = 1 sin x 2 8. y = 3 sin 1 x 42 9. y = -4 sin 3x 10. y = -2 sin 3 x 2 B. Determine the period of the following functions. 1. y = 6 sin 2 x 3 2. y = 5 sinx 3. y = 4 sin 1 x 2

4. y = - 1 cos 3 x 24 5. y = 3 cos 1 x 2 6. y = cos 3x 7. y = -3 sin 2 x 3 8. y = -6 sin 2x 9. y = 2 sin 1 x 5 10. y = sin 4x Lesson 3 Finding the value of y in a circular function when an angle is givenExamples: 1. Find y = 3 sin x , if x = 5π 4 y = 3 sin 5π 4 y = 3 (- 2 ) 2 y=-3 2 2 2. Find y = 2 cos x, if x = 3π 2 y = 2 cos 1 ( 3π ) 22 = 2 cos 3π 4 y = 2 (- 2 ) 2

y=- 2Try this out Find the value of y: 1. y = 3 sin x, If x = π 42 2. y = 4 sin 1 x, if x = π 23 3. y = 3 tan x, if x = 5π 4 4. y = cos 2x, if x = π 6 5. y = cos x, if x = 3πLet’s Summarize1. The graph of the sine function is periodic. The period is 2π. Its domain is theset of real number and range is [-1, 1]. Its amplitude is 1 and the curvecrosses the x-axis at the odd multiples of π . It has a maximum value 1 and a 2minimum value -1. The graph is increasing in the interval 0, π  and  3π ,2π  2   2 while decreasing in the interval π , 3π  the period 2π.  2 2  over2. The graph of the cosine function is periodic with a period 2π. Its domain is the set of real number and range is [-1, 1]. Its amplitude is 1 and the curve crosses the x-axis at the multiples of π. It has a maximum value 1 and a minimum value -1. The graph is increasing in the interval [π, 2π] while decreasing over the interval [0, π].3. The graph of the tangent function is periodic with a period π. Its domain is theset of real numbers except the odd multiples of π where tangent is undefined. 2The range is the set of real numbers. It is an odd function and has vertical πasymptotes at odd multiples of 2 .

4. The function in the form of y = a sin bx and y a cos bx, the amplitude is /a/ and the period is 2π . bWhat have you learnedGiven the following function, identify the amplitude of:1. y = 1 cos x 22. y = 4 cos x3. y = sin 4xDetermine the period of the following functions.4. y = 3 sin x5. y = cos x6. y = 4 sin 4x7. What is the value of y = 1 cos x if x = 1 228. What is the value of y = 3 tan 2x , it x = 7π . 69. The period of the tangent function isa. 2 π c. πb. π d. 3π 2 2.10. What is the amplitude of the sine function?a. 1 c. 2b. -1 d. -2

11. Which of the following are zeros of y = cos Ө for the interval 0 ≤ Ө ≤ 2π?a. 0, π and 2π c. π and 3π 44b. π and 3π d. π and 5π 22 6612. which of the intervals is the sine function increasing over the period of 2π?a. [0, π] c. [π, 2π]b. 0, π  and 3π , 2π d.  π , 3π  2   2  2 2 

Answer KeyHow much do you know 1. a 2. d 3. y = 2 4. 2 5. 2 6. 3 4 7. -2 8. 4π 9. 3π 10. 10π 11. a 12. aTry this outLesson 1A. 1. Real Numbers2. [-1, 1]3. 0, -π, -2π[ ] [ ]4. increasing: 3π , π − 2π,− 2 − 2 ,0 [ ]decreasing: − 3π ,− π 2 2B. 1. Real Numbers2. [-1, 1]3. − π , - 3π 2 24. increasing: [π, 0]

decreasing: [-2π, -π]C. 1. Real numbers except the odd multiples of π 2 2. Real Numbers 3. all odd multiples of π ; ± π ,± 3π 2 22 4. all odd multiples of π ; ± π ,± 3π 2 22 5. -2π, - π, 0, π, 2 π Lesson 2 A. 1. amplitude : 3 2. amplitude 2 3. amplitude: 3 2 4. amplitude: 2 5 amplitude: 1 6. amplitude: 2 7. amplitude: 1 2 8. amplitude: 3 4 9. amplitude: 4 10. amplitude: 2 B. 1. Period: 3π 2. Period: 2π 3. Period 4π 4. Period: 8π 3 5. Period : 4π 6. Period : 2π 3

7 Period : 3π 8. Period: π 9. Period: 10π 10. Period: π 2Lesson 3 1. y = 3 4 2. y = 2 3. y = - 3 2 2 4. y = 1 2 5. y = -1What have you learned 1. 1 2 2. 4 3. 1 4. 2π 5. 2π 6. π 2 7. 0 8. 3 3 9. c 10. 1

Module 4Circular Functions and Trigonometry What this module is about This module is about the properties of the graphs of a circular functions.You will learn how the graphs of circular function look like and how they behavein the coordinate plane. What you are expected to learnThis module is designed for you to:1. describe the properties of the graphs of the functions: • sine • cosine • tangent2. graph the sine, cosine and tangent functions.3. solve trigonometric equations.How much do you know1. What is The period of the sine function y = sin x?a. 2 π b. π c. π d. 3π 2 22. What is the amplitude of a cosine function y = cos x? d. 1a. -2 b. -1 c. 23. What is the value of y = 4 sin 1 x, if x = 5π ? 23

