Logarithms and Equation Solving

Slides:



Advertisements
Similar presentations
Copyright © Cengage Learning. All rights reserved.
Advertisements

Solving Exponential and Logarithmic Equations
Properties of Logarithms
Chapter 3 Linear and Exponential Changes 3
Exponential and Logarithmic Equations. Exponential Equations Exponential Equation: an equation where the exponent includes a variable. To solve, you take.
5.1 Exponential Functions
Properties of Logarithms
Questions over 4.6 HW???. 4.7 (Green) Solve Exponential and Logarithmic Equations No School: Monday Logarithms Test: 1/21/10 (Thursday)
Chapt. 9 Exponential and Logarithmic Functions
College Algebra Sixth Edition James Stewart Lothar Redlin Saleem Watson.
Functions and Logarithms. One-to-One Functions A function f(x) is one-to-one if f(a) ≠ f(b) whenever a ≠ b. Must pass horizontal line test. Not one-to-one.
Unit 3: Exponential and Logarithmic Functions
College Algebra Fifth Edition
7.6 – Solve Exponential and Log Equations
Solving Exponential and Logarithmic Equations
Chapter Inverse Relations and Functions.
Objectives Solve exponential and logarithmic equations and equalities.
Solve each equation for x. 1. 3x – 12 = 45 x = x = 39.2 Algebra 3 Warm-Up 5.3.
8.6 Solving Exponential and Logarithmic Equations Goal: Solve exponential and logarithmic equations. Correct WS 8.5A.
4.4 Solving Exponential and Logarithmic Equations.
Exponential Equations Like Bases. Warm Up  The following quadratic equation has exactly one solution for x. Find the value of k. Explore more than one.
Solving Exponential and Logarithmic Equations Section 8.6.
Chapter Exponential and logarithmic equations.
6.6 The Natural Base, e Objectives: Evaluate natural exponential and
4.5 Apply Properties of Logarithms p. 259 What are the three properties of logs? How do you expand a log? Why? How do you condense a log?
Exponential and Logarithmic Functions Exponential and Logarithmic Equations: Problem Solving EXPONENTIAL AND LOGARITHMIC FUNCTIONS Objectives Solve.
5.1 LOGARITHMS AND THEIR PROPERTIES Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally.
Logarithmic Functions Recall that for a > 0, the exponential function f(x) = a x is one-to-one. This means that the inverse function exists, and we call.
Logarithms can be used in measuring quantities which vary widely. We use the log function because it “counts” the number of powers of 10. This is necessary.
Solving Logarithmic Equations
2.2.1 MATHPOWER TM 12, WESTERN EDITION 2.2 Chapter 2 Exponents and Logarithms.
LOGARITHMIC EQUATIONS AND IDENTITIES Patterns#12.
Example Example 2 - Extraneous Solution.
Solving Equations Exponential Logarithmic Applications.
Solving Log Equations 1. log 2 72 = log 2 x + log 2 12 log log 2 12 = log 2 x x = x = 8 log 2 x = log 8 xlog2 = log 8 x = 3 3. xlog7 = 2log40.
4.4 Exponential and Logarithmic Equations. Solve: 2 x = 7 3 x+3 = 5.
WARM UP Simplify USING A CALCULATOR Use a calculator or Table 2 1.Find log Find log Find antilog Find antilog
Write in logarithmic form Write in exponential form Write in exponential form Math
Trashketball Review Chapter 3 Exponents and Logarithms.
2.8.1 MATHPOWER TM 12, WESTERN EDITION 2.8 Chapter 2 Exponents and Logarithms.
Welcome to Interactive Chalkboard
Inverse Functions Inverse Operations ▪ One-to-One Functions ▪ Inverse Functions ▪ Equations of Inverses ▪ An Application of Inverse.
Exponential Equations
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Chapter 5: Inverse, Exponential, and Logarithmic Functions
Splash Screen.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
College Algebra Chapter 4 Exponential and Logarithmic Functions
Chapter 10.4 Common Logarithms Standard & Honors
5-3 Logarithmic Functions
Logarithms and Equation Solving
Properties of Logarithms
Exponential & Logarithmic Equations
5-3 Logarithmic Functions
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Warm Up Which plan yields the most interest? Invest $100 Plan A: A 7.5% annual rate compounded monthly for 4 years Plan B: A 7.2% annual rate compounded.
Logarithmic and Exponential Equations
Solve for x: log3x– log3(x2 – 8) = log38x
Logarithmic and Exponential Equations
Which plan yields the most interest. Invest $100 Plan A: A 7
Logarithmic Functions as Inverses.
Precalculus Essentials
PROPERTIES OF LOGARITHMS
Warm-Up Algebra 2 Honors 2/19/18
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Unit 3: Exponential and Logarithmic Functions
Chapter 10.3 Properties of Logarithms Standard & Honors
Solving Systems of Linear Equations by Elimination
Presentation transcript:

