LOGARITHMS

the logarithm of a number is the exponent to which another fixed value, the base, must be raised to produce that number. For example, the logarithm of 1000 to base 10 is 3, because 10 to the power 3 is 1000: 1000 = 10 × 10 × 10 = 10³.The logarithm to base 10 (b = 10) is called the common logarithm and has many applications in science and engineering. The natural logarithm has the irrational (transcendental) number e (≈ 2.718) as its base; its use is widespread in pure mathematics, especially calculus. The binary logarithm uses base 2 (b = 2) and is prominent in computer science.

Some properties of logarithms are given below:
 If a^m = x, then m = loga(x).
 Properties :
 a. log_b(b) = 1
 b.log_b(1) = 0
 c.log_b(mn) = log_b(m) + log_b(n)
 d. log_b(^m/_n) = log_b(m) – log_b(n)
 e. log_b(m) = 1/log_m(b)
 f.  log_b(mⁿ) = n · log_b(m)
 g.log_b(m)= log_a(m)/log_a(m)
 Logarithms for base 1 does not exist.Remember that, when base is not mentioned,it is taken as 10.

 In less formal terms, the log rules might be expressed as:

1) Multiplication inside the log can be turned into addition outside the log, and vice versa.
2) Division inside the log can be turned into subtraction outside the log, and vice versa.
3) An exponent on everything inside a log can be moved out front as a multiplier, and vice versa.

Warning: Just as when you're dealing with exponents, the above rules work only if the bases are the same. For instance, the expression "log_d(m) + log_b(n)" cannot be simplified, because the bases (the "d" and the "b") are not the same, just as x² × y³ cannot be simplified (because the bases x and y are not the same).

Expanding logarithms

Log rules can be used to simplify expressions, to "expand" expressions, or to solve for values. 

Expand log₄( ¹⁶/_x ). I have division inside the log, which can be split apart as subtraction outside the log, so

    log₄( ¹⁶/_x ) = log₄(16) – log₄(x)

The first term on the right-hand side of the above equation can be simplified to an exact value, by applying the basic definition of what a logarithm is: 

        log₄(16) = 2

Then the original expression expands fully as: 

      log₄( ¹⁶/_x ) = 2 – log₄(x)

Always remember to take the time to check to see if any of the terms in your expansion (such as the log₄(16) above) can be simplified.

Expand the following:  
The 5 is divided into the 8x⁴, so split the numerator and denominator by using subtraction:
        
Don't take the exponent out front yet; it is only on the x, not the 8, and you can only take the exponent out front if it is "on" everything inside the log. The 8 is multiplied onto the x⁴, so split the factors by using addition:      
       log₂(8x⁴) – log₂(5) = log₂(8) + log₂(x⁴) – log₂(5)
The x has an exponent (which is now "on" everything inside its log), so move the exponent out front as a multiplier: 
        log₂(8) + log₂(x⁴) – log₂(5)  =   log₂(8) + 4log₂(x) – log₂(5)
Since 8 is a power of 2, I can simplify the first log to an exact value:
       log₂(8) + 4log₂(x) – log₂(5) = 3 + 4log₂(x) – log₂(5)
Each log contains only one thing, so this is fully simplified. The answer is: 3 + 4log₂(x) – log₂(5)

some simple problems
A. Solve log₂(8) = x.
I can solve this by converting the logarithmic statement into its equivalent exponential form, using The Relationship: 

         log₂(8) = x  

         2 ^x = 8

  But 8 = 2³, so:2 ^x = 2³  x = 3

Note that this could also have been solved by working directly from the definition of a logarithm: What power, when put on "2", would give you an 8? The power 3, of course!

If you wanted to give yourself a lot of work, you could also do this one in your calculator, using the change-of-base formula:

             log₂(8) = ln(8) / ln(2)

Plug this into your calculator, and you'll get "3" as your answer. While this change-of-base technique is not particularly useful in this case, you can see that it does work. (Try it on your calculator, if you haven't already, so you're sure you know which keys to punch, and in which order.) You will need this technique in later problems.B. Solve log₂(x) + log₂(x – 2) = 3
I can't do anything yet, because I don't yet have "log equals a number". So I'll need to use log rules to combine the two terms on the left-hand side of the equation: 
                   log₂(x) + log₂(x – 2) = 3  
                   log₂((x)(x – 2)) = 3  
                   log₂(x² – 2x) = 3.Then I'll use The Relationship to convert the log form to the corresponding exponential form, and then I'll solve the result: 
                log₂(x² – 2x) = 3   
               2³ = x² – 2x   
               8 = x² – 2x  
              0 = x² – 2x – 8   
              0 = (x – 4)(x + 2)   
               x = 4, –2.But if x = –2, then "log₂(x)", from the original logarithmic equation, will have a negative number for its argument (as will the term "log₂(x – 2)"). Since logs cannot have zero or negative arguments, then the solution to the original equation cannot be x = –2.so,the solution is 4.
Keep in mind that you can check your answers to any "solving" exercise by plugging those answers back into the original equation and checking that the solution "works":
          log₂(x) + log₂(x – 2) = 3  
         log₂(4) + log₂(4 – 2) = 3  
          log₂(4) + log₂(2) = 3
Since the power that turns "2" into "4" is 2 and the power that turns "2" into "2" is "1", then we have:
           log₂(4) + log₂(2) = 3  
          log₂(2²) + log₂(2¹) = 3   
             2 + 1 = 3 ,The solution checks. Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved 
C. Solve log₂(log₂(x))   = 1.
This may look overly-complicated, but it's just another log equation. To solve this, I'll need to apply The Relationship twice: 
      log₂(log₂(x)) = 1
       2¹ = log₂(x)  
       2 = log₂(x) x = 2²   
       x = 4,Then the solution is x = 4.
D. Solve log₂(x²)  = (log₂(x))². First, I'll write out the square on the right-hand side: 
log₂(x²) = (log₂(x))²
log₂(x²) = (log₂(x)) (log₂(x))
Then I'll apply the log rule to move the "squared", from inside the log on the left-hand side of the equation, out in front of that log as a multiplier. Then I'll move that term to the right-hand side:
2log₂(x) = [log₂(x)] [log₂(x)]
0 = [log₂(x)] [log₂(x)]  –  2log₂(x)
This may look bad, but it's nothing more than a factoring exercise at this point. So I'll factor, and then I'll solve the factors by using The Relationship:
0 = [log₂(x)] [log₂(x) – 2]
log₂(x) = 0  or  log₂(x) – 2 = 0
2⁰ = x   or  log₂(x) = 2
1 = x  or  2² = x
1 = x  or  4 = x
The solution is x = 1, 4.