4. What is the value of y = 2 sin x, if x = π ? 4Given the following functions, identify the amplitude of :5. y = 2 cos x6. y = 3 sin x 47. y = -2 sin 3 x 2Given the following functions, determine the period of:8. y = 2 cos 1 x 29. y = 4 sin 2 x 310. y = sin 1 x 511. Which of the following are zeros of y = tan Ө for the interval 0 ≤ Ө ≤ 2π?a. 0, π and 2π c. π and 3π 44b. π and 3π d. π and 5π 22 6612. In which of the following intervals is the cosine function decreasing over the interval [0, 2π]?a. [0, π] c. [π, 2π]b. 0, π  and 3π , 2π d.  π , 3π  2   2  2 2 13. solve for the solution set of sin x – 1 = 0 in the interval 0 < θ < 2π.

What you will do Lesson 1 Graphs of Sine, Cosine and Tangent Circular functions can also be graphed just like the other functions youhave learned before. The difference is that the graphs of circular functions areperiodic. A function is said to be periodic if the dependent variable y takes on thesame values repeatedly as the independent variable x changes. Observe the changes in the values of y = sin θ and y = cos θ for arclengths from -2π to 2π. θ - 3π to -π to - - π to 0 to - 0 to π to π to 3π tosin θ 2 3π 2 π π 2 3π 2 -2π 2 -π 2 2 π 2 2π 1 to 0 0 to 1 -1 to 0 0 to -1 0 to 1 1 to 0 0 to -1 -1 to 0cos θ 0 to 1 -1 to 0 0 to -1 1 to 0 1 to 0 0 to -1 -1 to 0 0 to 1 Using the arc length, θ, as the independent variable and y = sin θ and y =cos θ as the dependent variables, the graphs of the sine and cosine functionscan be drawn. Below is the graph of y = sin θ for -2π ≤ θ ≤ 2π. This was done by plottingthe ordinates on the y-axis and the arc lengths on the x-axis. Observe the properties of this graph. y = sin θ Y maximum point 2 1 amplitude -2π −3π 0 ππ 3π 2π 2 2 -π − π 2 2 -1 period minimum point -2

You can see that the graph is a curve. Call this the sine curve. Observethat the graph contains a cycle. One complete cycle is the interval from -2π to 0and another cycle is the interval from 0 to 2π. This is called the period of thecurve. Hence, the period of y = sin θ is 2π.The amplitude of the graph of y = sin θ is 1. The amplitude is obtained bygetting the average of the maximum value and the minimum value of thefunction. The maximum point is  π ,1 and the minimum point is  3π ,−1 for the 2  2 interval [0, 2π]. The graph crosses the x-axis at (0, 0), (π, 0), and (2π, 0) for theinterval [0, 2π]. Observe also that the sine graph is increasing from 0 to π and 2from 3π to 2π, and decreasing from π to 3π for the interval [0, 2π]. 2 22The Graph of Cosine Function The graph of y = cos θ can be constructed in the same manner as thegraph of y = sin θ, that is, by plotting the abscissa along the y-axis and the arclengths along the x-axis. Observe the properties of the graph of y = cos θ for theinterval -2π ≤ θ ≤ 2π shown below. y = cos θ Y 2 1-2π 3π -π π 0 π π 3π 2π −2 2 −2 2 -1 -2 You will observe that just like the graph of y = sin θ, it is also a curve. Italso has a period of 2π and amplitude 1. For the interval [0, 2π], the minimumpoint is (π, -1), maximum points are (0, 1) and (2π, 1) and the graph crosses the

x-axis at  π , 0 and  3π , 0 . The graph is decreasing from 0 to π and 2  2 increasing from π to 2π over the interval [0, 2π].The Graph of the Tangent Function The graph of y = tan θ can be drawn in the same manner that the graphsof the sine and cosine functions. The value of the tangent of an angle is plottedalong the y-axis and the arc lengths on the x-axis. Observe that the tangent ofthe odd multiples of π (90o) are not defined so that the graph is discontinuous at 2those values. These are denoted by the broken lines (called asymptotes) thatseparate one complete cycle from the others. Thus, the domain of these functionexclude all odd multiples of π while the range is the set of real numbers. 2 Y - 5π - 2π − 3π -π − π 0 π π 3π 2π 5π 2 2 2 2 22 Notice that the period of the graph of the tangent function is π. This is shownby a complete curve in the interval − π , π  . See that the other curves are 2 2 repetitions of the curve for the given interval. The graph of the tangent function isalso said to be an odd function and that the graph is symmetrical with respect tothe origin. You will also see from the graph that it is an increasing function for thedifferent sets of intervals.Try this outA. Refer to the graph of the y = sin θ to answer the following. 1. What is the domain of the sine function?

