Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Logarithm functions
In section 4.1 we studied exponential functions y = ax with various bases a > 0. Here
we assume a 6= 1. Each such function y = ax is an increasing function with domain
(−∞, ∞) and range (0, ∞). It follows that each such function is a one-to-one function
and therefore has an inverse function.
Definition of logarithmic functions
Suppose a > 0 and a 6= 1.
Then the function y = ax has an inverse function y = loga (x), defined for x > 0.
For every real x, loga (ax ) = x .
For every x > 0, aloga (x) = x .
In particular, for every such a,
loga (1) = 0 and loga (a) = 1 .
Let’s look at this statement for specific bases a of interest.
Stanley Ocken
M19500 Precalculus Chapter 4.3 Logarithmic functions
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Find values of log functions
The inverse function of y = 2x is y = log2 (x), the base 2 logarithm of x. It is defined
by log2 (2x ) = x. To figure out log2 (K) :
• Try to rewrite K as K = 2x . Then log2 (K) = log2 (2x ) = x.
• If you can’t rewrite K as K = 2x , use a √
calculator.
Example 1: Find log2 (16), log2 ( 18 ), log2 (8 2), and log2 (24)
Solutions:
•
•
•
•
log2 (16) = log2 (24 ) = 4.
1
= 213 = 2−3 , and so log2 ( 18 ) = log2 (2−3 ) = −3.
8√
√
1
7
7
8 2 = 23 · 21/2 = 23+ 2 = 2 2 , and so log2 (8 2) = log2 (2 2 ) = 27 .
There is no obvious way to rewrite 24 as 2x (other than 24 = 2log2 24 ). Use a calculator.
The inverse function of y = 10x is y = log10 (x), the common logarithm of x.
It is defined by log10 (10x ) = x .
Example 2: Find log10 (1000000) and log10 (.0001).
Solutions:
• log10 (1000000) = log10 (106 ) = 6.
• log10 (.0001) = log10 ( 1014 ) = log10 (10−4 ) = −4.
Stanley Ocken
M19500 Precalculus Chapter 4.3 Logarithmic functions
Logarithmic functions
Transforming graphs and equations
Worked examples
The inverse function of y = ex is y = ln(x), the natural logarithm of x.
Its defining property is that ln(ex ) = x .
Example 3: Find ln(1), ln(e), and ln( √1e ).
Solutions: • ln(1) = ln(e0 ) = 0.
• ln(e) = ln(e1) =1.
1
• ln( √1e ) = ln 11 = ln(e− 2 ) = − 21 .
e2
Switching between log and exponential functions
• y = ax if and only if x = loga y.
• y = 2x if and only if x = log2 y.
• y = 10x if and only if x = log10 y = log y.
Example 4: Solve log2 (x + 2) = 3
Solution: Start with log2 (x + 2) = 3
Rewrite: 23 = x + 2
Solve: x = 23 − 2 = 8 − 2 = 6
Answer: x = 6.
Stanley Ocken
M19500 Precalculus Chapter 4.3 Logarithmic functions
Quiz Review
Logarithmic functions
Transforming graphs and equations
The graph of y = 2x consists of all points
(x, y) with y = 2x , or (using the inverse
function) x = log2 y.
For example, since 23 = 8, the point
(2, 23 ) = (2, 8) = (log2 8, 8) is on the graph of
y = 2x .
Worked examples
Quiz Review
Y
32
y = 2x
28
x = log2 y
24
20
16
12
8
4
0
0
Stanley Ocken
2
4
M19500 Precalculus Chapter 4.3 Logarithmic functions
6
X
Logarithmic functions
Transforming graphs and equations
The graph of y = 2x consists of all points
(x, y) with y = 2x , or (using the inverse
function) x = log2 y.
For example, since 23 = 8, the point
(2, 23 ) = (2, 8) = (log2 8, 8) is on the graph of
y = 2x .
• To see this, start at 3 on the x-axis.
