Common and Natural Logarithms
The common log is the logarithm with base 10, and is typically written log( x) .
The natural log is the logarithm with base e, and is typically written ln( x) .
Example 5
Evaluate log(
)
1000 using the definition of the
Values of the common log
common log.
number number as log(number)
exponential
To evaluate log(
)
1000 , we can say
1000
103
3
x = log(
)
1000 , then rewrite into exponential
100
102
2
form using the common log base of 10.
10
101
1
10 x = 1000
1
100
0
From this, we might recognize that 1000 is the
0.1
10-1
-1
cube of 10, so x = 3.
0.01
10-2
-2
We also can use the inverse property of logs to
0.001
10-3
-3
write log (103 =
10
) 3
Section 4.3 Logarithmic Functions 245
Try it Now
2. Evaluate l 100000
og(
)
0 .
Example 6
Evaluate ln( e).
We can rewrite ln( e) as ln( 1/2
e ). Since ln is a log base e, we can use the inverse
property for logs: ln( 1/2
e )=
.
e ( 1/ 2
e ) 1
log
= 2
Example 7
Evaluate log(500) using your calculator or computer.
Using a computer, we can evaluate log( )
500 ≈ .
2 69897
To utilize the common or natural logarithm functions to evaluate expressions like
log
, we need to establish some additional properties.
2
)
10
(
Properties of Logs: Exponent Property
log ( Ar = log
b
) r b( A)
To show why this is true, we offer a proof.
Since the logarithmic and exponential functions are inverses, b log b A = A.
So r
A = ( log b
b
) r
A
Utilizing the exponential rule that states ( ) q
p
pq
x
= x ,
r
A = ( log b
b
) r
A
r log b A
= b
So then log ( r
log b
A = log b
b
)
b ( r
A )
Again utilizing the inverse property on the right side yields the result
log ( Ar = log
b
) r b A
Example 8
Rewrite log
using the exponent property for logs.
3 (25)
Since 25 = 52,
log
=
=
3 (25)
log (52
3
) 2log3 5
246 Chapter 4
Example 9
Rewrite 4ln( x) using the exponent property for logs.
Using the property in reverse, 4ln( x) = ( 4
ln x )
Try it Now
3. Rewrite using the exponent property for logs: 1
ln
.
2
x
The exponent property allows us to find a method for changing the base of a logarithmic
expression.
Properties of Logs: Change of Base
c A
log
=
b ( A)
log ( )
log c ( b)
Proof:
Let log
= . Rewriting as an exponential gives bx = A. Taking the log base c of
b ( A)
x
both sides of this equation gives
log bx = log
c
c A
Now utilizing the exponent property for logs on the left side,
x log
= log
c b
c A
Dividing, we obtain
log
log c A
c A
x =
or replacing our expression for x, log
=
b A
log
log c b
c b
With this change of base formula, we can finally find a good decimal approximation to
our question from the beginning of the section.
Example 10
Evaluate log
using the change of base formula.
2
)
10
(
According to the change of base formula, we can rewrite the log base 2 as a logarithm
of any other base. Since our calculators can evaluate the natural log, we might choose
to use the natural logarithm, which is the log base e:
log e 10 ln10
log
=
=
2 10
log e 2
ln 2
Using our calculators to evaluate this,
Section 4.3 Logarithmic Functions 247
ln10
.
2 30259
≈
≈ .
3 3219
ln 2
.
0 69315
This finally allows us to answer our original question – the population of flies we
discussed at the beginning of the section will take 3.32 weeks to grow to 500.
Example 11
Evaluate log
using the change of base formula.
5
)
100
(
We can rewrite this expression using any other base. If our calculators are able to
evaluate the common logarithm, we could rewrite using the common log, base 10.
log10 100
2
log
=
≈
=
5
)
100
(
861
.
2
log10 5
69897
.
0
While we were able to solve the basic exponential equation 2 x = 10 by rewriting in
logarithmic form and then using the change of base formula to evaluate the logarithm, the
proof of the change of base formula illuminates an alternative approach to solving
exponential equations.
Solving exponential equations:
1. Isolate the exponential expressions when possible
2. Take the logarithm of both sides
3. Utilize the exponent property for logarithms to pull the variable out of the exponent
4. Use algebra to solve for the variable.
Example 12
Solve 2 x = 10 for x.
Using this alternative approach, rather than rewrite this exponential into logarithmic
form, we will take the logarithm of both sides of the equation. Since we often wish to
evaluate the result to a decimal answer, we will usually utilize either the common log or
natural log. For this example, we’ll use the natural log:
ln(2 x ) = ln( )
10
Utilizing the exponent property for logs,
x ln(2) = ln( )
10
Now dividing by ln(2),
ln(10)
x = ( ) ≈ 2.861
ln 2
Notice that this result matches the result we found using the change of base formula.
248 Chapter 4
Example 13
In the first section, we predicted the population (in billions) of India t years after 2008
by using the function
t
f t() =
1
(
14
.
1
+ 0
)
0134
.
. If the population continues following
this trend, when will the population reach 2 billion?
We need to solve for the t so that f(t) = 2
t
2 =
)
0134
.
1
(
14
.
1
Divide by 1.14 to isolate the exponential expression
2
t
= 0134
.
1
Take the logarithm of both sides of the equation
14
.
1
2
ln
= ln(
t
0134
.
1
) Apply the exponent property on the right side
14
.
1
2
ln
= t ln( .