2. What is its range? 3. Give the intercepts of y = sin θ for the interval [-2π, 0] 4. Determine the interval where the graph of y = sin θ is (a) increasing, (b) decreasing for the interval [-2π, 0].B. Refer to the graph of y = cos θ to answer the following. 1. What is the domain of the cosine function? 2. What is its range? 3. Give the intercepts of y = cos θ for the interval [-2π, 0] 4. Determine the interval where the graph of y = cos θ is (a) increasing, (b) decreasing for the interval [-2π, 0].C. Refer to the graph of y = tan θ 1. What is the domain of the graph of y = tan θ? 2. What is its range? 3. At what values of θ in the graph is tangent not defined? 4. Give the vertical asymptotes of the graph? 5. What are the zeros of y = tan θ? Lesson 2 Properties of Sine and Cosine functions The two properties of Sine and Cosine functions are amplitude and aperiod of a function. This can be determined from a given equations. Thefunction in the form of y = a sin bx and y a cos bx, the amplitude is /a/ and theperiod is 2π . bExamples:Determine the amplitude and the period of the given function: 1. y = 3 sin 2x Solution:

a = 3; b = 2 a. amplitude is = / a / =/3/ =3 b. period is P = 2π b = 2π 2 =π2. y = 1 cos x 2 Solution: a= 1; b=1 2 a. amplitude is = /a/ =/1 / 2 =1 2 b. period is P = 2π b = 2π 1 = 2π3. y = -3 sin 4x Solution: a = -3; b = 4 a. amplitude is = /a /

= /-3/ =3 b. period is P = 2π b = 2π 4 =π 24. y = cos 4x Solution: a = 1; b = 4 a. amplitude is = /a / = /1/ =1 b. period is P = 2π b = 2π 4 =π 25. y = - 2 sin x 3 Solution: a=-2; b=1 3 a. amplitude is = /a/ = /- 2 / 3

=2 3 b. period is P = 2π b = 2π 1 = 2πTry this out A. Determine the amplitude of the following functions. 1. y = 3 sin x 2. y = 2 sin 1 x 2 3. y = 3 cos 2x 2 4. y = -2 cos 2x 5. y = cos 4 x 6. y = 2 sin 2x 7. y = 1 sin x 2 8. y = 3 sin 1 x 42 9. y = -4 sin 3x 10. y = -2 sin 3 x 2 B. Determine the period of the following functions. 1. y = 6 sin 2 x 3 2. y = 5 sinx 3. y = 4 sin 1 x 2

4. y = - 1 cos 3 x 24 5. y = 3 cos 1 x 2 6. y = cos 3x 7. y = -3 sin 2 x 3 8. y = -6 sin 2x 9. y = 2 sin 1 x 5 10. y = sin 4x Lesson 3 Trigonometric Equations In this section we will solve trigonometric equations using your knowledgein solving algebraic equations. We will also find values which are true for thedomain of the variables under some given conditions.Examples: 1. Find θ in 3 cosθ - 2 = 0 in the interval 0 < θ < 2π. Solution: 2cosθ - 3 = 0 2cosθ = 3 cosθ = 3 2 θ = π the reference angle 6In the interval 0 < θ < 2π, θ = π , 11π . 66

Since the cosine function has a period of 2π, we can obtain the general solutionby adding multiple of 2π. We have, θ = π + 2nπ , 11π + 2nπ , where n is an int eger. 66 or θ = 30o + 360on, 330o + 360on wnere n is an integer. 2. sec x = - 2 in the interval 0 < θ < 2π. Solution: sec x = - 2 Since sec x is 1 , then cos x 1 cos x = - 2 = 1⋅ 2 22 cos x = - 2 2 x = π the reference angle 4 Since x is in the interval interval 0 < x < 2π and cos x is negative, then x = 3π 4 and 5π . 4 3. Determine the solution set of tan2 θ = tan θ in the interval 0 < θ < 2π. Solution: tan2 θ = tan θ tan2 θ - tan θ = 0

tan θ ( tan θ - 1) = 0tan θ = 0 tanθ - 1 = 0θ = 0, π tan θ = 1 θ = π , 5π 44The solution of the tan2 θ = tan is {0, π , π, 5π } 444. Determine the solution set of sin 2x = 0 in the interval 0 < x < 2π.Solution:cos 2x = 0 Since 0 < x < 2π then 0 < 2x < 4π.. Then 2x = π , 3π , 5π , 7π 22 2 2and x = π , 3π , 5π , 7π 44 4 4The solution set of the given equation is { π , 3π , 5π , 7π }. 44 4 4Try this outDetermine the solution of the following equation in the interval 0 < x < 2π. 1. tan x sin x = 0 2. 2 cos2x + cos x = 0 3. 2 sin2 x + 5 cos x – 3 = 0 4. tan 2x = 1 5. 4 sin2 x = 3

6. cot 2 x – 1 = 07. 4sinx cos x = - 38. 3 cos x = -69. cot2 x – 1 = 010. (cos x – 1) (cos x + 1) = 0Let’s Summarize1. The graph of the sine function is periodic. The period is 2π. Its domain is the set of real number and range is [-1, 1]. Its amplitude is 1 and the curvecrosses the x-axis at the odd multiples of π . It has a maximum value 1 and a 2minimum value -1. The graph is increasing in the interval 0, π  and  3π ,2π  2   2 while decreasing in the interval π , 3π  the period 2π.  2 2  over2. The graph of the cosine function is periodic with a period 2π. Its domain is the set of real number and range is [-1, 1]. Its amplitude is 1 and the curve crosses the x-axis at the multiples of π. It has a maximum value 1 and a minimum value -1. The graph is increasing in the interval [π, 2π] while decreasing over the interval [0, π].3. The graph of the tangent function is periodic with a period π. Its domain is theset of real numbers except the odd multiples of π where tangent is undefined. 2The range is the set of real numbers. It is an odd function and has vertical πasymptotes at odd multiples of 2 .4. The function in the form of y = a sin bx and y a cos bx, the amplitude is /a/ and the period is 2π . b

