Inverse Relations and
Functions
OBJ:  Find the inverse of a relation
 Draw the graph of a function
and its inverse
 Determine whether the
inverse of a function
is a function
FINDING INVERSES OF LINEAR FUNCTIONS
An inverse relation maps the output values back to their
original input values. This means that the domain of the
inverse relation is the range of the original relation and
that the range of the inverse relation is the domain of the
original relation.
Original relation
x
y
– 2 DOMAIN
–1 0 1
4
Inverse relation
2
x
RANGE
2 0 –2 –4
y
4 DOMAIN
2 0 –2 –4
– 2 RANGE
–1 0
1
2
FINDING INVERSES OF LINEAR FUNCTIONS
Original relation
x –2 –1 0
y
4
2
1
Inverse relation
2
0 –2 –4
Graph of original
relation
Reflection in y = x
Graph of inverse
relation
x
4
2
0 –2 –4
y –2 –1 0
1
2
y=x
FINDING INVERSES OF LINEAR FUNCTIONS
To find the inverse of a relation that is given by an
equation in x and y, switch the roles of x and y and
solve for y (if possible).
Finding an Inverse Relation
Find an equation for the inverse of the relation y
= 2 x – 4.
SOLUTION
y=2x–4
x =2y – 4
x + 4 = 2y
1x+2=y
2
Write original relation.
Switch x
x and yy.
Add 4
4 to each side.
Divide each side by 2.
2
The inverse relation is y = 1 x + 2.
2
If both the original relation and the inverse relation happen to be
functions, the two functions are called inverse functions.
Introduction to Logarithmic Functions
FINDING THE INVERSE OF AN EXPONENTIAL
x
y
b
yx = log
Inverse
Logarithmic
of the Exponential
Form
Exponential
Function
Function
Writing Exponential form to Logarithmic form
First we must learn how to read logarithmic form:
The expression log b y is read as “log base b of y”
Examples:
log 5 125
log of base 5 of 125
log 6 36
log of base 6 of 36
1
log 3
5
log of base 3 of 1/5
Logarithms
Index
Power
Exponent
Logarithm
Base
102 = 100
Number
“10 raised to the power 2 gives 100”
“The power to which the base 10 must be raised to give 100 is 2”
“The logarithm to the base 10 of 100 is 2”
Log10100 = 2
Logarithms
y = bx
Logby = x
Logarithm
2
10
Base
= 100
Number
logby = x
is the inverse of
y = bx
Base
Logarithm
Log10100 = 2
Number
23 = 8
Log28 = 3
34 = 81
Log381 = 4
Log525 =2
52 = 25
Log93 = 1/2
91/2 = 3
Rewriting Logarithmic Equations
Exponential Form
2 1 
1
2
Logarithmic Form
log 2
1
 1
2
2 4  16
log 2 16  4
5 x  125
log 5 125  x
6 y  36
log 6 36  y
1
x
3 
9
1
log 3  x
9
103 = 1000
log101000 = 3
p = q2
logqp = 2
24 = 16
log216 = 4
xy = 2
logx2 = y
104 = 10,000
log1010000 = 4
pq = r
logpr = q
32 = 9
log39 = 2
logxy = z
xz = y
42 = 16
log416 = 2
loga5 = b
ab = 5
10-2 = 0.01
log100.01 = -2
logpq = r
pr = q
log464 = 3
43 = 64
c = logab
b = ac
log327 = 3
33 = 27
log366 = 1/2
361/2 = 6
log121= 0
120 = 1
103 = 1000
log101000 = 3
p = q2
logqp = 2
24 = 16
log216 = 4
xy = 2
logx2 = y
104 = 10,000
log1010000 = 4
pq = r
logpr = q
32 = 9
log39 = 2
logxy = z
xz = y
42 = 16
log416 = 2
loga5 = b
ab = 5
10-2 = 0.01
log100.01 = -2
logpq = r
pr = q
log464 = 3
43 = 64
c = logab
b = ac
log327 = 3
33 = 27
log366 = 1/2
361/2 = 6
log121= 0
120 = 1
103 = 1000
log101000 = 3
p = q2
logqp = 2
24 = 16
log216 = 4
xy = 2
logx2 = y
104 = 10,000
log1010000 = 4
pq = r
logpr = q
32 = 9
log39 = 2
logxy = z
xz = y
42 = 16
log416 = 2
loga5 = b
ab = 5
10-2 = 0.01
log100.01 = -2
logpq = r
pr = q
log464 = 3
43 = 64
c = logab
b = ac
log327 = 3
33 = 27
log366 = 1/2
361/2 = 6
log121= 0
120 = 1
103 = 1000
log101000 = 3
p = q2
logqp = 2
24 = 16
log216 = 4
xy = 2
logx2 = y
104 = 10,000
log1010000 = 4
pq = r
logpr = q
32 = 9
log39 = 2
logxy = z
xz = y
42 = 16
log416 = 2
loga5 = b
ab = 5
10-2 = 0.01
log100.01 = -2
logpq = r
pr = q
log464 = 3
43 = 64
c = logab
b = ac
log327 = 3
33 = 27
log366 = 1/2
361/2 = 6
log121= 0
120 = 1
103 = 1000
log101000 = 3
p = q2
logqp = 2
24 = 16
log216 = 4
xy = 2
logx2 = y
104 = 10,000
log1010000 = 4
pq = r
logpr = q
32 = 9
log39 = 2
logxy = z
xz = y
42 = 16
log416 = 2
loga5 = b
ab = 5
10-2 = 0.01
log100.01 = -2
logpq = r
pr = q
log464 = 3
43 = 64
c = logab
b = ac
log327 = 3
33 = 27
log366 = 1/2
361/2 = 6
log121= 0
120 = 1
Introduction to Logarithmic Functions
CHANGING FORMS
Example 1) Write the following into logarithmic form:
a) 33 = 27
b) 45 = 256
c) 27 = 128
d) (1/3)x=27
ANSWERS
Introduction to Logarithmic Functions
CHANGING FORMS
Example 1) Write the following into logarithmic form:
a) 33 = 27
log327=3
b) 45 = 256
log4256=5
c) 27 = 128
log2128=7
d) (1/3)x=27
log1/327=x
Simplifying Logarithmic Equations
Logarithmic Form
Exponential Form
Solution
log 4 16
4 x  16
x2
log 3 1
3y  1
y0
1
log 2
8
1
2 
8
z
z  3
log 4 2
4a  2
1
a
4
log 27 3
27  3
b
b
1
6
Introduction to Logarithmic Functions
CHANGING FORMS
Example 2) Write the following into exponential form:
a) log264=6
b) log255=1/2
c) log81=0
d) log1/31/9=2
ANSWERS
Introduction to Logarithmic Functions
CHANGING FORMS
Example 2) Write the following into exponential form:
a) log264=6
2^6=64
b) log255=1/2
25^1/2=5
c) log81=0
8^0=1
d) log1/31/9=2
1/3^2=1/9