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

Original relationInverse relation y420– 2– 2– 4– 4 x210– 1– 1– 2– 2 RANGE F INDING I NVERSES OF L INEAR F UNCTIONS x420– 2– 2– 4– 4 y210– 1– 1– 2– 2 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. RANGE DOMAIN

F INDING I NVERSES OF L INEAR F UNCTIONS x y420– 2– 4 210– 1– 2 Original relation x420– 2– 4 y210– 1– 2 Inverse relation Graph of original relation Reflection in y = x Graph of inverse relation y = x – – – – 4 2

F INDING I NVERSES OF L INEAR F UNCTIONS 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. y = 2 x – 4 Write original relation. S OLUTION Divide each side by 2. 2 x + 2 = y 1 2 The inverse relation is y = x If both the original relation and the inverse relation happen to be functions, the two functions are called inverse functions. Switch x and y. x y Add 4 to each side. 4 x + 4 = 2 y x = 2 y – 4

Introduction to Logarithmic Functions FINDING THE INVERSE OF AN EXPONENTIAL y = b x Exponential Function Inverse of the Exponential Function x y log Logarithmic Form

Writing Exponential form to Logarithmic form First we must learn how to read logarithmic form: The expression is read as “log base b of y” Examples:

Logarithms 10 2 = 100 “10 raised to the power 2 gives 100” Base Index Power Exponent Logarithm “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” Log = 2 Number

Logarithms 10 2 = 100 Base Logarithm Log = 2 Number Logarithm Number Base y = b x Log b y = x 2 3 = 8Log 2 8 = = 81Log 3 81 = 4 Log 5 25 =25 2 = 25 Log 9 3 = 1 / 2 9 1/2 = 3 log b y = x is the inverse of y = b x

Rewriting Logarithmic Equations Exponential Form Logarithmic Form

10 3 = 1000log = = 16log 2 16 = = 10,000log = = 9log 3 9 = = 16log 4 16 = = 0.01log = -2 log 4 64 = 34 3 = 64 log 3 27 = 33 3 = 27 log 36 6 = 1 / /2 = 6 log 12 1= = 1 p = q 2 log q p = 2 x y = 2log x 2 = y p q = rlog p r = q log x y = zx z = y log a 5 = ba b = 5 log p q = rp r = q c = log a bb = a c

10 3 = 1000log = = 16log 2 16 = = 10,000log = = 9log 3 9 = = 16log 4 16 = = 0.01log = -2 log 4 64 = 34 3 = 64 log 3 27 = 33 3 = 27 log 36 6 = 1 / /2 = 6 log 12 1= = 1 p = q 2 log q p = 2 x y = 2log x 2 = y p q = rlog p r = q log x y = zx z = y log a 5 = ba b = 5 log p q = rp r = q c = log a bb = a c

10 3 = 1000log = = 16log 2 16 = = 10,000log = = 9log 3 9 = = 16log 4 16 = = 0.01log = -2 log 4 64 = 34 3 = 64 log 3 27 = 33 3 = 27 log 36 6 = 1 / /2 = 6 log 12 1= = 1 p = q 2 log q p = 2 x y = 2log x 2 = y p q = rlog p r = q log x y = zx z = y log a 5 = ba b = 5 log p q = rp r = q c = log a bb = a c

10 3 = 1000log = = 16log 2 16 = = 10,000log = = 9log 3 9 = = 16log 4 16 = = 0.01log = -2 log 4 64 = 34 3 = 64 log 3 27 = 33 3 = 27 log 36 6 = 1 / /2 = 6 log 12 1= = 1 p = q 2 log q p = 2 x y = 2log x 2 = y p q = rlog p r = q log x y = zx z = y log a 5 = ba b = 5 log p q = rp r = q c = log a bb = a c

10 3 = 1000log = = 16log 2 16 = = 10,000log = = 9log 3 9 = = 16log 4 16 = = 0.01log = -2 log 4 64 = 34 3 = 64 log 3 27 = 33 3 = 27 log 36 6 = 1 / /2 = 6 log 12 1= = 1 p = q 2 log q p = 2 x y = 2log x 2 = y p q = rlog p r = q log x y = zx z = y log a 5 = ba b = 5 log p q = rp r = q c = log a bb = a c

Introduction to Logarithmic Functions CHANGING FORMS Example 1) Write the following into logarithmic form: a) 3 3 = 27 ANSWERS b) 4 5 = 256 c) 2 7 = 128 d) (1/3) x =27

Introduction to Logarithmic Functions CHANGING FORMS Example 1) Write the following into logarithmic form: a) 3 3 = 27 b) 4 5 = 256 c) 2 7 = 128 d) (1/3) x =27 log 3 27=3 log 4 256=5 log 2 128=7 log 1/3 27=x

Simplifying Logarithmic Equations Logarithmic Form Exponential Form Solution

Introduction to Logarithmic Functions CHANGING FORMS Example 2) Write the following into exponential form: a) log 2 64=6 b) log 25 5=1/2 c) log 8 1=0 d) log 1/3 1/9=2 ANSWERS

Introduction to Logarithmic Functions CHANGING FORMS Example 2) Write the following into exponential form: a) log 2 64=6 b) log 25 5=1/2 c) log 8 1=0 d) log 1/3 1/9=2  2^6=64  25^1/2=5  8^0=1  1/3^2=1/9