Exponent and Logarithm

 

Motivation:

1.      3  3  3  3  3  can be denoted as  3⁵ .

2.        = 2 .  Thus,     could be the candidate for  2^1/2 .

 

What if the exponent is not a real number?  For example,

                

               2 .  Or the case for 3 = 10^x ; what is the value of x?

 

Definition:   If  a  1,  a>0, b > 0 ,  the notation

                                         c = log­­_ab

                                     represents the relationship  

                                                                    a^c₌ b _.

                                     _{And a is called the base of the logarithm.}

                  _{Sometimes, if a=10, we just write it as   c = log b.}

                  _{However,  sometimes you might see people use  c = log b to represent “natural logarithm”; that is}

                                          c= log­_eb  or  c = ln b

 

                        where e=2.71828….  ( known as “the base of natural logarithm” ).

 

                  Anyway, before you touch the course “Calculus”,  log b should be treated as that it is base-10.

 

 

Properties:

1.      log_aAB = log_aA + log_aB

2.      log­_a  = log­­_aA - log_aB

3.      log_AB = ( log­_bB ) / ( log_bA)   for  legitimate b ( namely, b>0 and b is not equal to 1 ).

 

           Proof:

 

                      1.   Let     p = log_aAB , q = log_aA ,  r= log_aB .

                            Then   by definition,

                                                       a^p ₌ AB,  a^q = A,  a^r = B

                      

                                                                       

                               a^(q+r) = a^q a^r = AB = a^p

                             And a is a legitimate base for the logarithm; that mean it is not equal to 1

                             

                 q+r = p ;  in other words,  

                                    log_aAB = log_aA + log_aB

 

2.      Let  p =  log­_a  ,  q = log_aA ,  r= log_aB .

Then a^p =   ,  a^q = A,  a^r = B .

    a^p =   ₌  a^q­ / a­^r = a^(q-r)

 

           p = q – r .  In other words,

                       log­_a  = log­­_aA - log_aB

 

3.      Let p =  log_AB,  q=  log­_bB , r = log_bA

Then  A^p =B ,  b^q = B,  b^r = A .

   B =  b^q  = A^p  = (b^r)^p  = b^rp .   In other words,

           b^q = b^rp

 Since b > 0 , b  1

     Then  q = rp

  p =

   log_AB = ( log­_bB ) / ( log_bA)