Chapter 3 Notes
Section 3.3: Logarithmic Functions and Their Graphs
Section 3.4: Properties of Logarithmic Functions
We know how to find 42. You simply take 2 4’s and multiply them together to get 16. However, if you
know that 4x = 10, how do you find x? You find the value of x using what is called a logarithm.
Definition of logarithm with base b:
If b and y are positive numbers and b ≠ 1, the logarithm of y with
base b is shown by logby = x and means that:
logby = x if and only if xb = y.
Notice that a logarithm is an exponent.
Example:
log5
1
= !1 :
5
log14 196 = 2 :
Rewrite the equation in exponential form.
log5
1
1
= !1 is the same thing as saying 5 !1 = .
5
5
log14 196 = 2 is the same thing as saying 142 = 196.
Note that there are special logarithm values. If b is a positive real number and b ≠ 1,
logb1 = 0 because b0 = 1
and
logbb = 1 because b1 = b
logbby = y because by = by
and
b
logb x
= x because logbx = logbx
Example: Evaluate the expression without using a calculator.
log7 343 :
To solve this without a calculator, we want to solve x if 7 x = 343 . To do this, you need to
think of which power of 7 would give 343. Since taking 7 to the power of 3 would give
343, log7 343 = 3 .
log4 4 !0.38 :
To solve this without a calculator, we want to solve x if 4 x = 4 !0.38 . Since the bases are
already the same, the exponents must be the same. This means that log4 4 !0.38 = !0.38 .
x
log1/5 25 :
7
log7 3
:
! 1$
To solve this without a calculator, we want to solve x if # & = 25 . To do this, you need
" 5%
to flip the equation, meaning the exponent x will be negative. Since taking 5 to the power
of 2 would give 25, log1/5 25 = !2 .
To solve this without a calculator, we can rewrite the expression as log7x = log73. Since
the logarithms have the same base, the value being taken in the logarithm must be the
same. Therefore, x = 3.
log5
1
3
25
:
To solve this without a calculator, we want to solve x if 5 x =
that
( )
3
25
!1
1
3
25
, or 5 x =
( )
3
25
!1
. Note
is the same as 25-1/3. This is (52)-1/3, or 5-2/3. Since 5x = 5-2/3, x = - 2/3.
The most common logarithms are the common logarithm, or log x which is log10x, and the natural
logarithm, or ln x which is logex. Your calculator can do both of these.
Example: Use a calculator to evaluate the expression. Round the result to three decimal places.
log 0.3:
In your calculator, press the “log” button to see “log(“ and enter 0.3 and then close the
parentheses. The answer is – 0.523.
ln 150:
In your calculator, press the “ln” button to see “ln(“ and enter 150 and then close the
parentheses. The answer is 5.011.
We can also graph logarithmic functions just like exponential growth and decay functions. Although you
know most all of these already from similar graphs, here are some reminders about the characteristics of
the graph of y = logb(x – h) + k:
•
•
•
The line x = h is a vertical asymptote since plugging in h for x will make the log of 0. Since b > 0, you
can never get what is in the parentheses to be zero.
The domain is always x > h, and the range is always all real numbers.
If b > 1, the graph moves up to the right. If 0 < b < 1, the graph moves down to the right.
Example: Graph the function y = log5 (x + 4) . State the domain and range.
Before finding points, realize that you can’t put numbers equal to or less than – 4 for x since – 4 will
make it a log of 0. Also remember some of the transformations we have done before in Chapter 7 will
apply here.
•
Because we are subtracting 4 from the x in the expression, we move the normal graph of log5x 4
units to the left.
Now, let us find some points. You don’t have to plot a lot of points to find it; simply plot some convenient
points that you can calculate. For example, you know 51 = 5, so x + 4 = 5, or x = 1. Also, 50 = 1, so
x + 4 = 1, or x = – 3. Finally, 5–1 = 0.2, so x + 4 = 0.2, or x = – 3.8.
x
y
5–1 = 0.2, so x + 4 = 0.2, or x = – 3.8
–1
50 = 1, so x + 4 = 1, or x = – 3
0
51 = 5, so x + 4 = 5, or x = 1
1
Since the graph will never move beyond the line x = – 4, the line x = – 4 is a vertical asymptote.
If we graph these points, we get a graph that looks like the following:
Notice that the graph has x-values greater than – 4 . Also notice that the
graph can have all y-values, positive and negative.
This means the following is true:
Domain: x > !4
Range: All real numbers
For all logarithms (including natural logarithms that are of the form loge,) the following properties are true:
Characteristics of Logarithmic Functions of the Form
f (x ) = a logb (x ! h) + k
Domain
Range
x-intercepts
y-intercepts
Continuity
Extrema
Horizontal asymptotes
Vertical asymptotes
One-to-one?
