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38

LOGARITHMS

Definition

The three laws of logarithms

Common logarithms


A LOGARITHM is an exponent.

Since

23  =  8,

then 3 is called the logarithm of 8 with base 2.  We write

3  =  log28.

3 is the exponent to which 2 must be raised to produce 8.

We write the base 2 as a subscript.

Thus a logarithm is the exponent to which the base must be raised to produce a given number.

Since

104 = 10,000,

then

log1010,000 = 4.

"The logarithm of 10,000 with base 10 is 4."

4 is the exponent to which the base 10 must be raised to produce 10,000.

"104 = 10,000" is called the exponential form.

"log1010,000 = 4" is called the logarithmic form.

Here is the definition:

logbx = n   means   bn = x.

That  base  with that  exponent  produces x.

Example 1.   Write in exponential form:   log232 = 5

 Answer.   25 = 32

   Example 2.   Write in logarithmic form:  4−2  =    1
16		 .
   Answer.   log4  1
16		   =  −2.

Problem 1.   Which numbers have negative logarithms?

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Do the problem yourself first!

Proper fractions.

Example 3.   Evaluate  log81.

 Answer.   8 to what exponent produces 1?   80 = 1.

log81 = 0.

We can observe that in any base b, the logarithm of 1 is 0.

logb1 = 0

Example 4.   Evaluate  log55.

 Answer.   5 to what exponent will produce 5?   51 = 5.

log55 = 1.

In any base, the logarithm of the base itself is 1.

logbb = 1

Example 5.   log22m = ?

 Answer.   2 raised to what exponent will produce 2m ?   m, obviously.

log22m = m.

This is an important formal rule, valid for any base b:

logbbx = x

This rule embodies the very meaning of a logarithm.  x -- on the right -- is the exponent to which the base b must be raised.

   Example 6 .   Evaluate  log3  1
9		 .
  Answer.    1
9		  is equal to 3 with what exponent?    1
9		   =  3−2.
log3  1
9		   =   log33−2  =  −2

Example 7.   log2 .25 = ?

 Answer.   .25 = ¼ = 2−2.  Therefore,

log2 .25 = log22−2 = −2

Example 8.   log3 = ?

 Answer..    = 3.  (Definition of a fractional exponent.)  Therefore,

log3 = log33 = 1/5

Problem 2.   Write each of the following in logarithmic form.

   a)     bn = x    logbx = n      b)     23 = 8    log28 = 3
 
   c)     102 = 100    log10100 = 2      d)     5−2 =  1/25.    log51/25 = −2.

Problem 3.   Write each of the following in exponential form.

  a)   logbx = n    bn = x   b)   log232 = 5    25 = 32
 
  c)   2 = log864    82 = 64   d)   log61/36 = −2    6−2 = 1/36

Problem 4.   Evaluate the following.

  a)   log216   = 4   b)   log416   = 2
 
  c)   log5125   = 3   d)   log81   = 0
 
  e)   log88   = 1   f)   log101   = 0

Problem 5.   What number is n?

  a)   log10n = 3   1000   b)   5 = log2n   32
 
  c)   log2n = 0    1    d)   1 = log10n    10 
  e)   logn   1
16		  = −2    4      f)   logn  1
5		  = −1    5 
 
  g)   log2   1
32		  = n   −5      h)   log2 1
2		  = n    −1 

Problem 6.   logbbx  =  x

Problem 7.   Evaluate the following.

  a)   log9 1
9		    = log99−1 = −1
  b)   log9  1
81		  = −2     c)   log2 1
4		  = −2
 
  d)   log2 1
8		  = −3     e)   log2  1
16		  = −4
  f)   log10 .01  = −2     g)   log10 .001  = −3
 
  h)   log6   = 1/3     i)   logb   = 3/4

The three laws of logarithms

1.    logbxy  =  logbx  +  logby

"The logarithm of a product is equal to the sum
of the logarithms of each factor."

2.    logb x
y		   =  logbx  −  logby

"The logarithm of a quotient is equal to the logarithm of the numerator
minus the logarithm of the denominator."

3.    logb x n  =  n logbx

"The logarithm of a power of x is equal to the exponent of that power
times the logarithm of x."

For a proof of these laws, see Topic 20 of Precalculus.

   Example 1.    Use the laws of logarithms to rewrite  log 
  z5		 .