• Move vertically (up) to the graph to (3, 23 )
• Move horizontally (left) to the y-axis to see
23 = 8, and so the point on the graph is (2, 8).
Worked examples
Quiz Review
Y
32
y = 2x
28
x = log2 y
24
20
16
12
23 = 8
(2, 23 )
y = 2x
4
0
0
Stanley Ocken
2
4
M19500 Precalculus Chapter 4.3 Logarithmic functions
6
X
Transforming graphs and equations
The graph of y = 2x consists of all points
(x, y) with y = 2x , or (using the inverse
function) x = log2 y.
For example, since 23 = 8, the point
(2, 23 ) = (2, 8) = (log2 8, 8) is on the graph of
y = 2x .
• To see this, start at 3 on the x-axis.
• Move vertically (up) to the graph to (3, 23 )
• Move horizontally (left) to the y-axis to see
23 = 8, and so the point on the graph is (2, 8).
To figure out log2 8, proceed in reverse.
• Start at y = 8 on the y-axis.
• Move horizontally to the graph, to (log2 8, 8).
• Move vertically down to the x-axis.
Since we arrive at x = 3, conclude log2 8 = 3
Worked examples
Quiz Review
Y
32
y = 2x
28
x = log2 y
24
20
16
12
8
(log2 8, 8) = (2, 23 )
x = log2 y
4
0
2
4
lo
g
2
8
=
0
3
Logarithmic functions
Stanley Ocken
M19500 Precalculus Chapter 4.3 Logarithmic functions
6
X
Transforming graphs and equations
Stanley Ocken
Quiz Review
Y
32
y = 2x
28
x = log2 y
24
(log2 24, 24)
20
16
12
8
x = log2 y
4
0
4
4.
6
2
24
≈
0
2
The graph of y = 2x consists of all points
(x, y) with y = 2x , or (using the inverse
function) x = log2 y.
For example, since 23 = 8, the point
(2, 23 ) = (2, 8) = (log2 8, 8) is on the graph of
y = 2x .
• To see this, start at 3 on the x-axis.
• Move vertically (up) to the graph to (3, 23 )
• Move horizontally (left) to the y-axis to see
23 = 8, and so the point on the graph is (2, 8).
To figure out log2 8, proceed in reverse.
• Start at y = 8 on the y-axis.
• Move horizontally to the graph, to (log2 8, 8).
• Move vertically down to the x-axis.
Since we arrive at x = 3, conclude log2 8 = 3
Similarly, to find log2 (24),
• Start at y = 24 on the y-axis.
• Move horizontally to the graph.
• Move vertically to the x-axis, where you find
x that makes x = log2 (24) ≈ 4.6 .
Worked examples
lo
g
Logarithmic functions
M19500 Precalculus Chapter 4.3 Logarithmic functions
6
X
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Y
8
To change the graph of
y = 2x to the graph of
y = log2 x, reflect each
point (x, 2x ) across the
diagonal line y = x to
obtain point
(2x , x) = (2x , log2 2x ).
(x, 2x ) (x, log2 (x))
y = 2x
4
y = log2 (x)
X
0
-2
-2
0
Stanley Ocken
4
M19500 Precalculus Chapter 4.3 Logarithmic functions
8
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Y
8
To change the graph of
y = 2x to the graph of
y = log2 x, reflect each
point (x, 2x ) across the
diagonal line y = x to
obtain point
(2x , x) = (2x , log2 2x ).
(x, 2x ) (x, log2 (x))
(−2, 41 )
( 14 , −2)
y = 2x
4
y = log2 (x)
(−2, 1
)
4
X
0
, −2)
(1
4
-2
-2
0
Stanley Ocken
4
M19500 Precalculus Chapter 4.3 Logarithmic functions
8
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Y
8
To change the graph of
y = 2x to the graph of
y = log2 x, reflect each
point (x, 2x ) across the
diagonal line y = x to
obtain point
(2x , x) = (2x , log2 2x ).
