Chapter 05 Exponential and Logarithmic Functions Notes ...

Chapter 5: Exponential and Logarithmic Functions

Algebra 2

Chapter 5: Exponential and Logarithmic Functions

5-1 Exponential Functions

Exponential Functions: - a function where the input (x) is the exponent of a numerical base, a.

Exponential Function

f(x) = ax

where a > 0 and a ¡Ù 1

0

Note: because a = 1, the y-intercept of f(x) = ax is 1.

Example 1: Graph the following fucntions by creating a small table of values. Generalize yor graph using

transformation rules.

a. f(x) = 2x

x

f(x)

?3

2(?3) = ?

?2

2(?2) = ?

?1

2(?1) = ?

0

2(0) = 1

1

2(1) = 2

2

2(2) = 4

3

2(3) = 8

f(x) = ?2x

x

f(x)

?3

?2(?3) = ??

?2

?2(?2) = ??

?1

?2(?1) = ??

0

?2(0) = ?1

1

?2(1) = ?2

2

?2(2) = ?4

3

?2(3) = ?8

b.

(3, 8)

(?1, ?)

(2, 4)

(1, 2)

(0, 1)

(?1, ??)

(0, ?1)

(1, ?2)

(2, ?4)

(3, ?8)

This graph represents the basic exponential

function where the y-intercept is at 1. The

graph hugs the x-axis on the left side but

increases drastically as x increases.

c.

f(x) = 2?x = (?)x

x

f(x)

?3

2?(?3) = 8

?2

2?(?2) = 4

?1

2?(?1) = 2

0

2?(0) = 1

1

2?(1) = ?

2

2?(2) = ?

3

2?(3) = ?

d.

(?3, 8)

(?2, 4)

(?1, 2)

(0, 1)

(1, ?)

Using y = af(bx + h) + k, the graph reflects

horizontally against the y-axis. (Note: a negative

exponent can be rewritten where the base, a is

between 0 and 1.) The y-intercept remains at 1.

The graph hugs the x-axis on the right side as it

decreases drastically as x increases.

Page 132.

According to the transformation rules

from y = af(bx + h) + k, the graph

reflects vertically against the x-axis.

The y-intercept is reflected to ?1.

f(x) = 2x + 1

x

f(x)

?3 2(?3) + 1 = 1?

?2 2(?2) + 1 = 1?

?1 2(?1) + 1 = 1?

0

2(0) + 1 = 2

1

2(1) + 1 = 3

2

2(2) + 1 = 5

3

2(3) + 1 = 9

(3, 9)

(?1, 1?)

(2, 5)

(1, 3)

(0, 2)

Using y = af(bx + h) + k, the graph shifted

up by 1 unit. The graph is now hugging the

horizontal line, y = 1 on the left hand side.

The y-intercept is moved to 2.

Copyrighted by Gabriel Tang B.Ed., B.Sc.

Algebra 2

e.

Chapter 5: Exponential and Logarithmic Functions

f(x) = 2(x + 2)

x

f(x)

?3

2(?3 + 2) = ?

?2

2(?2 + 2) = 1

?1

2(?1 + 2) = 2

0

2(0 + 2) = 4

1

2(1 + 2) = 8

2

2(2 + 2) = 16

3

2(3 + 2) = 32

(1, 8)

(?1, 2)

(?2, 1)

Using y = af(bx + h) + k, the graph

shifted left by 2 unit. The graph still

hugs the x-axis on the left hand side.

The y-intercept is now at 4.

(0, 4)

Graphs of Exponential Functions

y

x

f(x) = a

for a > 1

y

(0, 1)

f(x) = ax

for 0 < a < 1

(0, 1)

x

x

Example 2: Graph the following fucntions by creating a small table of values. Generalize any changes.

a.

f(x) = 2x, g(x) = 3x, h(x)

= 10x

g(x) = 3x

h(x) = 10x

f(x) = 2x

x

?3

?2

?1

0

1

2

3

f(x)

2 =?

