An exponential function is a function of the form

where a is a positive real number (a > 0) and . The domain of f is the set of all real numbers.

3 2 1 0 1 2 3

2

4

6

(0, 1)

(1, 3)

(1, 6)

(-1, 1/3) (-1, 1/6)

Summary of the characteristics of the graph of

a >1

• The domain is all real numbers. Range is set of positive numbers.

• No x-intercepts; y-intercept is 1.

• The x-axis (y=0) is a horizontal asymptote as

• With a>1, is an increasing function and is one-to-one.

• The graph contains the points (0,1); (1,a), and (-1, 1/a).•The graph is smooth continuous with no corners or gaps.

3 2 1 0 1 2 3

2

4

6

(-1, 3)

(-1, 6)

(0, 1) (1, 1/3) (1, 1/6)

Summary of the characteristics of the graph of

0 <a <1

• The domain is all real numbers. Range is set of positive numbers.

• No x-intercepts; y-intercept is 1.• The x-axis (y=0) is a horizontal asymptote as

• With 0<a<1, is a decreasing function and is one-to-one.

• The graph contains the points (0,1); (1,a), and (-1, 1/a).

•The graph is smooth continuous with no corners or gaps.

0

5

10

(0, 1)

(1, 3)

y x3

0

5

10

(0, 1)(-1, 3)

y x 3

0

5

10

(0, 3)(-1, 5)y = 2

y x 3 2Domain: All real numbersRange: { y | y >2 } or

Horizontal Asymptote: y = 2

More Exponential Functions (Shifts) :

An equation in the form f(x) = ax.

Recall that if 0 < a < 1 , the graph represents exponential decay

and that if a > 1, the graph represents exponential growth

Examples: f(x) = (1/2)x f(x) = 2x

Exponential Decay Exponential Growth

We will take a look at how these graphs “shift” according to changes in their equation...

Take a look at how the following graphs compare to the original graph of f(x) = (1/2)x :

f(x) = (1/2)x f(x) = (1/2)x + 1 f(x) = (1/2)x – 3

Vertical Shift: The graphs of f(x) = ax + k are shifted vertically by k units.

Take a look at how the following graphs compare to the original graph of f(x) = (2)x :

f(x) = (2)x f(x) = (2)x – 3 f(x) = (2)x + 2 – 3

Horizontal Shift: The graphs of f(x) = ax – h are shifted horizontally by h units.

Notice that f(0) = 1

(0,1)

Notice that this graphis shifted 3 units to theright.

(3,1)

Notice that this graphis shifted 2 units to theleft and 3 units down.

(-2,-2)

Take a look at how the following graphs compare to the original graph of f(x) = (2)x :

f(x) = (2)x f(x) = –(2)x f(x) = –(2)x + 2 – 3

Notice that f(0) = 1

(0,1)

This graphis a reflection of f(x) = (2)x . The graph isreflected over the x-axis.

(0,-1)

Shift the graph of f(x) = (2)x ,2 units to the left. Reflect the graph over the x-axis. Then, shift the graph 3 units down

(-2,-4)

A logarithmic function is the inverse of an exponential function.

For the function y = 2x, the inverse is x = 2y.

In order to solve this inverse equation for y, we write it in logarithmic form.

x = 2y is written as y = log2x and is read as “y = the logarithm of x to base 2”.

x -3 -2 -1 0 1 2 3 4

y 1

8

1

4

1

21 2 4 8 16

x

y -3 -2 -1 0 1 2 3 4

1

8

1

4

1

21 2 4 8 16

y = 2x

y = log2x

(x = 2y)

y = 2x

y = x

y = log2x

Graphing the Logarithmic Function

The y-intercept is 1.

There is no x-intercept.

The domain is All Reals

The range is y > 0.

There is a horizontal asymptoteat y = 0.

There is no y-intercept.

The x-intercept is 1.

The domain is x The range is All Reals

There is a vertical asymptoteat x = 0.

y = 2x y = log2x

The graph of y = 2x has been reflected in the line of y = x, to give the graph of y = log2x.

Comparing Exponential and Logarithmic Function Graphs

Logarithms

Consider 72 = 49.