(x, 2x ) (x, log2 (x))
(−2, 41 )
( 14 , −2)
1
(−1, 2 )
( 12 , −1)
y = 2x
4
y = log2 (x)
)
(−1, 1
2
(−2, 1
)
4
X
0
(1
, −1)
2
, −2)
(1
4
-2
-2
0
Stanley Ocken
4
M19500 Precalculus Chapter 4.3 Logarithmic functions
8
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Y
8
To change the graph of
y = 2x to the graph of
y = log2 x, reflect each
point (x, 2x ) across the
diagonal line y = x to
obtain point
(2x , x) = (2x , log2 2x ).
(x, 2x ) (x, log2 (x))
(−2, 41 )
( 14 , −2)
1
(−1, 2 )
( 12 , −1)
(0, 1)
(1, 0)
y = 2x
4
y = log2 (x)
(−2,
(0, 1)
)
(−1, 1
2
1)
4
X
0
(1, 0)
(1
, −1)
2
, −2)
(1
4
-2
-2
0
Stanley Ocken
4
M19500 Precalculus Chapter 4.3 Logarithmic functions
8
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Y
8
To change the graph of
y = 2x to the graph of
y = log2 x, reflect each
point (x, 2x ) across the
diagonal line y = x to
obtain point
(2x , x) = (2x , log2 2x ).
(x, 2x ) (x, log2 (x))
(−2, 41 )
( 14 , −2)
1
(−1, 2 )
( 12 , −1)
(0, 1)
(1, 0)
(1, 2)
(2, 1)
y = 2x
4
y = log2 (x)
(1, 2)
(−2,
(0, 1)
)
(−1, 1
2
(2, 1)
1)
4
X
0
(1, 0)
(1
, −1)
2
, −2)
(1
4
-2
-2
0
Stanley Ocken
4
M19500 Precalculus Chapter 4.3 Logarithmic functions
8
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Y
8
To change the graph of
y = 2x to the graph of
y = log2 x, reflect each
point (x, 2x ) across the
diagonal line y = x to
obtain point
(2x , x) = (2x , log2 2x ).
(x, 2x ) (x, log2 (x))
(−2, 41 )
( 14 , −2)
1
(−1, 2 )
( 12 , −1)
(0, 1)
(1, 0)
(1, 2)
(2, 1)
(2, 4)
(4, 2)
y = 2x
(2, 4)
4
y = log2 (x)
(1, 2)
(4, 2)
(−2,
(0, 1)
)
(−1, 1
2
(2, 1)
1)
4
X
0
(1, 0)
(1
, −1)
2
, −2)
(1
4
-2
-2
0
Stanley Ocken
4
M19500 Precalculus Chapter 4.3 Logarithmic functions
8
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Y
8
(3, 8)
To change the graph of
y = 2x to the graph of
y = log2 x, reflect each
point (x, 2x ) across the
diagonal line y = x to
obtain point
(2x , x) = (2x , log2 2x ).
(x, 2x ) (x, log2 (x))
(−2, 41 )
( 14 , −2)
1
(−1, 2 )
( 12 , −1)
(0, 1)
(1, 0)
(1, 2)
(2, 1)
(2, 4)
(4, 2)
(8, 3)
(3, 8)
y = 2x
(2, 4)
4
y = log2 (x)
(8, 3)
(1, 2)
(4, 2)
(−2,
(0, 1)
)
(−1, 1
2
(2, 1)
1)
4
X
0
(1, 0)
(1
, −1)
2
, −2)
(1
4
-2
-2
0
Stanley Ocken
4
M19500 Precalculus Chapter 4.3 Logarithmic functions
8
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Is there a preferred base for the logarithm functions y = loga x?
The answer is....maybe. First, we usually use base
a > 1.
Y
4
y = log2 x
2
0
0
2
4
6
-2
-4
Stanley Ocken
M19500 Precalculus Chapter 4.3 Logarithmic functions
8
X
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Is there a preferred base for the logarithm functions y = loga x?
