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38

LOGARITHMS

Definition

WHEN WE ARE GIVEN the base 2, for example, and exponent 3, then we can evaluate 23.

23 = 8.

Inversely, if we are given the base 2 and its power 8 --

2? = 8

-- then what is the exponent that will produce 8?

That exponent is called a logarithm.  We call the exponent 3 the logarithm of 8 with base 2.  We write

3 = log28.

The base 2 is written as a subscript.

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

A logarithm is an exponent.

Since

104 = 10,000

then

log1010,000 = 4.

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

4 is the exponent to which 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, the logarithm of 1 is 0.

logb1 = 0

Example 4.   Evaluate  log55.

 Answer.   5 with what exponent will produce 5?   51 = 5.  Therefore,

log55 = 1.

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

logbb = 1

Example 5.   log22m = ?

 Answer.   2 with what exponent will produce 2m ?   m, obviously.

log22m = m.

The following 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  to produce bx.

   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.

Compare the previous rule.

Example 7.   log2 .25 = ?

 Answer.   .25 = ¼ = 2−2.  Therefore,

log2 .25 = log22−2 = −2.

Example 8.   log3logarithms = ?

 Answer.   logarithms = 31/5.  (Definition of a rational exponent.)  Therefore,

log3logarithms = log331/5 = 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)   log6logarithms   = 1/3  

Write logarithms in exponential form, and apply Problem 6.

  i)   logblogarithms   = 3/4

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?

The log of what number is 1?  Since 10 is the base, log 10 = 1. (See above.)  Therefore, log (log x) = 1 implies log x = 10.  And therefore since 10 is the base:

x = 1010 = 10,000,000,000

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 logarithms   =  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 9.    Apply the laws of logarithms to  log  abc2
 d 3
.

Answer.   According to the first two laws,

log  abc2
 d 3
 =   log (abc2) − log d 3
 
     =  log a + log b + log c2 − log d 3
 
     =  log a + log b + 2 log c − 3 log d,

according to the third law.

The Answer above shows the complete theoretical steps.  In practice, however, it is not necessary to write the line

log  abc2
 d 3
 =  log (abc2) − log d 3 .

The student should be able to go immediately to the next line --

log  abc2
 d 3
 =  log a + log b + log c2 − log d 3

-- if not to the very last lineexclamation
log  abc2
 d 3
 =  log a + log b + 2 log c − 3 log d.
 Example 10.    Use the laws of logarithms to rewrite  log  logarithms
  z5
.
  Answer.   log  logarithms
  z5
  =  log x  +  log logarithms  −  log z5

Now, square root  =  y½. (Lesson 29.)  Therefore, according to the third law,

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

Example 11.   Use the laws of logarithms to rewrite   log (sin x log x)

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

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

Example 12.   Use the laws of logarithms to rewrite  log logarithms.

 Solution.   

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

Problem 10.   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  logarithms
     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 logarithms  =  log (sin x cos x)1/2
 
   =  ½ log (sin x cos x)
 
   =  ½ (log sin x  +  log cos x).

Example 13.   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 11.   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 14.   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 12.   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 square root of 3 = ½ log 3 = .2386

k)   log square root of 5 = ½ log 5 = .3495

  l)   log square root of 27 =  3
2
 log 3 = .7157
  m)   log radical =  1
3
 log 2 = .1003

n)   log radical = 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 of Precalculus.

For logarithmic and exponential "functions," see Topic 21.

end

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