What have you learnedGiven the following function, identify the amplitude of: 1. y = 1 cos x 2 2. y = 4 cos x3. y = sin 4xDetermine the period of the following functions.4. y = 3 sin x5. y = cos x6. y = 4 sin 4x7. What is the value of y = 1 cos x if x = 1 228. What is the value of y = 3 tan 2x , it x = 7π . 69. The period of the tangent function isa. 2 π c. πb. π d. 3π 2 2.10. What is the amplitude of the sine function?a. 1 c. 2b. -1 d. -2

11. Which of the following are zeros of y = cos Ө for the interval 0 ≤ Ө ≤ 2π?a. 0, π and 2π c. π and 3π 44b. π and 3π d. π and 5π 22 6612. which of the intervals is the sine function increasing over the period of 2π?a. [0, π] c. [π, 2π]b. 0, π  and 3π , 2π d.  π , 3π  2   2  2 2 13. Determine the solution of sin 1 x = 0 in the interval 0 ≤ Ө ≤ 2π. 3

Answer KeyHow much do you know1. a2. d3. y = 24. 25. 26. 3 47. -28. 4π9. 3π10. 10π11. a12. a13. π 2Try this outLesson 1A. 1. Real Numbers2. [-1, 1]3. 0, -π, -2π[ ] [ ]4. increasing: 3π , π − 2π,− 2 − 2 ,0[ ]decreasing: − 3π ,− π 2 2B. 1. Real Numbers 2. [-1, 1]

3. − π , - 3π 2 24. increasing: [π, 0] decreasing: [-2π, -π]C. 1. Real numbers except the odd multiples of π 2 2. Real Numbers 3. all odd multiples of π ; ± π ,± 3π 2 224. all odd multiples of π ; ± π ,± 3π 2 225. -2π, - π, 0, π, 2 πLesson 2A. 1. amplitude : 3 2. amplitude 2 3. amplitude: 3 2 4. amplitude: 2 5 amplitude: 1 6. amplitude: 2 7. amplitude: 1 2 8. amplitude: 3 4 9. amplitude: 4 10. amplitude: 2B. 1. Period: 3π 2. Period: 2π 3. Period 4π 4. Period: 8π 3

5. Period : 4π 6. Period : 2π 3 7 Period : 3π 8. Period: π 9. Period: 10π10. Period: π 2Lesson 31. tan x sin x = 0Solution: tan x sin x = 0 tan x = 0 sin x = 0 x = π, 2π, x = π, 2πThe solution set of tan x sin x = 0 is π and 2π.2. 2 cos2 x + cos x = 0Solution: 2 cos2 x + cos x = 0 cos x(2 cos x – 1) = 0 cos x = 0 2 cos x – 1 = 0 x = π , 3π cos x = 1 22 2 x = π , 5π 33The solution of 2 cos2 x + cos x = 0 is { π , π , 3π , 5π }. 23 2 3

3. 2 sin2 x + 5 sin x – 3 = 0Solution:2 sin2 x + 5 sin x – 3 = 0(2 sin x – 1)( sin x + 3) = 02 sin x – 1 = 0 sin x + 3 = 0 2 sin x = 1 sin x = -3 sin x = 1 x = no solution 2 x = π , 5π 664. tan 2x = 1Solution:tan 2x = 10 < 2x < 4π 0 < x < 2π2x = π , 5π x = π , 5π 44 88The solution set is π , 5π . 885. 4 sin2 x = 3Solution:4 sin2 x = 3sin2 x = 3 4sin x = 3 2

x = π , 2π 33The solution set is { π , 2π }. 336. cot2 x – 1 = 0Solution:cot2 x – 1 = 0(cot x – 1)(cot x + 1) = 0cot x – 1 = 0 cot x + 1 = 0cot x = 1 cot x = -1 x = π , 5π x = 3π , 7π 44 44The solution set is { π , 3π , 5π , 7π } 44 4 47. 2 cos x - 3 = 0Solution:2 cos x - 3 = 02 cos x = 3 cos x = 3 2 x = π , 11π 66The solution is { π , 11π } 66

8. 3 cos x = -6Solution:3 cos x = -6cos x = -2No solution, because all values of cos x are between 1 and -1.9. 4 cos2 x = 1Solution: 4 cos2 x = 1cos2 x = 1 4 cos x = 1 2 x = π , 5π 33The solution set is { π , 5π } 3310. (cos x – 1) (cos x + 1) = 0Solution:(cos x – 1) (cos x + 1) = 0cos x – 1 = 0 cos x + 1 = 0cos x = 1 cos x = -1 x = 0, 2π x=πThe solution set is {0, π, 2π}

What have you learned 1. 1 2 2. 4 3. 1 4. 2π 5. 2π 6. π 2 7. 0 8. 3 3 9. c 10. 1 11. a 12. b 13. { π , 2π } 33