The answer is....maybe. First, we usually use base
a > 1.
Y
4
• The decimal system uses the base 10 function
y = 10x . For example, 10−3 = 1013 = 0.001 = one
thousandth; and 109 = 1, 000, 000, 000 = one billion.
Therefore log10 (.001) = −3, and
log10 (1, 000, 000, 000) = 9. This function was always
used for scientific computation, and so it is called the
common logarithm of x. It is usually written without a
base: log x means log10 x.
y = log2 x
2
y = log x = log10 x
0
0
2
4
6
-2
-4
Stanley Ocken
M19500 Precalculus Chapter 4.3 Logarithmic functions
8
X
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Is there a preferred base for the logarithm functions y = loga x?
The answer is....maybe. First, we usually use base
a > 1.
Y
4
• The decimal system uses the base 10 function
y = 10x . For example, 10−3 = 1013 = 0.001 = one
thousandth; and 109 = 1, 000, 000, 000 = one billion.
Therefore log10 (.001) = −3, and
log10 (1, 000, 000, 000) = 9. This function was always
used for scientific computation, and so it is called the
common logarithm of x. It is usually written without a
base: log x means log10 x.
• In calculus, we use the base e (about 2.718), The
inverse of the function exp(x) = ex is loge x and is
always written as ln x, the natural logarithm of x.
y = log2 x
2
y = ln x
y = log x = log10 x
0
0
2
4
6
-2
-4
Stanley Ocken
M19500 Precalculus Chapter 4.3 Logarithmic functions
8
X
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Is there a preferred base for the logarithm functions y = loga x?
The answer is....maybe. First, we usually use base
a > 1.
• The decimal system uses the base 10 function
y = 10x . For example, 10−3 = 1013 = 0.001 = one
thousandth; and 109 = 1, 000, 000, 000 = one billion.
Therefore log10 (.001) = −3, and
log10 (1, 000, 000, 000) = 9. This function was always
used for scientific computation, and so it is called the
common logarithm of x. It is usually written without a
base: log x means log10 x.
• In calculus, we use the base e (about 2.718), The
inverse of the function exp(x) = ex is loge x and is
always written as ln x, the natural logarithm of x.
• What’s special about base e? The tangent line to
the graph of y = loga x at (1, 0) is a diagonal line if
and only if a = e, that is, when y = ln x.
Stanley Ocken
y =x−1
Y
4
2
y = ln x
(1, 0)
0
0
2
4
6
-2
-4
M19500 Precalculus Chapter 4.3 Logarithmic functions
8
X
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming graphs and equations of logarithmic functions
• This slide shows the graphs of y = log2 x and its various reflections. They share the vertical
asymptote line (abbreviated VA) x = 0. Start with y = log2 x, with domain x > 0.
Y
4
3
2
y = log2 x
1
X
0
-1
-2
-3
-4
-16
Stanley Ocken
-12
-8
-4
0
4
8
12
M19500 Precalculus Chapter 4.3 Logarithmic functions
16
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming graphs and equations of logarithmic functions
• This slide shows the graphs of y = log2 x and its various reflections. They share the vertical
asymptote line (abbreviated VA) x = 0. Start with y = log2 x, with domain x > 0.
• Substituting −x for x in y = ln x reflects its graph across the y-axis to yield the graph of
y = log2 (−x), with domain −x > 0, i.e., x < 0.
Y
4
(−8, 3)
(8, 3)
3
2
1
y = log2 x
y = log2 (−x)
X
0
-1
-2
-3
-4
-16
Stanley Ocken
-12
-8
-4
0
4
8
12
M19500 Precalculus Chapter 4.3 Logarithmic functions
16
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming graphs and equations of logarithmic functions
• This slide shows the graphs of y = log2 x and its various reflections. They share the vertical
asymptote line (abbreviated VA) x = 0. Start with y = log2 x, with domain x > 0.
