Jumat, 19 September 2008

Exponentials and Logarithms,




OCW is pleased to make this textbook available online. Published in 1991 and still in print from Wellesley-Cambridge Press, the book is a useful resource for educators and self-learners alike. It is well organized, covers single variable and multivariable calculus in depth, and is rich with applications. There is also an online Instructor's Manual and a student Study Guide.


Textbook Components

  • Table of Contents (PDF)
  • Answers to Odd-Numbered Problems (PDF)
  • Equations (PDF)


6: Exponentials and Logarithms, pp. 228-282

6.1 An Overview, pp. 228-235
6.2 The Exponential e^x, pp. 236-241
6.3 Growth and Decay in Science and Economics, pp. 242-251
6.4 Logarithms, pp. 252-258
6.5 Separable Equations Including the Logistic Equation, pp. 259-266
6.6 Powers Instead of Exponentials, pp. 267-276
6.7 Hyperbolic Functions, pp. 277-282

Chapter 6 - complete (PDF - 4.9 MB)

Chapter 6 - sections:

6.1 - 6.4 (PDF - 3.0 MB)
6.5 - 6.7 (PDF - 2.2 MB)

Exponential and Log Equations

  1. Equations that Involve Logs

    Step by Step Method

    Step 1: Contract to a single log.

    Step 2: Get the log by itself.

    Step 3: Exponentiate both sides with the appropriate base.

    Step 4: Solve.

    Step 5: Check your solution for domain errors.


    Example:

    Solve

    log5 x + log5 (x + 2) = log5 (x + 6)

    1. log5 x + log5 (x + 2) - log5 (x + 6) = 0

      log5 x (x + 2) - log5 (x + 6) = 0

      log5 x (x + 2)/(x + 6) = 0

    2. Already done.

    3. x(x + 2)/(x + 6) = 50 = 1

    4. x(x + 2) = x + 6

      x2 + 2x - x - 6 = 0

      x2 + x - 6 = 0

      (x - 2)(x + 3) = 0

      x = 2 or x = -3

    5. Note that -3 is not in the domain of the first log hence the only solution is x = 2.

    Exercises: Solve

    1. log(x + 6) + 1 = 2log(3x - 2)

    2. 1/2 log(x + 3) + log2 = 1


  2. Exponential Equations

    Step 1: Isolate the exponential

    Step 2: Take the appropriate log of both sides.

    Step 3: Solve


    Example: Solve

    4e-7x = 15

    1. e-7x = 15/4

    2. lne-7x = ln(15/4)

    3. -7x = ln(15/4)

    4. x = ln(15/4)/-7

    Exercises:

    Solve

    1. 1 + 2ex = 9

    2. (10x - 4)/e2x - 4 = 0

    3. (lnx)2 = ln(x2)

    4. 23x + 4(2-3x) = 5

  3. Application

    All living beings have a certain amount of radioactive carbon C14 in their bodies. When the being dies the C14 slowly decays with a half life of about 5600 years. Suppose a skeleton is found in Tahoe that has 42% of the original C14. When did the person die?

    Solution:

    We can use the exponential decay equation:

    y = Cekt

    After 5600 years there is

    C/2

    C14 left. Substituting, we get:

    C/2 = Cek(5600)

    Dividing by C,

    1/2 = e5600k

    Take ln of both sides,

    ln(.5) = 5600k

    so that

    k = [ln(.5)]/5600 = -.000124

    The equation becomes

    y = Ce-.000124t

    To find out when the person died, substitute

    y = .42C

    and solve for t:

    .42C = Ce-.000124t

    Divide by C,

    .42 = e-.000124t

    Take ln of both sides,

    ln(.42) = -.000124t

    Divide by -.000124

    t = [ln(.42)]/(-.000124) = 6995

    The person died about 7,000 years ago.


Source:

OCW MIT

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