Module 4 Logarithmic Functions What this module is about This module deals with the definition, graph, properties, laws and application ofthe laws of the logarithmic function; and how to solve simple logarithmic equation. Asyou go over the discussion and exercises, you will appreciate the importance of thisfunction. Find enjoyment in learning this module and go over the discussion andexamples if you have not yet mastered a concept. What you are expected to learn This module is designed for you to:1. define the logarithmic function f(x) = logax as the inverse of the exponential function f(x) = ax,2. draw the graph of the logarithmic function f(x) = logax,3. describe some properties of the logarithmic function from its graph;4. apply the laws of logarithms; and5. solve simple logarithmic equations. How much do you know1. What is the inverse of the exponential function? a. Linear function c. Polynomial function b. Quadratic function d. Logarithmic function2. Which of the following is equivalent to log216 = 4? a. 24= 16 c. 162 = 4 b. 42 = 16 d. 164 = 22. Which of the following is equivalent to (125 ) 1 = 5? 3 a. log1255 = 1 c. log 1 5 = 125 3 3 b. log5125 = 1 d. log5 1 = 125 3 34. The graph of y = log5x is asymptotic with which of the following lines?

a. y = x c. y - axisb. x - axis d. y = 15. The graphs of y = 10x and y = log10x are symmetric with respect to whatline?a. y = x + 1 c. y = xb. y = 2x d. y = 2x6. What is the point common to the graphs of functions in the form y = logax?a. (0, 0) c. (0, 1)b. (1, 1) d. (1, 0)7. Write log3x2 + log3y3 – log3z as a single logarithm.a. log3  x2 • y3  c. log3  x2 + y3   z   z b. log3(x2 + y3 - z) d. none of the above8. Simplify log318 + log32 – log34 as a single number.a. 16 c. 3b. 9 d. 29. Given: log102 = 0.3010, log103 = 0.4771, and log105 = 0.6990, what is log101.2?a. 0.2054 c. 1.1132b. 0.0791 d. -0.555310. Determine the value of x if log7x = - 2. c. 49 a. 14 d. 1 b. 1 14 49 What you will do Lesson 1 Logarithmic Function In the previous module, you have learned about inverse functions. Recallthat when the domain of one function is the range and the range is the domain ofthe other then they are inverses. Remember also that to determine the inverse ofa function given an equation you have to interchange x and y then solve for y.

Look at the illustration below on how to find the inverse of the exponentialfunction, y = ax.Exponential Function Inverse function y = ax x = ay You will notice that the inverse of the exponential function shows that “y isthe exponent to which the base a is raised in order to obtain the power x”. The inverse of the exponential function above is called logarithmicfunction. The function is defined by the equation - Exponent x = ay or y = logax (a>0, a ≠ 1, x > 0) Base PowerThe equation of a logarithmic function is read as “y is the logarithm of x to thebase a”. Take note that in the notation, a is the base, x is the power and y is theexponent to which a is raised in order to obtain x.Example 1. The logarithmic equation 2 = log749 is read as “2 is the logarithm of 49 to the base 7” or “the logarithm of 49 to the base 7 is 2” which means that the exponent of 7 in order to get 49 is 2.Example 2. The logarithmic equation log6108 = 3 is read as “the logarithm of 108 to the base 6 is 3” or “3 is the logarithm of 108 to the to the base 6” meaning the exponent of 6 is 3 to get 108. Notice from the notation above that y = logax is equivalent to x = ay. Thus,an equation in exponential form can be expressed in logarithmic form and vice-versa. Study the examples that follow.Example 3. Transform the following equations in logarithmic form. 1. 26 = 64 2. 3-2 = 1 3. (128 ) 1 = 2Solutions: 7 91. In 26 = 64, the base is 2, the exponent is 6 and the power is 64. Thus, 26 = 64 is equivalent to 6 = log264 or log264 = 6.

2. In 3-2 = 1 , the base is 3, the exponent in -2 and the power is 1 . 99 Hence, 3 -2 = 1 is equivalent to -2 = log3 1 or log3 1 = -2. 9 9 9 .3. In (128 ) 1 = 2, the base is 128, the exponent is 1 and the power is 2. 7 7 Therefore, (128 ) 1 =2 is equivalent to 1 = log1282 or log1282 = 1. 7 77Example 4. Transform the following logarithmic equations to exponential form.1. log51 = 0 2. log10.0001 = -4 3. logca = -bSolutions:1. In log51 = 0, the base is 5, the exponent is 0, and the power is 1. Therefore, log51 = 0 is equivalent to 50 = 1.2. In log100.0001 = -4, the base is 10, the exponent is -4, and the power is 0.0001. Thus, log100.0001 = -4 is equivalent to 10-4 = 0.0001.3. In logca = -b, the base is c, the exponent is -b, and the power is a. Hence, logca = -b is equivalent to c-b = a. Logarithms can be obtained by considering the corresponding exponentialform of the expression.Example 5. Evaluate the logarithms of the following: 1. log464 2. log927 3. log5 1Solutions: 6251. Let log464 = x. Transform it in exponential form then solve for x. 4x = 64 4x = 43 x=3 ∴log464 = 32. Let log927 = x. Transform it in exponential form then solve for x. log927 = x 9x = 27 (32)x = 33 32x = 33

2x = 3 x=3 2 ∴ log927 = 3 23. Let log5 1 = x. Transform it in exponential form then solve for x. 625 log5 1 =x 625 5x = 1 625 5x = 5-4 x = -4 ∴ log5 1 = -4 625Try this outA. Write the equivalent exponential form of the following. 1. log232 = 5 7. log2 1 = -4 2. log381 = 4 16 3. log1212 = 1 4. log 10100000 = 5 8. logrt = p 5. log1/5125 = -3 9. logap = s 6. log464 = 3 10. logqp = mB. Write the equivalent logarithmic form. 1. 35 =243 6. 6 −2 = 1 2. 90 = 1 36 3. 112 = 121 7. a-5 = 1 a5 ( ) ( )4. 1 =2 4 16 8. 16 2 = 4 3 81 9. 1252/3 = 25 ( )5. 1 =16 2 4 10. 642/3 = 16 3 9C. Evaluate the logarithm of each of the following 1. log981 7. log644 2. log2128 3. log7343 8. log6 1 4. log25625 216 5. log81 6. log48 9. log4 1 256 10. log8 5 64

