SOLUTION: Solve each logarithmic equation. Check each solution. 1. log 3x=2 2. 4log x=4 3. log(3x-2)=3 4. 2logx -log5=-2 5. log 8 - log 2x=-1 6. log(x+21)+l

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Solve each logarithmic equation. Check each solution. 1. log 3x=2 2. 4log x=4 3. log(3x-2)=3 4. 2logx -log5=-2 5. log 8 - log 2x=-1 6. log(x+21)+l      Log On


   

Question 25293: Solve each logarithmic equation. Check each solution.
1. log 3x=2 2. 4log x=4 3. log(3x-2)=3 4. 2logx -log5=-2

5. log 8 - log 2x=-1 6. log(x+21)+logx=2

I have a hard time understanding algebra period because of lack of training that was given to me. So you coul be my permanet helper if you like!.! It's six problems if you wouldn't mind.
Answer by AnlytcPhil(1807) About Me  (Show Source):
You can put this solution on YOUR website!
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1. log(3x) = 2

For that you need to know that an equation of the form

 log(A) = B  can be written as A = 10B.  You should memorize this rule.

here A = 3x and B = 2, so rewrite log(3x) = 2 as 3x = 102

3x = 102
3x = 100
 x = 100/3 or 33 1/3
 
==================================================

 2.  4log(x) = 4 

First divide both sides by 4

      log(x) = 1

Now use the same rule used in problem 1 to rewrite that equation as

      x = 101 
      x = 10

==========================================

3. log(3x - 2) = 3

Use the same rule to rewrite that as

       3x - 2 = 103
       3x - 2 = 1000
           3x = 1002
            x = 1002/3
            x = 334         

=============================================

 4. 2log(x) - log(5) = -2

  Here you need the rule

 A·log(B) = log(BA)

 to rewrite 2log(x) as log(x2)

   log(x2) - log(5) = -2

 Now you need the rule

   log(A) - log(B) = log(A/B)

 to rewrite the left side as log(x2/5) = -2

  log(x2/5) = -2

Now us the rule we used in previous problems to write that equation as

  x2/5 = 10-2

Now use the rule A-B = 1/AB to rewrite the right side

  x2/5 = 1/102

  x2/5 = 1/100

Multiply both sides by 100

 100·x2/5 = 100·1/100

     20x2 = 1

       x2 = 1/20
              ____
        x = ±Ö1/20 
                __ 
        x = ±1/Ö20 
                 ___
        x = ±1/(Ö4·5)
                  _ 
        x = ±1/(2Ö5)
                   _    _  _
        x = [±1/(2Ö5)][Ö5/Ö5]
              _
        x = ±Ö5/(2·5)
              _
        x = ±Ö5/10

We discard the negative answer, because we cannot take logarithms of
negative numbers.  So the answer is
             _
        x = Ö5/10

===================================================== 

5. log(8) - log(2x) = -1 

   Use the rule

   log(A) - log(B) = log(A/B) to rewrite the LHS

   log[8/(2X)] = -1

      log(4/X) = -1

   Use the rule used earlier to rewrite this equation as 

      4/x = 10-1

   Use the rule A-B = 1/AB to rewrite the RHS

      4/x = 1/101

      4/x = 1/10

   Multiply both sides by 10x

     10x·4/x = 10x·1/10

          40 = x   

========================================

6.  log(x + 21) + log(x) = 2

  Use the rule 

    log(A) + log(B) = log(AB) to rewrite the LHS

        log[(x + 21)x] = 2

         log(x2 + 21x) = 2

  Now use the first rule to rewrite that equation as

              x2 + 21x = 102

        x2 + 21x - 100 = 0
 
       (x + 25)(x - 4) = 0

        x + 25 = 0;  x - 4 = 0

             x = -25;  x = 4

  We discard the negative answer because we cannot take logs 
  of negative numbers.

  The answer is x = 4  

Edwin
AnlytcPhil@aol.com