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How do I condense complex logarithms?

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3 Answers

 What is important to realize is that:
 
  Log ( AB) = log A + Log B
 
   Log A/ B = Log A - Log B
 
   Log2 ( 2X +6) - Log 2 ( X - 2) = Log2 3
 
     Log2 [ ( 2X - 6) / X -1 )] = Log2 3
 
 
          (2X + 6 / X - 1) = 3
 
          3X -3 = 2 X + 6
 
              X = 9 
 
        It will be Good Exercise to do the following calculations:
 
          Log6 9 + Log6 4 =
 
          Log3 27 + Log3 3 =
 
            Log 2 8 + Log2 2 =
                  
 
 
             
1. Use the relationship logxa - logxb = logx(a/b)
log2(2x+6) - log2(x-1) = log23
⇒ log2((2x+6)/(x-1)) = log23
 
2. Use the relationship that logxa = logxb implies a = b
log2((2x+6)/(x-1)) = log23
⇒ (2x+6)/(x-1) = 3
 
3. Multiply both sides by x-1
(2x+6)/(x-1) = 3
⇒ 2x+6 = 3(x-1)
 
4. Solve for x
2x+6 = 3(x-1)
⇒ 2x+6 = 3x-3
⇒ 9 = x
 
So x=9.
Hi Jensen;
log2(2x+6)-log2(x-1)=log23
Our first priority is to verify that all logs are to the same base.  These are all to the base of 2.  We may now proceed.
The logarithm-principle we need to apply is...
log a/b=log a - log b
Therefore...
log2[(2x+6)/(x-1)]=log23
We can eliminate log2 on both sides...
(2x+6)/(x-1)=3
(2x+6)/(x-1)=3/1
Let's cross-multiply...
1(2x+6)=3(x-1)
(2x+6)=3(x-1)
2x+6=3x-3
Let's combine like terms...
Let's subtract 2x from both sides...
-2x+2x+6=3x-3-2x
6=x-3
Let's add 3 to both sides...
3+6=x-3+3
9=x
 
Let's check our work by returning to the original equation...
log2(2x+6)-log2(x-1)=log23
log2[(2*9)+6)-log2(9-1)=log23
log2(18+6)-log2(8)=log23
log2(24)-log2(8)=log23
log a/b=log a - log b
log2[(24)/(8)]=log23
log23=log23