Lesson 2Graphs of Logarithmic Function From a previous module, you have learned about the graphs of inverse functions,that is, the graphs of inverse functions are reflections of each other and that they aresymmetrical about the line y = x. Thus, the graph of the logarithmic function y = logaxcan be obtained from the graph of the exponential function y = ax. To do this, simply flipthe graph of y = ax along the line y = x.Example 1. Draw the graph of y = log2x .Solution: To draw the graph of y = log2x, recall the graph of y = 2x. Flip the graph of y = 2xabout the line y = x. You should be able to observe that the two graphs contain thefollowing integral values.x -3 -2 -1 0 1 2 3 x 1 1 1 1248 Fory = 2x 1 1 1 1 2 4 8 8 4 2 842 y = log2x -3 -2 -1 0 1 2 3 example, the point (1, 2) is on the graph of y = 2x and the point (2, 1) is on the graph of y =log2x. Observe now that the graph of y = log2x is a reflection of the graph of theexponential function y = 2x along the line y = x which is the axis of symmetry. This canbe seen in the figure below. y = 2x y=x y = log2x Figure 1

To draw the graph of y = ( )1 x and its inverse y = log 1 x, follow the steps in the 2 2previous example, that is, draw the graph of y = ( )1 x then flip it along the line y = x. 2Observe that the graph of y = log 1 x contain the following integral values shown below. 2 x -3 -2 -1 0 1 2 3 x 84211 1 1 y = 2x 8 4 2 1 1 1 1 248 248 y = log2x -3 -2 -1 0 1 2 3 ( )y = 1x y=x 2 y = log 1 x 2 Figure 2 Notice that just like in Figure 1, the graph of y = log 1 x is a reflection of the graph 2of y = ( )1 x . The two graphs are also symmetrical about the line y = x. 2Try this out A. Sketch the graph of each of the following logarithmic functions using the graph of its inverse. 1. f(x) = log3x 2. f(x) = log4x 3. f(x) = log5x 4. f(x) = log1/3x 5. f(x) = log1/4x 6. f(x) = log1/5x B. Compare and contrast the graphs of numbers 1 and 4, numbers 2 and 5, and numbers 3 and 6. Lesson 3 Properties of the Graph of a Logarithmic Function

From Figure 1 in Lesson 2, you can see the properties of the graph of thelogarithmic function y = log2x. Observe that the domain is the set of all positive realnumbers and the range are all real numbers. It has an x-intercept (1. 0) and itsasymptote is the y-axis. Notice also that the function is positive for all x greater than 1and negative for all x less than 1. Thus, this function is increasing. 4 3 y = log2x 2 1 0 0 1 23 4 5 67 8 9 -1 -2 -3 -4 Figure 1 The properties of the graph in figure 1 illustrates the properties of logarithmicfunctions in the form y = ax, a > 1.1. The domain is the set of all positive real numbers and the range are all realnumbers.2. The x-intercept is (1. 0) and its asymptote is the y-axis.3. The function is positive for all x greater than 1 and negative for all x less than 1.4. The function is increasing. 5 From Figure 2 in Lesson 2, the 4properties of the graph of thelogarithmic function y = log1/2x can be 3 y = log 1 xobserved. The domain is the set of all 2 2positive real numbers and the range are 1all real numbers. It has an x-intercept (1. 00) and its asymptote is the y-axis. -2 -1-1 0 1 2 3 4 5 6 7 8 9Notice also that the function is negativefor all x greater than 1 and positive for -2all x less than 1. Thus, this function is -3decreasing. -4 The properties of the graph in figure 1 illustrates the following properties oflogarithmic functions in the form y = ax, 0 < a < 1.

1. The domain is the set of all positive real numbers and the range are all real numbers. 2. The x-intercept is (1. 0) and its asymptote is the y-axis. 3. The function is negative for all x greater than 1 and positive for all x less than 1. 4. The function is decreasing.Try this outA. Enumerate the properties of the graphs of the following logarithmic functions. Refer to the graphs drawn in Lesson 2. 1. f(x) = log3x 2. f(x) = log4x 3. f(x) = log5x 4. f(x) = log1/3x 5. f(x) = log1/4x 6. f(x) = log1/5xB. Give the properties common to the graphs in A. Lesson 4 Laws of Logarithms Since logarithmic function and exponential function are inverse functions, thelaws of exponents will be used to derive the laws of logarithms. The laws of logarithmswere important tools in shortening complicated computations long before the use ofscientific calculators and computers. Nowadays, logarithms are used for differentpurposes specifically in sciences.A. The Logarithm of a Product Let M = ax and N = ay. By Law of Exponents for Products, MN = ax•ay = ax + y. By definition of logarithmic function, M = ax ↔ logaM = x, N = ay ↔ logaN = y, and MN = ax + y ↔ logaMN = x + y. By substitution, logaMN = logaM + logaN.

