Exponential and
Logarithmic Functions
Presenter: Njabulo “Mr-N” Nkabinde
Grade 12
Content
One-to-One Functions
• A function is a one-to-one function if each value in the range corresponds with
exactly one value in the domain.

• For a function to be one-to-one, it must not only pass the vertical line test, but
also the horizontal line test.
y
y
x
Function

x
Not a one-to-one
function

y
x
One-to one
function
Inverse Functions
If f(x) is a one-to-one function with ordered pairs of
the form (x,y), its inverse function, f -1(x), is a oneto-one function with ordered pairs of the form (y,x).
Function:
Inverse Function:

{(2, 6), (5,4), (0, 12), (4, 1)}
{(6, 2), (4,5), (12, 0), (1, 4)}

• Only one-to-one functions have inverse functions.

• Note that the domain of the function becomes the
range of the inverse function, and the range
becomes the domain of the inverse function.
To Find the Inverse Function of a One-to-One Function
1. Replace f(x) with y.

2. Interchange the two variables x and y.
3. Solve the equation for y.
4. Replace y with f –1(x). (This gives the inverse function
using inverse function notation.)

Example:
Find the inverse function of f  x   x  1, x  1.
Graph f(x) and f(x) –1 on the same axes.
f  x 

x  1,  1

y

x 1

Replace f(x) with y.

x

y 1

Interchange x and y.

x 
2



y 1



2

Solve for y.

x2  y  1
x2  1  y
f 1( x)  x 2  1, x  0

Replace y with f –1(x) .
f 1( x)  x 2  1, x  0

f  x   x  1, x  1

Note that the symmetry is about the line y = x.
If two functions f(x) and f –1(x) are inverses of each
other,
.
( f f 1)( x)  x and ( f 1 f )( x)  x
Example:
f  x   x  1, x  1 and f 1( x)  x 2  1, x  0
Show that ( f
f  x 

(f

1

f 1)( x)  x and ( f 1 f )( x)  x..
x 1

f )( x)  x  1  1
2

 x2  x

f 1( x)  x 2  1
1

f ( x) 





2

x 1 1

 x 1 1  x
For any real number a > 0 and a  1,
f(x) = ax
is an exponential function.

For all exponential functions of this form,
1. The domain of the function is (, ).
2. The range of the function is (0, ).

3. The graph passes through the points
(1, 1 ),  0,1 , 1, a  .
a
Example:

Graph the function f(x) = 3x.

1, 3 

 -1, 1 
3

 0,1

Domain: (-∞,∞)
Range: {y|y > 0}
Example:



x

Graph the function f(x) = 1 .
3

Notice that each
graph passes
through the
point (0, 1).

 -1, 3

 0,1

 1, 1 
3

Domain:
Range: {y|y > 0}
Logarithmic Functions

For all positive numbers a, where a  1,
y = logax means x = ay.
logarithm
(exponent)

exponent

number

y = logax
base

x = ay

means
number

base
Exponential Form

Logarithmic Form

50 = 1

log101= 0

23 = 8

log28= 3

4

1
  = 16

log1 2 1 = 4
16

6-2 = 1
36

log6 1 = -2
36

1
2
For all logarithmic functions of the form y = logax or
f(x) = logax, where a > 0, a  1, and x > 0,
1. The domain of the function is (0, ) .
2. The range of the function is (, ) .
3. The graph passes through the points
( 1 , 1), 1,0  ,  a,1 .
a
Logarithmic Graphs
Example:
Graph the function f(x) = log10x.
10,1
1, 0 
1
 10 , -1

Notice that the
graph passes
through the
point (1,0).

Domain: {x|x > 0}
Range:
Exponential Function

Logarithmic Function

y = ax (a > 0, a  1)

y = logax (a > 0, a  1)

Domain:
Range:

Points on
Graph:

 0,  
 ,  

 ,  
 0,  

1
 a , 1

1
 1, a 

 0,1
1, a 

x becomes y
y becomes x

1,0
 a,1
Exponential vs.
Logarithmic Graphs
f(x)
f(x) = 10x

f(x) = log10x

f -1(x)

Notice that the two graphs are inverse functions.
Product Rule for Logarithms

For positive real numbers x, y, and a, a  1,
loga xy  loga x  loga y

Example:
log5(4 · 7) = log54 + log57

log10(100 · 1000) = log10100 + log101000 = 2 + 3 = 5
Quotient Rule for Logarithms

