Section 5-6:  Laws of Logarithms
 
 

   1)  Logb MN = Logb M + Logb N
   2)  Logb M/N = Logb M - Logb N
   3)  Logb M = Logb N if and only if M = N
        4)  Logb Mk = k Logb M
   5)  Logb b = 1
   6)  Logb 1 = 0
   7)  Logb bk = k
   8)  bLogb x x

 Sample problems
 
Write each log in expanded form.
 
1)  Log5 xy2 =
                         Solution:  Log5 x + Log5 y2 = Log5 x + 2 Log5 y
 
2)  Log7 (xy/z2) =
 
                        Solution:  Log7 x + Log7 y - 2 Log7 z
 
3) 
 
Express each as a single log.
 
1)  Log x + Log y - Log z =
 
                     Solution:  Log (xy)/z
 
2)  2 Ln x + 3 Ln y =
 
                     Solution:  Ln x2y3
 
3)  (1/2) Ln x - (1/3) Ln y =
 
                     Solution: 
 

Writing logs as single logs can be helpful in solving many log equations.
1)  Log2 (x + 1) + Log2 3 = 4
                     Solution: First combine the logs as a single log.
                                         Log2 3(x + 1) = 4
                                      Now rewrite as an exponential equation.
                                         3(x + 1) = 24
                                       Now solve for x.
                                         3x + 3 = 16
                                         3x = 13
                                         x = 13/3  Since this doesn't make the number inside the log zero or negative, the answer is acceptable.
 
2)  Log (x + 3) + Log x = 1
                     Solution:  Again, combine the logs as a single log.
                                         Log x(x + 3) = 1
                                       Rewrite as an exponential.
                                           x(x + 3) = 10
                                       Solve for x.
                                           x2 + 3x = 10
                                           x2 + 3x - 10 = 0
                                           (x + 5)(x - 2) = 0
                                           x = -5 or x = 2  We have to throw out 5.  Why?  Because it makes (x + 3) negative and we can't take the log of a negative number.  So the only answer is x = 2.
 
3)  Ln (x - 4) + Ln x = Ln 21
                     Solution:  Notice, this time we have a log on both sides.  If we write the left side as a single log, we can use the rule that if the logs are equal, the quantity inside must be equal.
                                     Ln x(x - 4) = Ln 21
                                   Since the logs are equal, what is inside must be equal.
                                     x(x - 4) = 21
                                   Solve for x.
                                     x2 - 4x = 21
                                     x2 - 4x - 21 = 0
                                     (x - 7)(x + 3) = 0
                                      x = 7 or x = -3  Again, we need to throw out one of the answers because it makes both quantities negative.  Throw out -3 and keep 7.  Thus, the answer is x = 7.

Simplify each log
 
1)  ln e5
                 Solution:  This is rule number 7.  The answer is 5!

2)  Log 10-3
                 Solution:  This is again rule #7.  The answer:  -3  (This answers the question:  what power do you raise 10 to get 10 to the third?

3)  eln 7
                 Solution:  This is rule #8.  The answer is 7.
 
4)  e2ln 5
                 Solution:  We can use rule #8 as soon as we simplify the problem.  Rewrite as:  eln 25 = 25  The 25 came from 52.
 
5)  10Log 6
                 Solution:  Rule #8 again.  Answer:  6
 
6)  102 + log 5
                 Solution:  We need to simplify before we can apply one of the rules.  Rewrite as:  (102)(10log 5)  Adding exponents means you are multiplying the bases.
                                = 100(5)  Use rule #8 again.
                                = 500


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