1 0134)
Divide both sides by ln(1.0134)
1 14
.
2
ln
14
.
1
t =
≈
years
ln(
)
23
.
42
0134
.
1
If this growth rate continues, the model predicts the population of India will reach 2
billion about 42 years after 2008, or approximately in the year 2050.
Try it Now
4. Solve (
5 .
0 )
93 x = 10 .
In addition to solving exponential equations, logarithmic expressions are common in
many physical situations.
Example 14
In chemistry, pH is a measure of the acidity or basicity of a liquid. The pH is related to
the concentration of hydrogen ions, [ H+], measured in moles per liter, by the equation
pH = −log( H +
) .
If a liquid has concentration of 0.0001 moles per liber, determine the pH.
Determine the hydrogen ion concentration of a liquid with pH of 7.
To answer the first question, we evaluate the expression −log(
)
0001
.
0
. While we could
use our calculators for this, we do not really need them here, since we can use the
inverse property of logs:
− log( .
0
)
0001 = −log(10 4− ) = −(− )
4 = 4
Section 4.3 Logarithmic Functions 249
To answer the second question, we need to solve the equation 7 = −log( H +
) . Begin
by isolating the logarithm on one side of the equation by multiplying both sides by -1:
7
− = log( H +
)
Rewriting into exponential form yields the answer
7
H +
10−
=
= 0.0000001
moles per liter.
Logarithms also provide us a mechanism for finding continuous growth models for
exponential growth given two data points.
Example 15
A population grows from 100 to 130 in 2 weeks. Find the continuous growth rate.
Measuring t in weeks, we are looking for an equation
rt
P t() = ae so that P(0) = 100 and
P(2) = 130. Using the first pair of values,
0
100
r
ae ⋅
=
, so a = 100.
Using the second pair of values,
2
130 100 r
e ⋅
=
Divide by 100
130
r 2
= e
Take the natural log of both sides
100
ln( )
3
.
1 = ( r 2
ln e )
Use the inverse property of logs
ln( .
1 )
3 = 2 r
ln( .
1 )
3
r =
≈ 1312
.
0
2
This population is growing at a continuous rate of 13.12% per week.
In general, we can relate the standard form of an exponential with the continuous growth
form by noting (using k to represent the continuous growth rate to avoid the confusion of
using r in two different ways in the same formula):
x
kx
a 1
( + r) = ae
x
kx
1
( + r) = e
k
1+ r = e
Using this, we see that it is always possible to convert from the continuous growth form
of an exponential to the standard form and vice versa. Remember that the continuous
growth rate k represents the nominal growth rate before accounting for the effects of
continuous compounding, while r represents the actual percent increase in one time unit
(one week, one year, etc.).
250 Chapter 4
Example 16
A company’s sales can be modeled by the function
12
.
0 t
S t() =
e
5000
, with t measured in
years. Find the annual growth rate.
Noting that
k
1+ r = e , then
12
.
0
r = e
−1 = 1275
.
0
, so the annual growth rate is 12.75%.
The sales function could also be written in the form
t
S t() =
1
(
5000 + .
0
)
1275 .
Important Topics of this Section
The Logarithmic function as the inverse of the exponential function
Writing logarithmic & exponential expressions
Properties of logs
Inverse properties
Exponential properties
Change of base
Common log
Natural log
Solving exponential equations
Try it Now Answers
1. log
= =
=
4 (16)
2 log 42
4
2log4 4
2. 6
3. − 2ln( x)
4. ln( )
2 ≈ − 5513
.
9
ln(
)
93
.
0
Section 4.3 Logarithmic Functions 251
Section 4.3 Exercises
Rewrite each equation in exponential form
1. log ( q) = m
2. log ( t) = k
3. log b = c
4. log z = u
p ( )
a ( )
4
3
5. log( v) = t
6. log( r) = s
7. ln ( w) = n
8. ln ( x) = y
Rewrite each equation in logarithmic form.
9. 4 x = y
10. 5 y = x
11. d
c = k
12. z
n = L
13. 10 a = b
14. 10 p = v
15. k
e = h
16. y
e = x
Solve for x.
17. log x = 2
18. log ( x) = 3
19. log ( x) = 3
−
20. log ( x) = 1
−
3 ( )
4
2
5
21. log( x) = 3
22. log( x) = 5
23. ln ( x) = 2
24. ln ( x) = 2
−
Simplify each expression using logarithm properties.
25. log 25
26. log 8
27.
1
log
28.
1
log
2 ( )
5 (
)
3 27
6 36
29. log
6
30. log ( 3 5
31. log(10,000)
32. log(100)
5
)
6 (
)
33. log(
)
0.001
34. log(
)
0.00001
35. ( 2
ln e− )
36. ( 3
ln e )
Evaluate using your calculator.
37. log(0.04)
38. log(1045)
39. ln (15)
40. ln (0.02)
Solve each equation for the variable.
41. 5 x =14
42. 3 x = 23
43. x
1
7 =
44. x 1
3 =
15
4
45. 5 x
e =17
46. 3 x
e =12
47. 4 x−5
3
= 38
48. 2 x−3
4
= 44
49. 1000(1.03) t = 5000
50. 200(1.06) t = 550
51. (
)3
3 1.04 t = 8
52. (
)4
2 1.08 t = 7
53.
0.12
50
t
e−
= 10
54.
0.03
10
t
e−
= 4
x
x
55.
1
10 8