From the derivation, the logarithm of the product of two numbers is the sum ofthe logarithms of the two factors.B. The Logarithm of a QuotientUse the given in A and the Law of Exponents for Quotients, M = ax = ax - y . N ayBy definition of logarithmic function, M = ax ↔ logaM = x, N = ay ↔ logaN = y, and M = ax-y ↔ loga  M  = x - y N  N By substitution, loga  M  = logaM - logaN  N  From the derivation, the logarithm of the quotient of two numbers is thedifference of the logarithms of the dividend and the divisor.C. The logarithm of a PowerLet M = ax.By the Law of Exponents for a Power, Mk = (ax)k. By definition of logarithmic function, M = ax ↔ logaM = x, and Mk = (ax)k ↔ logaMk = xk. By substitution, logaMk = klogaM From the derivation, the logarithm of the kth power of a number is k times thelogarithm of the number.Below are examples of the application of the laws of logarithms.Example 1. Simplify the following using the laws of logarithms.1. loga5PQ 4. logb 3 x7 y2. loga  32  53. logb(24)3Solutions:1. Apply the logarithm of a product.

loga5PQ = loga5 + logaP + logaQ2. Apply the logarithm of a quotient. loga  32  = loga32 – loga5  5 3. Apply the logarithm of a power. logb(24)3 = 3logb244. Apply the logarithm of a power and logarithm of a quotient and distributiveproperty. 1 logb 3 x7 = logb  x7  3 y y = 1 logb  x7  3 y = 1 ( logbx7 – logby) 3 = 1 logbx7 - 1 logby 33 = 1 (7logbx) - 1 logby 33 = 7 logbx - 1 logby 33Example 2: Write the following as a single logarithm. Simplify if possible.1. log412p + 9log4q2. logd(h2 – 16) – log4(h + 4)Solution:1. The expression is an application of the logarithm of a power and logarithm of a product. log412p + 9log4q = log412p + log4q9 = log4(12pq9)2. The expression is an application of the logarithm of a quotient. logd(h2 – 16) – logd(h + 4) = log  h2 - 16  h +4 d = log  (h- 4)(h+ 4)   h+4 d    = logd(h – 4)

Example 3: Given that log 2 = 0.3010, log 3 = 0.4771, and log 5 = 0.6990, determine-1. log 15 4. log 4 625 5. log 3 302. log 0.4 23. log 3.6Solution: Observe that no base was indicated in example 3. When no base is indicated, itmeans that the base is 10. By convention, when base 10 is used we write log x insteadof log10x. We call this common logarithm. Thus, we will use this convention from nowon.1. log 15 = log (3•5) = log 3 + log 5 = 0.4771 + 0.6990 = 1.17612. log 0.6 = log 3 = log 3 – log 5 = 0.4771 – 0.6990 = -0.2219 53. log 3.6 = log18 5 = log 2•32 5 = log 2 + log 32 – log 5 = log 2 + 2log3 – log 5 = 0.3010 + 2(0.4771) – 0.6990 = 0.3010 + 0.9542 – 0.6990 = 0.5562 ( )14. log 4 625 = log125 4 ( )1 = log 53 4 3 = log5 4 = 3 log5 4 = 3 (0.6990) 4 = 0.524255. log 3 30 = log3 30 - log 2 2 ( )1 = log 2•3•5 3 - log 2 = 1(log2+log3+log5) - log 2 3

= 1(0.3010+0.4771+0.6990) - 0.3010 3 = 1(1.4771)−0.3010 3 = 0.4924 – 0.3010 = 0.1914Try this outA. Express the following as a single logarithm and simplify:1. log2192 + log262. log336 + log34 – log3163. log5 3 6254. log798 + 2log77 – log725. 1 log3 27 + log3 18 - log3 2 66. log2x + 3 log2y – log2z7. log (3x2 + 11x – 20) – log (3x – 4)8. logb(2x -5) + logb(x + 1)9. 2logax3 + 3logay + 4logaz - 3logaw10. 1 log(x2 - 4) + 3 log(x + 2) 2B. Given log 3 = 0.4771, log 5 = 0.6990, and log 7 = 0.8451, evaluate the following applying the laws of logarithms:1. log 21 8. log 754 212. log 63 53. log 4.24. log 492 9. log (21)25. log 4 45 276. log 75 7 10. log147 57. log 49 9C. If z = log 5, write the following expressions in terms of z:1. log 25 4. log 252. log56 5. log 50003. log 1 125

Lesson 5 Logarithmic Equations Logarithmic equations are equations involving logarithmic functions. From previous lesson, you learned that exponential functions andlogarithmic functions are inverses. Hence, the properties of these functions canbe used to solve equations involving these two functions. Study the following examples illustrating how logarithmic equations aresolved.Example 1: Solve the missing terms in the following logarithmic equations.1. log6216 = x2. logx81 = 2 33. log25x = 3 24. log4(5x + 4) = 35. log (2x2 + 4x – 3) = log (x2 + 2x + 12)Solutions:1. Transform log6216 = x in exponential form, then solve for x. 6x = 216 6x = 63 x=3 Check: If x = 3, then 63 = 216. ∴log6216 = 3 and x = 3.2. Transform logx81 = 2 in exponential form, then solve for x using the 3 properties of exponents. 2 x 3 = 81 2 = 34 x3 3 ( )x 3 2  2 = 34 2 Raise both sides to the same power and 3  x = 36 simplify x = 729Check:

