Algebra 2

Unit 5 - Logarithm and Exponential Functions

Teacher Notes

Unit 5 Logarithm and Exponential Functions 3 ....................

Day 1 - Introduction to Logarithms 3 .............................................

Day 2 - Expanding and Condensing Logs 8 ...................................

Day 3 - Solving Exponential Equations with and without Logs 14 ....

Day 4 - Logarithmic Equations 16 .................................................

Day 5 - Natural Log, e, and Equations 19 .....................................

Day 6 - Graphing Exponential Functions 22 ...................................

Day 7 - Graphing Log and Natural Logs 30 ..................................

Day 8 - Application of Exponentia Growth, Decay and Compound Interest 34...................................................................................

U N I T 5 L O G A R I T H M A N D E X P O N E N T I A L F U N C T I O N S

DAY 1 - I N T R O D U C T I O N T O L O G A R I T H M S

Solve Evaluate Solve

Solve Solve Solve

When we have the VARIABLE in the EXPONENT position and the answer is a “perfect” number (perfect square, perfect cube, etc). We use mental math to solve. When the answer is NOT a perfect number, we don’t have a way to solve it, until now.

We can rewrite an exponential equation as a logarithmic equation.

Exponential Form is the same as Logarithmic Form

x equals base b to the power of y y equals log base b of x

x2 = 81 3−2 24 = 3x − 5

x = ± 919

x = 7

3x = 9 2x = 16 2x = 9

x = 2 x = 4 3 < x < 4

x = by y = logbx

Convert to logarithmic form

Convert to Exponential Form

23 = 8 512 = 5 ( 1

2 )x

=116

log28 = 3 log5 5 =12

log 12

116

= x

32 = 9 712 = 7 ( 1

3 )x

=127

log39 = 2 log7 7 =12

log 13

127

= x

43 = 64 4 13 = 3 4 ( 1

2 )x

=132

log264 = 3 log4 4 =13

log 12

132

= x

2 = log864 0 = log41 −3 = log101

1000

82 = 64 40 = 1 10−3 =1

1000

Evaluate Logarithmic Functions

We can solve and evaluate logarithmic functions by converting the equation to it’s exponential form.

Find the value of x

3 = log464 0 = logx1 −2 = log101

100

43 = 64 x0 = 1 10−2 =1

100

logx36 = 2 log4x = 3 log 12

18

= x

x = ± 6 x = 64 x = 3

logx64 = 2 log5x = 3 log 12

14

= x

x = 8 x = 125 x = 2

logx81 = 2 log3x = 5 log 13

127

= x

x = 9 x = 243 x = 3

Find the exact value of each log without using a calculator

log525 log93 log2116

212

−4

log12144 log42 log2132

212

−5

log981 log82 log319

213

−2

A common log is one that we use all the time with a base 10, . Since it is used so often, it can be written without the base.

can be rewritten as , but since 35 is not a perfect 4th, I need another way to solve. This is where we will use the Change of Base Formula.

Change of Base REMEMBER BASE GOES TO THE

BOTTOM

Calculate by using the change of base formula

� �

log10

log10x = logx

log435 4x = 35

logbM =logMlogb

log435

log435 =log35log4

=1.544.6021

= 2.565

log342 log46

3.402 1.292

DAY 2 - E X PA N D I N G A N D C O N D E N S I N G L O G S

Exponential expressions have rules to simplify them, logarithms have similar rules

Exponential Rule Logarithmic Rule

Evaluate using the properties of logarithms

b0 = 1 logb1 = 0

b1 = a logbb = 1

log81 log66 log131

0 1 0

log99 log51 log77

1 0 1

The next property

Evaluate using the properties of logarithms

Product Property Rule

Exponential Rule Logarithmic Rule

Use the Product Property of Logarithms to write each logarithm as a sum of logarithms. Simplify if possible.

blogbx = x logbbx = x

4log49 log335 5log515

9 5 15

log774 2log28 log2215

4 8 15

bm ⋅ bn = bm+n logbM ⋅ N = logbM + logbN

log37x log464xy log33x

log37 + log3x 3 + log4x + log4y 1 + log3x

Quotient Property Rule

Exponential Rule Logarithmic Rule

Use the Quotient Property of Logarithms to write each logarithm as a difference of logarithms. Simplify, if possible.

log28xy log99x log327xy

3 + log2x + log2y 1 + log9x 3 + log3x + log3y

bm

bn= bm−n logb

MN

= logbM − logbN

log557

logx

100log4

34

1 − log57 logx − 2 log43 − 1

logx

1000log2

54

log10y

logx − 3 log25 − 2 1 − logy

Power Property Rule

Exponential Rule Logarithmic Rule

Use the Power Property Rule to write each logarithm as a product of logarithms. Simplify if possible.