Logarithms and Equation Solving Chapter 2 Exponents and Logarithms 2.7A Logarithms and Equation Solving MATHPOWERTM 12, WESTERN EDITION 2.7A.1

Solving Log Equations 1. log272 = log2x + log212 2. 2x = 8 log272 - log212 = log2x 2. 2x = 8 log 2x = log 8 xlog2 = log 8 x = 3 x = 6 3. x = 3.79 xlog7 = 2log40 2.7A.2

4. log7(2x + 2) - log7(x - 1) = log7(x + 1) Solving Log Equations 4. log7(2x + 2) - log7(x - 1) = log7(x + 1) 2x + 2 = (x- 1)(x + 1) 2x + 2 = x2 - 1 0 = x2 - 2x - 3 0 = (x - 3)(x + 1) x - 3 = 0 or x + 1 = 0 x = 3 x = -1 Therefore, x = 3. Check: log7(2x + 2) - log7(x - 1) = log7(x + 1) log7(2x + 2) - log7(x - 1) = log7(x + 1) log7(2(3) + 2) - log7(3 - 1) = log7(3 + 1) log7(2(-1) + 2) - log7(-1 - 1) = log7(-1 + 1) log70 - log7(-2) = log7(0) Negative logarithms and logs of 0 are undefined. log74 = log74 2.7A.3

Solving Log Equations 5. log7(x + 1) + log7(x - 5) = 1 log7[(x + 1)(x - 5)] = log77 (x + 1)(x - 5) = 7 x2 - 4x - 5 = 7 x2 - 4x - 12 = 0 (x - 6)(x + 2) = 0 x - 6 = 0 or x + 2 = 0 x = 6 x = -2 6. log4(4x) - log4(x - 2) = 3 64(x - 2) = 4x 64x - 128 = 4x 60x = 128 x = 2.13 x = -2 is extraneous. Therefore, x = 6. 2.7A.4

Solve: log5(x - 6) = 1 - log5(x - 2) Solving Log Equations Express 12 as a power of 2: Solve: log5(x - 6) = 1 - log5(x - 2) log5(x - 6) + log5(x - 2) = 1 log5(x - 6)(x - 2) = 1 log5(x - 6)(x - 2) = log551 (x - 6)(x - 2) = 5 x2 - 8x + 12 = 5 x2 - 8x + 7 = 0 (x - 7)(x - 1) = 0 x = 7 or x = 1 xlog2 = log12 x = 3.58 23.58 = 12 Since x > 6, the value of x = 1 is extraneous. Therefore, the solution is x = 7. 2.7A.5

Solving Log Equations 7. 3x = 2x + 1 log(3x) = log(2x + 1) x log 3 = x log 2 + 1 log 2 x log 3 - x log 2 = log 2 x(log 3 - log 2) = log 2 8. 23x - 1 = 32x - 1 (3x - 1) log 2 = (2x - 1) log 3 3x log 2 -1 log2 = 2x log3 - log3 3xlog2 - 2xlog3 = log2 - log3 x(3log2 - 2log3) = log2 - log3 x = 3.44 x = 1.71 2.7A.6