Answer.   According to the first two laws,

log 
  z5		   =  log x  +  log  −  log z5

Now,  =  y½.  Therefore, according to the third law,

log 
  z5		   =  log x  +  ½ log y  −  5 log z

Example 2.   Use the laws of logarithms to rewrite   log (sin x log x)

 Solution.    This has the form  log ab.  a = sin x,  b = log x.  Therefore,

log (sin x log x) = log sin x+ log log x

Example 3.   Use the laws of logarithms to rewrite  log .

 Solution.   

log   =   log (x cos x)
 
    =   ½ log (x cos x),   3rd Law
 
    =   ½ (log x  +  log cos x),   1st Law.

Problem 8.   Use the laws of logarithms to rewrite the following.

   a)  log  ab
 c		   = log a  +  log b  −  log c
 
   b)  log  ab²
 c4		   = log a  +  2 log b  −  4 log c
 
   c)  log 
   z		     = 1/3 log x  +  1/2 log y  −  log z
  d)   log (sin²x log x) = log sin²x  +  log log x
 
  = 2 log sin x  +  log log x
  e)   log   =   log (sin x cos x)1/2
 
    =   ½ log (sin x cos x)
 
    =   ½ (log sin x  +  log cos x).

Common logarithms

The system of common logarithms has 10 as its base.  When the base is not indicated:

log 100 = 2

then the system of common logarithms -- base 10 -- is implied.

Here are the powers of 10 and their logarithms:

Powers of 10:        1   
1000		     1  
100		    1
10		   1   10   100   1000   10,000
 
Logarithms:     −3   −2   −1   0    1     2      3      4

Logarithms replace a geometric series with an arithmetic series.

Problem 8.

a)   log 105 = 5.  10 is the base.

b)   log 10n = n

c)   log 58 = 1.7634.   Therefore, 101.7634 = 58

1.7634 is the common logarithm of 58.  When 10 is raised to that exponent, 58 is produced.

Problem 9.   log (log x) = 1.  What number is x?

log a = 1, implies a = 10. (See above.)  Therefore, log (log x) = 1 implies log x = 10.  Since 10 is the base,

x = 1010 = 10,000,000,000

Example 4.   Given:  log 3 = .4771   Evaluate

a)   log 3000

Solution.  Write 3000 in scientific notation:

log 3000 = log (3 × 103)
 
  = log 3 + log 103
 
  = .4771 + 3
 
  = 3.4771

b)   log .003

  Solution. log .003 = log (3 × 10−3)
 
  = log 3 + log 10−3
 
  = .4771 − 3
 
  = −2.5229

Problem 10   Given:  log 6 = .7781   Use the laws of logarithms to evaluate the following.

  a)   log 600 = log (6 × 102)
 
  = log 6 + log 102
 
  = .7781 + 2
 
  = 2.7781
  b)   log 60 = log (6 × 10)
 
  = log 6 + log 10
 
  = .7781 + 1
 
  = 1.7781
  c)   log .06 = log (6 × 10−2)
 
  = log 6 + log 10−2
 
  = .7781 − 2
 
  = −1.2219

Example 5.   Given:  log 2 = .3010,  log 3 = .4771   Evaluate log 18.

Solution.   18 = 2· 3².  Therefore,

log 18 = log (2· 3²)
 
  = log 2 + log 3²
 
  = log 2 + 2 log 3
 
  = .3010 + 2(.4771)
 
  = .3010 + .9542
 
  = 1.2552

Problem 11.   Given:  log 2 = .3010    log 3 = .4771    log 5 = .6990

Use the laws of logarithms to find the following.

a)   log 6 = log 2 + log 3 = .7781

b)   log 15 = log 3 + log 5 = 1.1761

c)   log 4 = log 2² = 2 log 2 = .6020

d)   log 8 = log 2³ = 3 log 2 = .9030

e)   log 30 = log 3 + log 10 = 1.4771

f)   log 300 = log 3 + log 100 = 2.4771

g)   log 3000 = log 3 + log 1000 = 3.4771

h)   log 12 = log 3 + log 4 = 1.0791

  i)   log  3
5		   =   log 3 − log 5 = −.2219

j)   log  = ½ log 3 = .2386

k)   log  = ½ log 5 = .3495

  l)   log  =  3
2		  log 3 = .7157
  m)   log  =  1
3		  log 2 = .1003

n)   log  = ½(log 2 − log 3) = −.0881

o)   log 1500 = log 3 + log 5 + log 100 = 3.1761

For the system of natural logarithms, see Topic 20 in Precalculus.


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