Concavity
Boundedness
Limits
(h, ∞)
All real numbers
When f(x) = 0
When x = 0
Continuous on (h, ∞)
No local or absolute extrema
None
x=h
Yes (hence why it is a function)
Concave down if a > 0, concave up if a < 0
Not bounded above or below
lim f (x) = " if a > 0 or lim f (x) = #" if a < 0
Increasing/Decreasing
Symmetry
Transformations
Increasing on domain if a > 0, decreasing on domain if a < 0
No symmetry
h moves the graph left or right from where it would normally be.
k moves the graph up or down from where it would normally be.
Making x negative reflects the graph across the y-axis.
Making a negative reflects the graph across the x-axis.
x!"
x!"
You learned how to solve logarithms in the last section. Try these out to get ready for this section:
log10 10 = log 10 = 1
log10 1000 = log 1000 = 3
log10 0.01= log 0.01= !2
Now, take a look at some of the properties that come from these.
log10 1000 = log 1000 = log (10 • 10 • 10) = log 10 + log 10 + log 10 = 1+ 1+ 1= 3
( )
log10 1000 = log 1000 = log 103 = 3log 10 = 3(1) = 3
! 10 $
log10 0.01= log #
= log 10 ' log 1000 = 1' 3 = '2
" 1000 &%
The examples above illustrate three properties of logarithms.
Properties of Logarithms
Let b, u, and v be positive numbers where b ≠ 1.
Property
Name
Product
Property
Quotient
Property
Power
Property
Property
logb uv = logb u + logb v
logb
u
= logb u ! logb v
v
logb u n = n logb u
Example
log5 125 = log5 5 • 25 = log5 5 + log5 25 = 1+ 2 = 3
(
)
1
= log 1! log 10 = 0 ! 1= !1
10
log2 43 = 3 log2 4 = 3(2) = 6
log
Example: Use log 5 ≈ 0.699 and log 15 ≈ 1.176 to approximate the value of the expression.
log 25 :
log
1
:
3
log 25 = log 5 2 = 2 log 5 = 2(0.699) = 1.398
log
1
5
= log
= log 5 ! log 15 = 0.699 ! 1.176 = !0.477
3
15
Sometimes it is helpful, when calculating, to simply expand a term so there is no fraction, exponent, or
combined expression – only the simplest number – inside the logarithmic function itself.
Example: Expand the expression.
ln 22x :
ln 22x = ln 22 + ln x
log3 25 :
log3 25 = log3 5 2 = 2 log3 5
ln
3y 4
:
x3
ln
3y 4
= ln 3y 4 ! ln x 3 = ln 3 + ln y 4 ! ln x 3 = ln 3 + 4 ln y ! 3 ln x
3
x
On the other hand, sometimes it is helpful to condense an expression (make it into one expression).
Example: Condense the expression.
4 log16 12 ! 4 log16 2 :
4 log16 12 ! 4 log16 2 = log16 124 ! log16 24
4
! 124 $
! 12 $
= log16 # 4 & = log16 # & = log16 6
" 2%
" 2 %
()
log3 2 +
1
log y :
2 3
log3 2 +
4
= log16 1296
1
1
1
log3 y = log3 2 + log3 y 2 = log3 2y 2 = log3 2 y
2
1
1
"
%
"
%
1
1
log5 81! $ 2 log5 6 ! log5 4' = log5 814 ! $ log5 6 2 ! log5 4 2 '
4
2
#
&
#
&
"
%
1
1
log5 81! $ 2 log5 6 ! log5 4' :
4
2
#
&
(
)
(
= log5 4 81 ! log5 36 ! log5 4 = log5 3 ! log5 36 ! log5 2
= log5 3 ! log5
)
36
3
1
= log5 3 ! log5 18 = log5
= log5
2
18
6
Up to this point, the only way you have been able to work with logarithmic bases other that 10 or e has
been to have numbers that come out perfect with exponents. Given you will deal with not-so-nice
numbers and ones that are not so easy, you need to find a way to find logarithms of different bases.
Change-Of-Base Formula
Let u, b, and c be positive numbers with b ≠ 1 and c ≠ 1. Then,
logc u =
Formula
log u
logc u =
log c
ln u
logc u =
ln c
logb u
logb c
Example
log7 12 =
log 12
= log(12) / log(7) ! 1.277
log 7
! 5$
ln # &
" 16 %
5
log9
=
= ln(5 / 16) / ln(9) ' (0.529
16
ln 9