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.

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