For positive real numbers x, y, and a, a  1,
log a x  log a x  log a y
y

Property 1

Example:
log 7 10  log 7 10  log 7 2
2
log10 1  log10 1  log10 1000  0  3  3
1000
Power Rule for Logarithms

If x and y are positive real numbers, a  1, and n is any
real number, then
log a x n  n log a x
Property 2
Example:
log 9 34  4 log 9 3
log10 1002  2 log10 100  2  2  4
Additional Properties of Logarithms
If a > 0, and a  1,

log a a x  x

aloga x  x (x  0)
Example:
log 9 9 4  4
log10 106  6

Property 4
Property 5
Example:
Write the following as the logarithm of a single
expression.
5log6(x  3)  [2log 6(x  4)  3log 6 x] 
log 6(x  3)5  [log 6(x  4) 2  log 6 x3] 

Power Rule

log 6(x  3)5  [log 6(x  4) 2  x3] 

Product Rule

 (x  3)5 
log 6 
(x  4)2 x3 



Quotient Rule
Properties for Solving Exponential and
Logarithmic Equations
a. If x = y, ax = ay.

Properties 6a-6d

b. If ax = ay, then x = y.

c. If x = y, then logbx = logby (x > 0, y > 0).
d. If logbx = logby, then x = y (x > 0, y > 0).
Solving Equations
Example:
Solve the equation 4 x  256.

2 
2

x

 28

Rewrite each side with the same base.

22 x  28

2x  8

Property 6b.

x4

Solve for x.
Solving Equations
Example:
Solve the equation log(x  3)  log x  log 4.
log(x  3)x  log 4

Product Rule

(x  3)x  4

Property 6d.

x 2  3x  4
x 2  3x  4  0

Check:
log(4  3)  log(4)  log 4.
Stop! Logs of negative numbers
are not real numbers.

(x  4)(x  1)  0

log(1  3)  log(1)  log 4.

x  4 or x  1

log 4  0  log 4
log 4  log 4 True
Reference
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Exponential and logarithmic functions

