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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?

To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
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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