Kamis, 10 Mei 2012

Logarithmic Properties


3.      Logarithmic Properties
Previously, we've talked about understanding the logarithm of a number, that is if  x = an , for a > 0 and a ≠ 1 then n = alog x. By considering the section G.1 , of course you can understand that

a alog x = x
alog an = n
In addition to the above two properties, the logarithm also apply the following properties.
For a > 0, a ≠ 1, x > 0, y > 0, b > 0, and c > 0 , applies
a.       alog a = 1
b.      alog (x × y) = alog x + alog y
c.       alog  = alog x - alog y
d.      alog xn = n × alog x
e.       1.  alog x = blog x/ blog a , b ≠ 1
2.  alog x = 1/ xlog a
f.       alog x × xlog y = alog y
g.       alog x = alog y if and only if x = y
h.      1.  a^nlog xm = alog xm/n =  × alog x
2.  a^nlog xn = alog x
i.        alog  = - alog
Proof of these properties are as follows:
a.  alog a = 1
    Proof: suppose alog a = x . Based on the definition of logarithms, is obtained
    ax = a  x = 1
    So, alog a = 1                                            (valid)
b.  alog (x × y) = alog x +  alog y
    Proof: a alog x × y = x × y
                         = a alog x × a alog y
                                  = a alog x + alog y
    Because a alog x × y = a alog x + alog y then alog x × y = alog x + alog y                              (valid)
c.  alog  = alog x - alog y
    Proof: a alog x/y =
                        = a alog x/a alog y = a alog x - alog y
     Because a alog x/y = a alog x - alog y so alog  = alog x - alog y                                           (valid)
d.  alog xn = n × alog x
    Proof: a alog x^n = xn
                         = (a alog x)n
 = a n × alog x
                      Because a alog x^n = a n × alog x then alog xn = n × alog x                                                     (valid)
            e. alog x = blog x/blog a and alog x = 1/ xlog a
                1. alog x = blog x/blog a
                    Proof:
                    Let blog x = m and blog a = n. Mean, x = bm or b = x1/m and a = bn or b = a1/n
                    Therefore, a blog x/blog a = am/n
                                                     = (a1/n)n
                                                     = bm
                                                     = x
                                                     = a alog x
                        Because a blog x/blog a = a alog x then alog x = blog x/blog a                                         (valid)
                 2. alog x = 1/ xlog a
                     Proof:
                     From the previous properties, has been demonstrated that
                     alog x = blog x/blog a .
                     If it is taken b = x then alog x = xlog x/xlog a
                                                                   = 1/ xlog a                                                  (valid)
           

f. alog x × xlog y = alog y
               Proof: alog x × xlog y             = log x/log a × log y/log x
                                                = log y/log a = alog y                                                    (valid)
            g. alog x = alog y if and only if x = y
               Proof: alog x = alog y  a alog x = a alog y
                                                  x = y                                                          (valid)
            h. a^nlog xm = alog xm/n =  × alog x and  a^nlog xn = alog x
               1. a^nlog xm = alog xm/n =  × alog x
                  Proof: let a^nlog xm = p. Consequently, (an)p = xm
                                                                         an × p = xm
                                                                         ap = x m/n
                                                                         alog xm/n = p
                                                                         alog xm/n = a^nlog xm                                                 (valid)
                 By using properties d, alog xn = n × alog x then alog xm/n =  × alog x
                 So, proved that  a^nlog xm = alog xm/n =  × alog x
              2. a^nlog xn = alog x
                 Proof: . a^nlog xn =  × alog x
                                         = alog x                                                        (valid)
i.        alog  = - alog
Proof:
                  alog = alog ()-1
                           = -1 × alog
                           = -alog
                 So, alog  = -alog                                                          (valid)


