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 + 2e^x = 9
      

   2. (10^{x - 4})/e^2x - 4 = 0
      

   3. (lnx)² = ln(x²)
      

   4. 2^3x + 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 = Ce^kt
   
   After 5600 years there is
   
   C/2
   
   C₁₄ left. Substituting, we get:
   
   C/2 = Ce^k(5600)
   
   Dividing by C,
   
   1/2 = e^5600k
   
   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