• Substituting −x for x in y = ln x reflects its graph across the y-axis to yield the graph of
y = log2 (−x), with domain −x > 0, i.e., x < 0.
• Multiplying the RHS of y = log2 x by −1 reflects its graph across the x-axis to yield the
graph of y = − log2 x.
Y
4
(8, 3)
3
2
y = log2 x
1
X
0
y = − log2 x
-1
-2
-3
(8, −3)
-4
-16
Stanley Ocken
-12
-8
-4
0
4
8
12
M19500 Precalculus Chapter 4.3 Logarithmic functions
16
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming graphs and equations of logarithmic functions
• This slide shows the graphs of y = log2 x and its various reflections. They share the vertical
asymptote line (abbreviated VA) x = 0. Start with y = log2 x, with domain x > 0.
• Substituting −x for x in y = ln x reflects its graph across the y-axis to yield the graph of
y = log2 (−x), with domain −x > 0, i.e., x < 0.
• Multiplying the RHS of y = log2 x by −1 reflects its graph across the x-axis to yield the
graph of y = − log2 x.
Y
4
(8, 3)
• Replacing x by −x in equation
y = − log2 x by −1 reflects its graph across
the y-axis to yield the graph of
y = − log2 (−x).
3
2
y = log2 x
1
X
0
-1
y = − log2 (−x)
y = − log2 x
-2
-3
(−8, −3)
-4
-16
Stanley Ocken
-12
-8
-4
(8, −3)
0
4
8
12
M19500 Precalculus Chapter 4.3 Logarithmic functions
16
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming graphs and equations of logarithmic functions
• This slide shows the graphs of y = log2 x and its various reflections. They share the vertical
asymptote line (abbreviated VA) x = 0. Start with y = log2 x, with domain x > 0.
• Substituting −x for x in y = ln x reflects its graph across the y-axis to yield the graph of
y = log2 (−x), with domain −x > 0, i.e., x < 0.
• Multiplying the RHS of y = log2 x by −1 reflects its graph across the x-axis to yield the
graph of y = − log2 x.
Y
4
(8, 3)
• Replacing x by −x in equation
y = − log2 x by −1 reflects its graph across
the y-axis to yield the graph of
y = − log2 (−x).
• Combine the last two steps: Substitute
−x for x in the original y = ln x, then
multiply the RHS by −1: the graph reflects
through the origin to give y = − ln(−x).
3
2
y = log2 x
1
X
0
-1
y = − log2 (−x)
-2
-3
(−8, −3)
-4
-16
Stanley Ocken
-12
-8
-4
0
4
8
12
M19500 Precalculus Chapter 4.3 Logarithmic functions
16
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming graphs and equations of logarithmic functions
Y
4
• Start over: Replacing x by x + 12 in
the equation y = ln x shifts its graph
left 12 units to yield the graph of
y = ln(x + 12), with domain
x + 12 > 0, i.e., x > −12.
3
2
y = log2 x
1
X
0
-1
-2
-3
-4
-16
-12
-8
-4
0
4
8
12
16
It’s easy to find the vertical asymptote of a log graph just by looking at the function
definition. The idea is that the VA of y = log x is x = 0.
If you have a more complicated log function, the vertical asymptote is obtained by
setting the input of the log function to zero.
For example, y = 4 + 3 log(2x + 5) looks complicated, but the input to the log
function is 2x + 5. Therefore the VA is given by setting 2x + 5 = 0. Solve for x to see
that the vertical asymptote is the line x = −5/2.
Stanley Ocken
M19500 Precalculus Chapter 4.3 Logarithmic functions
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming graphs and equations of logarithmic functions
Y
4
(8, 3)
(−4, 3)
3
y = log2 (x + 12)
2
y = log2 x
1
-1
-2
-3
-4
-16
-12
Asymptote x = 0
0
Asymptote x = −12
• Start over: Replacing x by x + 12 in
the equation y = ln x shifts its graph
left 12 units to yield the graph of
y = ln(x + 12), with domain
x + 12 > 0, i.e., x > −12.