Examples of application of the properties  of logarithms above are as follows.
Example :
1.      Specify the value of the logarithm 6log 9 + 6log 4.
Solution :
6log 9 + 6log 4    =  6log ( 9 × 4 )
                                     =     6log 36
                                     =     6log 62
                                          =      2
2.      If  2log 5 = p, declare logarithms below in p.
a.       2log 125                       b.    16log 25
Solution :
a.   2log 125                  =      2log 53
                                                   =   3 2log 5
                                                 =   3p
b.      16log 25                   =   2log 25/2log 16
                   =   2log 52/2log 24
                   =   2 2log 5/4 2log 2
                   =  
                   =  
3.      Determine the value of a given equation.
a.       5log (2a – 5) = 5log (3a – 8)
b.      3log (–2a – 1) –  3log (a – 7) = 1
Solution :
a.       5log (2a – 5) = 5log (3a – 8)                                                                                                                             2a – 5 = 3a – 8                                                                                                                                                 3a 2a = 8 –5                                                                                                                                                                  a = 3
b.      3log (–2a – 1) –  3log (a – 7) =  1                                                                                                                                      3log (–2a – 1) =  3log (a – 7) + 3log 3                                                                                                                      3log (–2a – 1) =  3log ((a – 7) × 3)                                                                                                                                                           –2a – 1 = 3 (a – 7)                                                                                                                                                                 3a + 2a = 21 – 1                                                                                                                                                                                   5a = 20                                                                                                                                                                               a =  = 4
Exercise
1.      Simplify the forms of the following logarithmic.
a.       Log 25 + log 4                                     d. log 50 – log
b.      6log 72 + 6log                                      e. (x + 1)log (x – 1) – (x + 1)log (x2 – 1)
c.       2log 3 + 2log 5                                                 f. alog x4 + 3 alog x – 7 alog x
2.      If 8log 5 = p, determine the value of the following logarithms.
a.       4log                                                     c. 64log 125
b.      2log                                                 d. 512log
3.      If 4log 3 = p, 9log 8 = q, 4log 6 = r, determine the value of the following logarithms.
a.       4log 18                                                 c. 4log 2log
b.      2log +  64                               d. 81log 64 : 64log 27


4.      Determine the Logarithm of a Number of More Than 10 Using the Logarithm Tables
As mentioned previously, to determine the logarithm of the number is more than 10 by using tables of logarithms, can use the properties of logarithms. At first, the number was stated earlier in the raw form a × 10n, with 1 ≤ a ≤ 10 and n are natural numbers.
Suppose b is a number that more than  10 then
            b = a × 10n                                                           (logarithm the both sides)
log b = log (a × 10n)
log b = log a + log 10n
log b = log a + n
So, obtained log a × 10n = log a + n
From the last result, log a is called the decimal (mantis / Mantissas) of the logarithm of b which can be read at the table of logarithms, while n is called the spherical (characteristic) logarithm b. Suppose that the logarithm of a number of more than 10 is 3.748 then it can be made with the following chart.
Flowchart: Data: Mantisa (the decimal)
sought from a table of logarithms
Flowchart: Data: Characteristic
(the spherical)
                         3                                                                                 748





In this case we have learned how to specify a logarithmic Mantissas by reading the table of logarithms (as shown in the chart above) then consider the following example. Next, consider the following example.
*      Example:
Determine the value of the following logarithms.
a.       Log 542
b.      Log 350000
Solution:
a.       Log 542 = log (5.42 × 102)
 = log 5.42 + log 102
  = 0.734 + 2
 = 2.734

b.      Log 350000 = log (3.5 × 105)
            = log 3.5 + log 105
= 0.544 + 5
= 5.544

5.      Determine the Logarithm of a Number Between 0 and 1 Using Logarithm Tables
Just as the numbers that more than 10, to determine the logarithm of the number between 0 and 1 with a logarithm table, the first number's change in raw form a × 10-n, with 1 0 < 10 and n are natural numbers. Suppose b is a number between 0 and 1 then b = a × 10-n (logarithm the both sides)
log b = log (a × 10-n)
                   log b = log a + log 10-n
                    log b = log a – n
So, obtained log (a × 10-n) = log a – n. From the last result, log a is called the decimal (mantis / Mantissas) of the logarithm of  b which can be read at the table of logarithms, while -n is called the spherical (characteristic) logarithm b. Consider the following example.
*      Example:
Determine the value of the following logarithms.
a.       Log 0.0761
b.      Log 0.003492
Solution:
a.       Log 0.0761 = log (7.61 × 10-2)
= log 7.61 + log 10-2
= 0.8814 + (-2)
= 0.8814 – 2 = -1.1186
b.      Log 0.003492 = log (3.492 × 10-3)
   = log 3.492 + log 10-3
     = 0.5428 + (-3)
   = 0.5428 – 3 = -2.4572
From the example of a, it appears that the spherical part of the logarithm of the number between 0 and 1 is -2. It can be seen in the chart below.
            Log 0.761 = 0,8814                                                                             -2





Flowchart: Data: Mantisa (the decimal)
sought from a table of logarithms
Flowchart: Data: Characteristic
(the spherical)

 

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