Applications of Logarithms 1. Carbon 14 has a half-life of 5760 years. Find the age of a specimen with 24% C-14 relative to living matter. t = 11 859.23 Therefore, the specimen is 11 859.23 years old. 2.7A.7

Applications of Logarithms Find the time period required for $7000 invested at 10%/a compounded semi-annually to grow to $10 000. A(t) = P(1 + i) 2n 10 000 = 7000(1.05)2n log10 - log7 = 2nlog1.05 7.31 = 2n 3.66 = n It would take 3.66 years for the investment to grow to $10 000. 2.7A.8

Applications of Logarithms 3. The value of an investment is given by f(x) = 237.50(1.052)x, where x is the number of 6-month periods. Find the number of complete periods until the investment is worth at least $600. f(x) = 237.50(1.052)x 600 = 237.50(1.052)x 2.5263 = (1.052)x log 2.5263 = x log 1.052 x = 18.28 Therefore, after 19 periods the investment would be worth at least $600. 2.7A.9

36 = t Applications of Logarithms 4. Cell population doubles every 3 h. How long would it take 4 cells to reach a count of 16 384? It would take 36 h to reach 16 384 cells. 36 = t 2.7A.10

Applications of Logarithms 5. For every metre below the water surface, light intensity is reduced by 5%. At what depth is light intensity 40% of that at the surface? Id = Io(1 - 0.05)d 40 = 100(0.95)d 0.4 = 0.95d log 0.4 = dlog0.95 d = 17.86 Therefore, at a depth of 17.86 m the light intensity would be 40%. 2.7A.11

More Applications - Comparing Intensities of Sound For any intensity, I, the decibel level, dB, is defined as follows: where Io is the intensity of a barely audible sound 6. The sound at a rock concert is 106 dB. During the break, the sound is 76 dB. How many times as loud is it when the band is playing? Louder Softer Comparison Thus, it would be 1000 times as loud. I = 107.6 Io I = 1010.6 Io 2.7A.12

More Applications - The Richter Scale I = Io(10)m where m is the measure on the scale 7. Compare the intensities of the Japan earthquake of 1933, which measured 8.9 on the Richter Scale, to the earthquake of Turkey in 1966, which measured 6.9 on the scale. Therefore, the earthquake in Japan is 100 times as intense as the one in Turkey. 2.7A.13

More Applications - The Richter Scale 8. The magnitude of earthquakes is given by where I is the quake intensity and Io is the reference intensity. How many times as intense is a quake of 8.1 compared to a quake with a magnitude of 6.4? Comparison Therefore, a quake of 8.1 is 50.1 times as great. 2.7A.14

More Applications - The Richter Scale 9. Earthquake intensity is given by I = Io (10)m, where m is the magnitude and Io is the relative intensity. A quake of magnitude 7.9 is 120 times as intense as a tremor. What is the magnitude of the tremor? Iq = Io (10)7.9 It = Io (10)m log 120 = (7.9 - m) log 10 log 120 = (7.9 - m) The magnitude of the tremor is 5.8. m = 7.9 - log 120 m = 5.8 2.7A.15

Assignment Suggested Questions: Pages 113 and 114 A 1-13 odd, 21-25, 40 a-e B 26-33, 40 f-h, 42-44 Page 99 46 2.7A.16

Logarithmic Identities Chapter 2 Exponents and Logarithms 2.7B Logarithmic Identities MATHPOWERTM 12, WESTERN EDITION 2.7B.17

Some equations are solved for all values of the variable for Proving Identities Some equations are solved for all values of the variable for Which both sides of the equation are defined. These equations are called identities. = logax -1 = -1logax -logax = -logax L.S. = R.S. Therefore, the identity is proved. L.S. = R.S. 2.7B.18

Proving Identities Prove: = Conclusion: This suggests that L.S. = R.S. 2.7B.19

Proving Identities Verify: L.S. = R.S. 2.7B.20

Assignment Suggested Questions: Page 106 44-46 Page 113 33-38 2.7B.21