• When the graph shifts left 12 units, so
does its VA x = 0, which becomes
x = −12.
-8
-4
X
0
4
8
12
16
It’s easy to find the vertical asymptote of a log graph just by looking at the function
definition. The idea is that the VA of y = log x is x = 0.
If you have a more complicated log function, the vertical asymptote is obtained by
setting the input of the log function to zero.
For example, y = 4 + 3 log(2x + 5) looks complicated, but the input to the log
function is 2x + 5. Therefore the VA is given by setting 2x + 5 = 0. Solve for x to see
that the vertical asymptote is the line x = −5/2.
Stanley Ocken
M19500 Precalculus Chapter 4.3 Logarithmic functions
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming logarithmic graphs and equations
Example 5: Transform the graph of y = log2 x step by
step to the graph of y = 4 − log2 (x + 4). Sketch the
resulting graph. Find and label its intercepts and
asymptote.
Solution: Start with the graph of y = log2 x, with
asymptote the y-axis.
Y
16
12
8
4
X
0
-4
-8
-12
-16
-16
Stanley Ocken
y = log2 x
-12
-8
-4
0
4
8
M19500 Precalculus Chapter 4.3 Logarithmic functions
12
16
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming logarithmic graphs and equations
Example 5: Transform the graph of y = log2 x step by
step to the graph of y = 4 − log2 (x + 4). Sketch the
resulting graph. Find and label its intercepts and
asymptote.
Solution: Start with the graph of y = log2 x, with
asymptote the y-axis.
• First substitute x + 4 for x in y = log2 x. The graph
shifts 4 units left to yield the graph of y = log2 (x + 4).
The asymptote shifts 4 left to x = −4.
Y
16
12
8
4
X
0
-4
-8
-12
-16
-16
Stanley Ocken
y = log2 x
y = log2 (x + 4)
-12
-8
-4
0
4
8
M19500 Precalculus Chapter 4.3 Logarithmic functions
12
16
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming logarithmic graphs and equations
Example 5: Transform the graph of y = log2 x step by
step to the graph of y = 4 − log2 (x + 4). Sketch the
resulting graph. Find and label its intercepts and
asymptote.
Solution: Start with the graph of y = log2 x, with
asymptote the y-axis.
• First substitute x + 4 for x in y = log2 x. The graph
shifts 4 units left to yield the graph of y = log2 (x + 4).
The asymptote shifts 4 left to x = −4.
• Multiply the RHS of the equation y = log2 (x + 4) by
−1. Its graph reflects across the x-axis to yield the
graph of y = − log2 (x + 4). The asymptote remains
x = −4.
Y
16
12
8
4
y = − log2 (x + 4)
X
0
-4
-8
-12
y = log2 (x + 4)
-16
-16
Stanley Ocken
-12
-8
-4
0
4
8
M19500 Precalculus Chapter 4.3 Logarithmic functions
12
16
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming logarithmic graphs and equations
Example 5: Transform the graph of y = log2 x step by
step to the graph of y = 4 − log2 (x + 4). Sketch the
resulting graph. Find and label its intercepts and
asymptote.
Solution: Start with the graph of y = log2 x, with
asymptote the y-axis.
• First substitute x + 4 for x in y = log2 x. The graph
shifts 4 units left to yield the graph of y = log2 (x + 4).
The asymptote shifts 4 left to x = −4.
• Multiply the RHS of the equation y = log2 (x + 4) by
−1. Its graph reflects across the x-axis to yield the
graph of y = − log2 (x + 4). The asymptote remains
x = −4.
• Adding 4 to the RHS of y = − log2 (x + 4) shifts its
graph up 4 units to give the requested graph of
y = 4 − log2 (x + 4). The asymptote remains x = −4.