2(?2) = ?

2(?1) = ?

2(0) = 1

2(1) = 2

2(2) = 4

2(3) = 8

(?3)

g(x)

3 = 271

3(?2) = 19

(?3)

3(?1) = ?

3(0) = 1

3(1) = 3

3(2) = 9

3(3) = 27

h(x)

1

10 = 1000

1

10(?2) = 100

10(?1) = 101

10(0) = 1

10(1) = 10

10(2) = 100

10(3) = 1000

(?3)

For bases, a > 1, as they increase, the graph

increases more sharply as x-increases.

b.

f(x) = 2?x = (?)x, g(x) = 3?x = (?)x , h(x) = 10?x =

g(x) = 3?x = (?)x

x

f(x)

f(x) = 2?x

= (?)x

h(x) = 10?x

x

= ( 101 )

?3

?2

?1

0

1

2

3

?(?3)

2

=8

2?(?2) = 4

2?(?1) = 2

2?(0) = 1

2?(1) = ?

2?(2) = ?

2?(3) = ?

(101 )x

g(x)

3

= 27

3?(?2) = 9

3?(?1) = 3

3?(0) = 1

3?(1) = ?

3?(2) = 19

?(?3)

3?(3) =

1

27

h(x)

10

= 1000

10?(?2) = 100

10?(?1) = 10

10?(0) = 1

10?(1) = 101

1

10?(2) = 100

?(?3)

10(3) =

1

1000

For bases, 0 < a < 1, as they decrease, the

graph decreases more sharply as x-increases.

Copyrighted by Gabriel Tang B.Ed., B.Sc.

Page 133.

Chapter 5: Exponential and Logarithmic Functions

Algebra 2

Since the y-int = 1,

there are no

transformations.

However, it is

decreasing as x

increases (0 < a < 1)

f(x) = ax

x = ?2,

16 = a?2

y = 16

1

16 = a2

Example 3: Determine the exponential function given the graph below.

a.

Since the y-int = 1,

there are no

transformations.

f(x) = ax

x = 2,

(2, 25) 25 = a2

y = 25

a=5

b.

(?2, 16)

f(x) = 5x

(0, 1)

a2 =

(0, 1)

1

16

a=?

f(x) = (?)x

Natural Exponential Number (e): - an irrational base number that occurs very frequent in nature

(e ¡Ö 2.71828182845904523¡­).

To access the natural number, e, on

the graphing calculator, press

2nd

x

e

or

2nd

LN

e

Graphs of Natural Exponential Functions

y

y

f(x) = ex

f(x) = e?x

¡Â

(0, 1)

(0, 1)

x

x

Example 4: Graph the following fucntions. Generalize yor graph using transformation rules.

a.

f(x) = ex

b.

f(x) = 3ex

This graph of y = ex has the

same shape as other

exponential graph where

the y-intercept is at 1. The

curve exists between the

graphs of y = 2x and y = 3x.

Page 134.

f(x) = e?x + 2

(1, 3e)

(2, e2)

(1, e)

(0, 1)

c.

(?1, 3e?1)

(0, 3)

Using y = af(bx + h) + k, the

graph stretched vertically up

by a factor of 3. The graph

still hugs the x-axis on the left

hand side. The y-interept is

stretched up to 3.

(?1, e + 2)

(0, 3)

(1, e?1 + 2)

From y = af(bx + h) + k, the

graph reflected horizontally

against the y-axis, and it has

moved up 2 units. The graph

still hugs the horizontal line

of y = 2 on the left hand

side. The y-interept has

moved up to 3.

Copyrighted by Gabriel Tang B.Ed., B.Sc.

Algebra 2

Chapter 5: Exponential and Logarithmic Functions

Compound Interest: - interests earned in every term are not withdrawn, but accumulated.

- the closing balance of each term is the opening balance of the next term.

Term: - the period of time spent before interest is calculated.

r?

?

A = P ?1 + ?

n?

