Question 25293
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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 = 10<sup>B</sup>.  You should memorize this rule.

here A = 3x and B = 2, so rewrite log(3x) = 2 as 3x = 10<sup>2</sup>

3x = 10<sup>2</sup>
3x = 100
 x = 100/3 or 33 1/3
 
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 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 = 10<sup>1</sup> 
      x = 10

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3. log(3x - 2) = 3

Use the same rule to rewrite that as

       3x - 2 = 10<sup>3</sup>
       3x - 2 = 1000
           3x = 1002
            x = 1002/3
            x = 334         

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

  Here you need the rule

 A·log(B) = log(B<sup>A</sup>)

 to rewrite 2log(x) as log(x<sup>2</sup>)

   log(x<sup>2</sup>) - log(5) = -2

 Now you need the rule

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

 to rewrite the left side as log(x<sup>2</sup>/5) = -2

  log(x<sup>2</sup>/5) = -2

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

  x<sup>2</sup>/5 = 10<sup>-2</sup>

Now use the rule A<sup>-B</sup> = 1/A<sup>B</sup> to rewrite the right side

  x<sup>2</sup>/5 = 1/10<sup>2</sup>

  x<sup>2</sup>/5 = 1/100

Multiply both sides by 100

 100·x<sup>2</sup>/5 = 100·1/100

     20x<sup>2</sup> = 1

       x<sup>2</sup> = 1/20
              ____
        x = ±<font face="symbol">Ö</font>1/20 
                __ 
        x = ±1/<font face="symbol">Ö</font>20 
                 ___
        x = ±1/(<font face="symbol">Ö</font>4·5)
                  _ 
        x = ±1/(2<font face="symbol">Ö</font>5)
                   _    _  _
        x = [±1/(2<font face="symbol">Ö</font>5)][<font face="symbol">Ö</font>5/<font face="symbol">Ö</font>5]
              _
        x = ±<font face="symbol">Ö</font>5/(2·5)
              _
        x = ±<font face="symbol">Ö</font>5/10

We discard the negative answer, because we cannot take logarithms of
negative numbers.  So the answer is
             _
        x = <font face="symbol">Ö</font>5/10

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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<sup>-1</sup>

   Use the rule A<sup>-B</sup> = 1/A<sup>B</sup> to rewrite the RHS

      4/x = 1/10<sup>1</sup>

      4/x = 1/10

   Multiply both sides by 10x

     10x·4/x = 10x·1/10

          40 = x   

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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(x<sup>2</sup> + 21x) = 2

  Now use the first rule to rewrite that equation as

              x<sup>2</sup> + 21x = 10<sup>2</sup>

        x<sup>2</sup> + 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</pre>