Y
16
12
y = 4 − log2 (x + 4)
8
4
y = − log2 (x + 4)
X
0
-4
-8
-12
-16
-16
Stanley Ocken
-12
-8
-4
0
4
8
M19500 Precalculus Chapter 4.3 Logarithmic functions
12
16
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming logarithmic graphs and equations
Example 5: Transform the graph of y = log2 x step by
step to the graph of y = 4 − log2 (x + 4). Sketch the
resulting graph. Find and label its intercepts and
asymptote.
Solution: Start with the graph of y = log2 x, with
asymptote the y-axis.
• First substitute x + 4 for x in y = log2 x. The graph
shifts 4 units left to yield the graph of y = log2 (x + 4).
The asymptote shifts 4 left to x = −4.
• Multiply the RHS of the equation y = log2 (x + 4) by
−1. Its graph reflects across the x-axis to yield the
graph of y = − log2 (x + 4). The asymptote remains
x = −4.
• Adding 4 to the RHS of y = − log2 (x + 4) shifts its
graph up 4 units to give the requested graph of
y = 4 − log2 (x + 4). The asymptote remains x = −4.
• Set x = 0 in y = 8 − log2 (x + 4).to find the
y-intercept: y = 4 − log2 (4) = 4 − log2 (22 )
= 4 − 2 = 2, and so the y-intercept is 2. The graph
meets the y-axis at (0, 2).
Stanley Ocken
Y
16
12
y = 4 − log2 (x + 4)
8
4
(0, 2)
X
0
-4
-8
-12
-16
-16
-12
-8
-4
0
4
8
M19500 Precalculus Chapter 4.3 Logarithmic functions
12
16
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming logarithmic graphs and equations
Example 5: Transform the graph of y = log2 x step by
Y
16
step to the graph of y = 4 − log2 (x + 4). Sketch the
resulting graph. Find and label its intercepts and
12
asymptote.
y = 4 − log2 (x + 4)
Solution: Start with the graph of y = log2 x, with
asymptote the y-axis.
8
• First substitute x + 4 for x in y = log2 x. The graph
shifts 4 units left to yield the graph of y = log2 (x + 4).
4
(0, 2)
The asymptote shifts 4 left to x = −4.
• Multiply the RHS of the equation y = log2 (x + 4) by
0
X
−1. Its graph reflects across the x-axis to yield the
(12, 0)
graph of y = − log2 (x + 4). The asymptote remains
-4
x = −4.
• Adding 4 to the RHS of y = − log2 (x + 4) shifts its
-8
graph up 4 units to give the requested graph of
y = 4 − log2 (x + 4). The asymptote remains x = −4.
-12
• Set x = 0 in y = 8 − log2 (x + 4).to find the
y-intercept: y = 4 − log2 (4) = 4 − log2 (22 )
Asymptote x = −4
= 4 − 2 = 2, and so the y-intercept is 2. The graph
-16
-16
-12
-8
-4
0
4
8
12
16
meets the y-axis at (0, 2).
4
• Set y = 0 to find the x-intercept: y = 0 = 4 − log2 (x + 4) ⇒ 4 = log2 (x + 4) ⇒ 2 = x + 4 ⇒ x = 12. The
graph meets the x-axis at (12, 0). The asymptote is the vertical line x = −4.
Stanley Ocken
M19500 Precalculus Chapter 4.3 Logarithmic functions
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming logarithmic graphs and equations
Example 6: Transform the graph of y = log3 x
step by step to the graph of y = log3 (−x) − 2.
Sketch the resulting graph. Find and label its
intercepts and asymptote.
Solution: Start with the graph of y = log3 x,
with asymptote the y-axis (the line x = 0.).
Y
9
6
y = log3 x
3
(3,1)
X
0
-3
-6
-9
-12
Stanley Ocken
-6
0
6
M19500 Precalculus Chapter 4.3 Logarithmic functions
12
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming logarithmic graphs and equations
Example 6: Transform the graph of y = log3 x
step by step to the graph of y = log3 (−x) − 2.