?

nt

A = Final Amount after t years

r = Interest Rate per year

P = Principal

n = Number of Terms per year

Compound Term

per Year (n)

Number of times interest

is calculated in a year (nt)

Interest Rate per term ( nr )

(r = interest rate quoted per annum)

Annually (n = 1)

1

r

Semi-annually (n = 2)

2

Quarterly (n = 4)

4

Monthly (n = 12)

12

Daily (n = 365)

365

r

2

r

4

r

12

r

365

Example 5: Mary invested $2000 compounded semi-annually for 3 years at 4%/a. Using the compound

interest formula and the table below, calculate the value of her investment and the total interest

earned at the end of the three years.

P = $2000

n = 2 (semi-annually)

r = 4% / annum (yr) = 0.04

t = 3 years

A = P (1 + nr )nt

Interest Earned = Final Amount ? Principal

0.04 2(3)

A = $2000(1 + 2 )

Interest = $2252.32 ? $2000

6

A = $2000(1.02)

A=?

A = $2252.32

Interest = ?

Interest = $252.32

Example 6: Using the compound interest formula, calculate the value of her investment and the total

interest earned at the end of the three years, if Mary was to invest $2000 for 3 years at 4%/a

a. compounded quarterly.

P = $2000

r = 4%/a = 0.04

t = 3 years

A=?

a. compounded quarterly (n = 4)

A = P (1 + nr )nt

b. compounded monthly.

b. compounded monthly (n = 12)

A = P (1 + nr )nt

A = $2000(1 + 0.404 )4(3)

A = $2000(1.01)12

A = $2000(1 + 012.04 )12(3)

A = $2253.65

A = $2254.54

A = $2000(1 + 012.04 )36

Note: As number of compound period (term) increases, the final amount increases.

5-1 Assignment: pg. 384?388 #15, 17, 19 to 24 (all), 25, 31, 35, 51b, 65, 67, 75

Honours: pg. 384?388 #39 and 69

Copyrighted by Gabriel Tang B.Ed., B.Sc.

Page 135.

Chapter 5: Exponential and Logarithmic Functions

Algebra 2

5-2 Logarithmic Functions

Logarithmic Function: - an inverse function of an exponential equation.

- most commonly use to isolate the unknown exponent of an exponential equation.

y = ax

x = logay

a > 0 and a ¡Ù 1

y>0

Example 1: Convert the following to logarithmic equations.

?3

a. 3x = 81

b. 10?2 = 0.01

10?2 = 0.01

3x = 81

log381 = x

c. x?3 = 64

216

?5?

d. ? ? =

125

?6?

x?3 = 64

log100.01 = ?2

logx64 = ?3

?3

216

?5?

? ? =

125

?6?

log 5

6

216

125

= ?3

Example 2: Convert the following to exponential equations.

a. log636 = x

log636 = x

6x = 36

b. ?5 = logb32

?5 = logb32

c. log101000 = 3

d. log

7

log101000 = 3

log

7

103 = 1000

b?5 = 32

49 = y

49 = y

( 7)

y

= 49

Strategy to Solve Simple Logarithmic Equations

1. If the logarithm is not in base 10, convert it into an exponential form. (Note: the log function of all

scientific and graphing calculators are in base 10.)

2. If y is easily recognized as the power of the base, a or some other base, then write both sides of the

exponential equation in the same base. Equate the exponents and solve.

Example 3: Solve the following logarithmic equations.

a. log8 x = 2

b. log 1 16 = x

c. logx

2

log8x = 2

82 = x

log 1 16 = x

2

x

x = 64

(?) = 16

(2?1)x = 24

2?x = 24

?x = 4

x = ?4

logx

1

=5

243

1

=5

243

1

243

1

x5 = 5

3

x5 = (?)5

x5 =

d. log5 5 5 = x

log5 5 5 = x

5x = 5 5

5x = (51)(5?)

5x = 51 + ?

5x = 53/2

3

x=

2

x=?

Page 136.

Copyrighted by Gabriel Tang B.Ed., B.Sc.

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