  • 1.
    Exponential and Logarithmic Functions Presenter:Njabulo “Mr-N” Nkabinde Grade 12
  • 2.
  • 3.
    One-to-One Functions • Afunction is a one-to-one function if each value in the range corresponds with exactly one value in the domain. • For a function to be one-to-one, it must not only pass the vertical line test, but also the horizontal line test. y y x Function x Not a one-to-one function y x One-to one function
  • 4.
    Inverse Functions If f(x)is a one-to-one function with ordered pairs of the form (x,y), its inverse function, f -1(x), is a oneto-one function with ordered pairs of the form (y,x). Function: Inverse Function: {(2, 6), (5,4), (0, 12), (4, 1)} {(6, 2), (4,5), (12, 0), (1, 4)} • Only one-to-one functions have inverse functions. • Note that the domain of the function becomes the range of the inverse function, and the range becomes the domain of the inverse function.
  • 5.
    To Find theInverse Function of a One-to-One Function 1. Replace f(x) with y. 2. Interchange the two variables x and y. 3. Solve the equation for y. 4. Replace y with f –1(x). (This gives the inverse function using inverse function notation.) Example: Find the inverse function of f  x   x  1, x  1. Graph f(x) and f(x) –1 on the same axes.
  • 6.
    f  x x  1,  1 y x 1 Replace f(x) with y. x y 1 Interchange x and y. x  2  y 1  2 Solve for y. x2  y  1 x2  1  y f 1( x)  x 2  1, x  0 Replace y with f –1(x) .
  • 7.
    f 1( x) x 2  1, x  0 f  x   x  1, x  1 Note that the symmetry is about the line y = x.
  • 8.
    If two functionsf(x) and f –1(x) are inverses of each other, . ( f f 1)( x)  x and ( f 1 f )( x)  x Example: f  x   x  1, x  1 and f 1( x)  x 2  1, x  0 Show that ( f f  x  (f 1 f 1)( x)  x and ( f 1 f )( x)  x.. x 1 f )( x)  x  1  1 2  x2  x f 1( x)  x 2  1 1 f ( x)    2 x 1 1  x 1 1  x
  • 10.
    For any realnumber a > 0 and a  1, f(x) = ax is an exponential function. For all exponential functions of this form, 1. The domain of the function is (, ). 2. The range of the function is (0, ). 3. The graph passes through the points (1, 1 ),  0,1 , 1, a  . a
  • 11.
    Example: Graph the functionf(x) = 3x. 1, 3   -1, 1  3  0,1 Domain: (-∞,∞) Range: {y|y > 0}
  • 12.
    Example:  x Graph the functionf(x) = 1 . 3 Notice that each graph passes through the point (0, 1).  -1, 3  0,1  1, 1  3 Domain: Range: {y|y > 0}
  • 13.
    Logarithmic Functions For allpositive numbers a, where a  1, y = logax means x = ay. logarithm (exponent) exponent number y = logax base x = ay means number base
  • 14.
    Exponential Form Logarithmic Form 50= 1 log101= 0 23 = 8 log28= 3 4 1   = 16 log1 2 1 = 4 16 6-2 = 1 36 log6 1 = -2 36 1 2
  • 15.
    For all logarithmicfunctions of the form y = logax or f(x) = logax, where a > 0, a  1, and x > 0, 1. The domain of the function is (0, ) . 2. The range of the function is (, ) . 3. The graph passes through the points ( 1 , 1), 1,0  ,  a,1 . a
  • 16.
    Logarithmic Graphs Example: Graph thefunction f(x) = log10x. 10,1 1, 0  1  10 , -1 Notice that the graph passes through the point (1,0). Domain: {x|x > 0} Range:
  • 17.
    Exponential Function Logarithmic Function y= ax (a > 0, a  1) y = logax (a > 0, a  1) Domain: Range: Points on Graph:  0,    ,    ,    0,   1  a , 1 1  1, a   0,1 1, a  x becomes y y becomes x 1,0  a,1
  • 18.
    Exponential vs. Logarithmic Graphs f(x) f(x)= 10x f(x) = log10x f -1(x) Notice that the two graphs are inverse functions.
  • 20.
    Product Rule forLogarithms For positive real numbers x, y, and a, a  1, loga xy  loga x  loga y Example: log5(4 · 7) = log54 + log57 log10(100 · 1000) = log10100 + log101000 = 2 + 3 = 5
  • 21.
    Quotient Rule forLogarithms For positive real numbers x, y, and a, a  1, log a x  log a x  log a y y Property 1 Example: log 7 10  log 7 10  log 7 2 2 log10 1  log10 1  log10 1000  0  3  3 1000
  • 22.
    Power Rule forLogarithms If x and y are positive real numbers, a  1, and n is any real number, then log a x n  n log a x Property 2 Example: log 9 34  4 log 9 3 log10 1002  2 log10 100  2  2  4
  • 23.
    Additional Properties ofLogarithms If a > 0, and a  1, log a a x  x aloga x  x (x  0) Example: log 9 9 4  4 log10 106  6 Property 4 Property 5
  • 24.
    Example: Write the followingas the logarithm of a single expression. 5log6(x  3)  [2log 6(x  4)  3log 6 x]  log 6(x  3)5  [log 6(x  4) 2  log 6 x3]  Power Rule log 6(x  3)5  [log 6(x  4) 2  x3]  Product Rule  (x  3)5  log 6  (x  4)2 x3    Quotient Rule
  • 26.
    Properties for SolvingExponential and Logarithmic Equations a. If x = y, ax = ay. Properties 6a-6d b. If ax = ay, then x = y. c. If x = y, then logbx = logby (x > 0, y > 0). d. If logbx = logby, then x = y (x > 0, y > 0).
  • 27.
    Solving Equations Example: Solve theequation 4 x  256. 2  2 x  28 Rewrite each side with the same base. 22 x  28 2x  8 Property 6b. x4 Solve for x.
  • 28.
    Solving Equations Example: Solve theequation log(x  3)  log x  log 4. log(x  3)x  log 4 Product Rule (x  3)x  4 Property 6d. x 2  3x  4 x 2  3x  4  0 Check: log(4  3)  log(4)  log 4. Stop! Logs of negative numbers are not real numbers. (x  4)(x  1)  0 log(1  3)  log(1)  log 4. x  4 or x  1 log 4  0  log 4 log 4  log 4 True
  • 29.