( )If 2 x = 729, then 2 = 36 3 = 34 = 81. 729 3∴x = 729.3. Transform log25x = 3 in exponential form, then solve for x using the 2 properties of exponents. 3 x =25 2 ( )3 x= 52 2 x = 53 x = 125Check: ( )3 = 52 3 = 53 = 125 225 2∴ x = 125.4. Transform log4(5x + 4) = 3 in exponential form, then solve for x. 5x + 4 = 43 5x + 4 = 64 5x = 64 – 4 5x = 60 x= 60 12 x=5Check: If x = 5, then log4[5(12) + 4] = log4(60 + 4) = log464 = 3. ∴x=55. log (2x2 - 7x – 9) = log (x2 - 9x + 6)Since both sides of the equation are in base 10, then2x2 - 7x – 9 = x2 - 9x + 6 Addition Property of Equality2x2 - 7x – 9 – (x2 - 9x + 6) = 0 Simplifyx2 + 2x - 15 = 0 Factoring(x + 5)(x - 3) = 0x + 5 = 0 or x - 3 = 0x = -5 or x = 3Check: If x = -5, then log (2x2 - 7x – 9) = log [2(-5)2 - 7(-5) – 9] = log [2(25) + 35 – 9] = log (50 + 35 – 9) = log 76 and log (x2 - 9x + 6) = log [(-5)2 - 9(-5) + 6]

= log (25 + 45 + 6) = log 76. Hence, x = -5 is a solution If x = 3, then log5 log (2x2 - 7x – 9) = log [2(3)2 - 7(3) – 9] = log [2(9) – 21 – 9] = log (18 – 21 – 9) = log (-12) Hence, x = 3 is not a root since logarithms is defined only for positivenumber. ∴x = -5 is the only solution. Notice that to solve problems 1 to 4 the logarithmic equations were firsttransformed to exponential form.Example 2: Solve for x in the following exponential equations:1. 5x = 9 2. 3x + 7 = 2xSolutions: Exponential equations involving different bases can be solved with the use oflogarithms and the aid of a scientific calculator or a table of logarithms. In the caseof the example 2, a scientific calculator will be used.1. 5x = 9 Get the logarithm of both sides. log 5x = log 9 Logarithm of a Power xlog 5 = log 9 Multiplication Property of Equality x= log9 log 5 Replacement of value from the table or calculator Division Fact 0.9542 x= 0.6990 x = 1.3651Check: Using a scientific calculator, 51.3651 = 9 (rounded to the nearest whole number)2. 3x + 2 = 2x Get the logarithm of both sides log 3x + 2 = log 2x Logarithm of a Power (x + 2)log 3 = x log 2 Distributive Property x log 3 +log 3 = x log 2 Addition Property of Equality x log 3 – x log 2 = -2log 3 Factoring x(log 3 – log 2) = -2log 3

x = −2log3 Multiplication Property of Equality log3- log 2 Replacement of value from the table orx = -2(0.4771) calculator 0.4771 - 0.3010 Multiplication Fact Division Factx = −0.9542 .1761x = -5.4185Check: Using a scientific calculator, 3-5.4185 + 2 = 2-5.4185 3-3.4185 = 2-5.4185 0.0234 = 0.0234Try this outA. Solve for the missing term:1. log7n = -32. log4512 = x3. logc36 = -24. log25r = 7 25. logp256 = -76. log8128 = y7. log x - log 12 = 38. log5x5 - log5x3 = 29. log p – log 3 = 110. log2z + log22z = 311. log7a + log7(a - 1) = log71212. log3y – log3(y – 2) = log3213. log x + log (x + 1) = log 2014. log (3x – 5) + log 2x = log 415. log 25x – log 5 = 216. log3(x2 – 16) – log3(x + 4) = 217. log5(m + 5) – l0g5(m + 1) = log5418. log7(q + 4) – log73 = log7(q – 4)19. log (14v2 + 35v) = log 3 + log (2v + 5)20. log (21h2 + 50h – 14) = log(3h2 + 25h + 14)B. Solve the following exponential equations using logarithms:1. 4x = 52. 32x = 18

3. 5x + 1 = 94. 5x – 2 = 4x – 15. 23x + 5 = 3xLet’s Summarize1. The inverse of the exponential function above is called logarithmic function. The function is defined by the equation, x = ay or y = logax (a>0, a ≠ 1, x > 0)read as “y is the logarithm of x to the base a” where a is the base, x is thepower and y is the exponent to which a is raised in order to obtain x.2. The graph of the logarithmic function y = logax can be obtained from the graph of the exponential function y = ax. To do this, flip the graph of y = ax along the line y = x.3. The properties of the graph logarithmic functions in the form y = ax, a > 1 are as follow:a. The domain is the set of all positive real numbers and the range are all real numbers.b. The x-intercept is (1. 0) and its asymptote is the y-axis.c. The function is positive for all x greater than 1 and negative for all x less than 1.d. The function is increasing.4. The properties of the graph of logarithmic functions in the form y = ax, 0 < a < 1are as follow:a. The domain is the set of all positive real numbers and the range are all real numbers.b. The x-intercept is (1. 0) and its asymptote is the y-axis.c. The function is negative for all x greater than 1 and positive for all x less than 1.d. The function is decreasing.5. The Law of Logarithms are as follows:a. Logarithm of a Product: logaMN = logaM + logaNb. Logarithm of a Quotient: loga  M  = logaM - logaN  N c. Logarithm of a Power: logaMk = klogaM