Sketch the resulting graph. Find and label its
intercepts and asymptote.
Solution: Start with the graph of y = log3 x,
with asymptote the y-axis (the line x = 0.).
• Substituting −x for x in y = log3 x reflects
its graph across the y-axis to yield the graph of
y = log3 (−x). The asymptote is still x = 0.
Y
9
6
3
y = log3 (−x)
y = log3 x
(-3,1)
(3,1)
X
0
-3
-6
-9
-12
Stanley Ocken
-6
0
6
M19500 Precalculus Chapter 4.3 Logarithmic functions
12
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming logarithmic graphs and equations
Example 6: Transform the graph of y = log3 x
step by step to the graph of y = log3 (−x) − 2.
Sketch the resulting graph. Find and label its
intercepts and asymptote.
Solution: Start with the graph of y = log3 x,
with asymptote the y-axis (the line x = 0.).
• Substituting −x for x in y = log3 x reflects
its graph across the y-axis to yield the graph of
y = log3 (−x). The asymptote is still x = 0.
• Subtracting 2 from RHS of the equation
y = log3 (−x) shifts its graph down 2 units to
yield the requested graph of y = log3 (−x) − 2.
The asymptote shifts down and is still x = 0.
Y
9
6
3
y = log3 x
(-3,1)
(3,1)
X
0
(-3,-1 )
-3
-6
y = log3 (−x) − 2
-9
-12
Stanley Ocken
y = log3 (−x)
-6
0
6
M19500 Precalculus Chapter 4.3 Logarithmic functions
12
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
Transforming logarithmic graphs and equations
Y
Example 6: Transform the graph of y = log3 x
9
step by step to the graph of y = log3 (−x) − 2.
Sketch the resulting graph. Find and label its
6
intercepts and asymptote.
Solution: Start with the graph of y = log3 x,
y = log3 x
3 y = log3 (−x)
with asymptote the y-axis (the line x = 0.).
(-3,1)
(3,1)
• Substituting −x for x in y = log3 x reflects
0
X
its graph across the y-axis to yield the graph of
(−9, 0)
y = log3 (−x). The asymptote is still x = 0.
(-3,-1 )
-3
• Subtracting 2 from RHS of the equation
y = log3 (−x) shifts its graph down 2 units to
yield the requested graph of y = log3 (−x) − 2.
-6
The asymptote shifts down and is still x = 0.
y = log3 (−x) − 2
• Setting x = 0 in y = log3 (−x) − 2 makes the
-9
-12
-6
0
6
12
RHS undefined. There is no y-intercept.
2
• Set y = 0 in that equation to get 0 = log3 (−x) − 2 ⇒ log3 (−x) = 2 ⇒ −x = 3 = 9 and so
x = −9 is the x-intercept, at point (−9, 0). The asymptote is the line x = 0 (the y-axis).
Stanley Ocken
M19500 Precalculus Chapter 4.3 Logarithmic functions
Logarithmic functions
Transforming graphs and equations
Worked examples
Quiz Review
√
Example 1: Find log2 (16), log2 ( 18 ), log2 (8 2), and log2 (24).
Example 2: Find log10 (1000000) and log10 (.0001).
Example 3: Find ln(1), ln(e), and ln( √1e ).
Example 4: Solve log2 (x + 2) = 3
Example 5: Transform the graph of y = log2 x step by step to the graph of
y = 4 − log2 (x + 4). Sketch the resulting graph. Find and label its intercepts and
asymptote.
Example 6: Transform the graph of y = log3 x step by step to the graph of
y = log3 (−x) − 2. Sketch the resulting graph. Find and label its intercepts and
asymptote.
Stanley Ocken
M19500 Precalculus Chapter 4.3 Logarithmic functions
Quiz Review