2 is the exponent of the power, to which 7 is raised, to equal 49.

The logarithm of 49 to the base 7 is equal to 2 (log749 = 2).

72 = 49 log749 = 2

Exponential notation

Logarithmic form

In general: If bx = N, then logbN = x.

State in logarithmic form:

a) 63 = 216

b) 42 = 16

log6216 = 3

log416 = 2

State in exponential form:

a) log5125 = 3

b) log2128= 7

53 = 125

27 = 128

Logarithms

State in logarithmic form:

y 1

2

x

1.4

log0.5 y x

1.4

1.4log0.5 y x

a) b) 23x2 32

log2 32 = 3x + 2

Evaluating Logarithms

1. log2128

log2128 = x 2x = 128 2x = 27

x = 7

2. log327

log327 = x 3x = 27 3x = 33

x = 3

Note:log2128 = log227

= 7 log327 = log333

= 3

3. log556 = 6logaam = m because logarithmic and exponential functions are inverses.

4. log816

log816 = x 8x = 16 23x = 24

3x = 4

5. log81

log81 = x 8x = 1 8x = 80

x = 0

loga1 = 0

x 4

3

Think – What power must you raise 2 to, to get 128?

6. log4(log338)

log48 = x 4x = 8 22x = 23

2x = 3

7. log 4 83

log 4 83 = x

4x 83

2 2x 23

3

2x = 1

8. 2 log2 8

2 log2 23

= 23

= 8

9. Given log165 = x, and log84 = y, express log220 in terms of x and y.log165 = x

16x = 5 24x = 5

log84 = y8y = 423y = 4

log220 = log2(4 x 5) = log2(23y x 24x) = log2(23y + 4x) = 3y + 4x

Evaluating Logarithms

x 3

2x

1

2

Logarithmic Functions

x = 2y is an exponential equation.

Its inverse (solving for y) is called a logarithmic equation.

Let’s look at the parts of each type of equation:

Exponential Equationx = ay

exponent

base

number

/logarithm

y = loga xLogarithmic Equation

It is helpful to remember: “The logarithm of a number is the power to which the base must be raised to get the given number.”

Example: Rewrite in exponential form and

solve loga64 = 2

a2 = 64

a = 8

Example: Solve log5 x = 3

Rewrite in exponential form:

53 = x

x = 125

base number exponent

Example: Solve

7y = 1 49

y = –2

log7

1

49y

An equation in the form y = logb x where b > 0 and b ≠ 1 is called a logarithmic function.

Logarithmic and exponential functions are inverses of each other

logb bx = x

blogb x = x

Examples. Evaluate each:a. log8 8

4

b. 6[log6 (3y – 1)]

logb bx = x

log8 84 = 4

blogb x = x

6[log6 (3y – 1)] = 3y – 1Here are some special logarithm values:

1. loga 1 = 0 because a0 = 1

2. loga a = 1 because a1 = a

3. loga ax = x because ax = ax

Laws of LogarithmsConsider the following two problems:

Simplify log3 (9 • 27)

= log3 (32• 33)

= log3 (32 + 3)

= 2 + 3

Simplify log3 9 + log3 27

= log3 32

+ log3 33

= 2 + 3

These examples suggest the Law:

Product Law of Logarithms:

For all positive numbers m, n and b where b ≠ 1, logb mn = logb m + logb n

Consider the following:

a. b.

= log3 3

4

33

= log3 34 – 3

= 4 – 3

log3

81

27

log3 81 log3 27

= log3 34 – log3 3

3

= 4 – 3

These examples suggest the following Law:Quotient Law of Logarithms:

For all positive numbers m, n and b where b ≠ 1, logb m = logb m – logb n

n

The Product and Quotient Laws

Product Law: logb(mn) = logbm + logbn

logb

m

n

logb m logb n Quotient Law:

Express log3

AB

C

as a sum and difference of logarithms:

log3

AB

C

= log3A + log3B - log3C

Evaluate: log210 + log212.8

= log2(10 x 12.8)= log2(128)= log2(27)= 7

The logarithm of a product equals the sum of the logarithms.

The logarithm of a quotient equals the difference of the logarithms.

Solve: x = log550 - log510

Given that log79 = 1.129, find the value of log763:

log763 = log7(9 x 7) = log79 + log77 = 1.129 + 1 = 2.129

Evaluate: x = log45a + log48a3 - log410a4

x log4

5a 8a3

10a4

x log4

40a4

10a4

x = log44

x = 1

Simplifying Logarithms

x = log55 = 1log5

50

10

Consider the following:Evaluate a. log3 9

4

b. 4 log3 9

= log3 (32)4

= log3 32 • 4

= 2 • 4

= (log3 32) • 4

= 2 • 4

These examples suggest the following Law:

Power Law of Logarithms:

For all positive numbers m, n and b where b ≠ 1, logb m

p = p • logb m

Power Law: logbmn = n logbm

logb mn

d n

dlogb m

Express as a single log: 3 log5 3 2 log5 2 1

2log5 4

log 5 33 log5 22 log5 41

2

log5 33 22 4

1

2

log 5 27 4 2 = log5216

The Power Law

The logarithm of a number to a power equals the power times the logarithm of the number.

Evaluate: log5 25 125 log3 81 2433

log 5 52 log51251

2 log 3 34 log3 2431

3

2 log5 5 1

2log5 53 4log3 3

1

3log3 35

= 2(1) 1

23 + 4(1)

1

35

Given that log62 = 0.387 and log65 = 0.898 solve log6 204 :

log6 204 1

4log6 2 2 5

1

4log6 2 log6 2 log6 5

1

40.387 0.387 0.898

= 0.418

Applying the Power Laws

55

6

Applying the Power Laws

Evaluate: 3 log5 2 3log5 4 3(log5 2 log5 4)

3 log5 (2 4)

3 log5 8

If log28 = x, express each in terms of x:

a) log2512

= log283

= 3log28= 3x

b) log22 log28 = xlog223 = x3log22 = xlog2 2

x

3

More examples: Given log12 9 = 0.884 and log12 18 = 1.163, find each:a.

b. log12 2

= log12 9 12

= log12 9 – log12 12

= 0.884 – 1

= –0.116

= log12 18 9

= log12 18 – log12 9

= 1.163 – 0.884

= 0.279

log12

3

4

Example: Solve 2 log6 4 – 1 log6 8 = log6 x 3

log6 42 – log6 8

1/3 = log6 x

log6 16 – log6 2 = log6 x

log6 (16/2) = log6 x

16/2 = x

x = 8

Natural Exponential Functions

The most commonly used base for exponential and logarithmic functions is the irrational number e.

• Exponential functions to base e are called natural exponential functions and are written y = ex.

• Natural exponential function follows the same rules as other exponential functions.

71828.2)1

1(lim

m

m me

Exponential Function

0

1

2

3

4

5

6

7

8

9

10

-3 -2 -1 0 1 2 3 4

xey

y > 0 for all x

passes through (0,1)

positive slope increasing

Natural Logarithms • logarithms to base e ( 2.71828)• loge x or ln x (Note: These mean the same thing!)• ln x is simply the exponent or power to which e

must be raised to get x.y = ln x x = ey

• Since natural exponential functions and natural logarithmic functions are inverses of each other, one is generally helpful in solving the other. Mindful that ln x signifies the power to which e must be raised to get x, for a > 0,

eln x = x [Let’s y = ln x and x = ey x = elnx]

ln ex = x [Let’s y = ln ex ey = ex y = x]

eln x = ln ex = x

Ex) the natural logarithm of x

• ln e =

• ln 1 =

• ln 2 =

• ln 40 =

• ln 0.1 =

Ex) the natural logarithm of x

• ln e = 1 since e1 = e

• ln 1 = 0 since e0 = 1

• ln 2 = 0.6931... since e0.6931... = 2

• ln 40 = 3.688... since e3.688.. = 40

• ln 0.1 = -2.3025 since e-2.3025. = -1

Natural Logarithmic Function

y > 0 for x > 1

y < 0 for 0 < x < 1

passes through (1,0)

positive slope (increasing)

-5

-4

-3

-2

-1

0

1

2

3

4

5

0 1 2 3 4 5 6 7 8 9 10

y = ln x