(bm)n = bm⋅n logbMp = p ⋅ logbM

log543 logx10 log754

3log54 10logx 4log75

logx100 log237 logx20

100logx 7log23 20logx

Property Base a

Inverse Properties

Product Properties

Quotient Properties

Power Properties

Change of Base Formula �logbM =logMlogb

�logb1 = 0

�logbMp = p ⋅ logbM

�logbMN

= logbM − logbN

�logb(M ⋅ N ) = logbM + logbN

�logbbx = x

�blogb x = x

�logbb = 1

Use the Properties to expand the logarithm, simplify if possible

log4(2x3y2) log2(5x4y2)

2 + log4x + 2log4y log25 + 4log2x + 2log2y

log3(7x5y3) log24 x3

3y2z

log37 + 5log3x + log3y34

log2x −14

log23 −12

log2y −14

log2z

lne3

3ln

e4

16

3 − ln3 4 − ln16

lnx 3 lnx3 4

3lnx 3 4lnx

log45 x4

2y3z2log3

3 x2

5yz

20log4x − log432 − 15log4y − 10log4z −13

log3z

Use the Properties of Logarithms to condense the logarithm

Use the Properties of Logarithms to condense the logarithm

� �

log43 + log4x − log4y log25 + log2x − log2y

log43xy

log25xy

log36 − log3x − log3y

log36xy

2log3x + 4log3(x + 1) 3log2x + 2logx(x − 1)

log3x2(x + 1)4 log2x3(x − 1)2

2logx + 2log(x + 1) 3lnx + 4lny − 2lnz

logx2(x + 1)2 lnx3y4

z2

DAY 3 - S O LV I N G E X P O N E N T I A L E Q U AT I O N S W I T H A N D W I T H O U T L O G S

Solve

If , then x = y

1. Create the Same Base

2. Set exponents equal to each other

3. Solve

logb49 = 2 loga64 = 3

b = 7 a = 4

log2(3x − 5) = 4

x = 7

bx = by

32x−5 = 27

33x−2 = 81 7x−3 = 7

If we can’t write the expression with the same base, then we can change it to exponential form and use the change of base formula.

1. Rewrite in logarithmic form

2. Use Change of Base

3. Calculate

4. Check for Extraneous Solutions

� �

5x = 11

log511 = x

log11log5

= x

x = 1.490

7x = 43 8x = 98

x = 1.933 x = 2.205

DAY 4 - L O G A R I T H M I C E Q U AT I O N S

One-to-One Property of Logarithms

Solve

1. Condense each side to a single log

2. Use One-to-One Property to solve

x = 9 and x = -9 3. Eliminate -9, we cannot take the

x=9 log of a negative number.

� �

� �

I f logbM = logbN, then M = N

2log5x = log581

log5x2 = log581

x2 = 81

2log3x = log336 3logx = log64

x = 6 x = 4

1. Condense to one log

2. Rewrite in exponential form

3. Simplify and Solve

4. Eliminate -1, we cannot take the

x=9 log of a negative number

Solve

and and

and

log3x + log3(x − 8) = 2

log3(x)(x − 8) = 2

32 = x(x − 8)

9 = x2 − 8x

0 = x2 − 8x − 9

0 = (x − 9)(x + 1)

x = 9 x = − 1

log2x + log2(x − 2) = 3 log2x + log2(x − 6) = 4

x ≠ − 2 x = 4 x ≠ − 2 x = 8

log4(x + 6) − log4(2x + 5) = − log4x

x ≠ − 5 x = 1

and

and

log(x + 2) − log(4x + 3) = − logx

x ≠ − 1 x = 3

log(x − 2) − log(4x + 16) = log1x

x ≠ − 2 x = 8

DAY 5 - N AT U R A L L O G , E , A N D E Q U AT I O N S

Common Log

When the base of a logarithm is 10, we call it the Common Logarithmic Function and the base is not shown.

Natural Base

The number is defined as the value 2.718…

Similar to , it is a value that is used often in the math, specifically in exponential growth and decay.

Natural Exponential Function is

Natural Log

Natural Logarithmic Function is and it can be written

can be written and is equivalent to

1. Rewrite ln as log base e

2. Rewrite using properties of logs

log10x = logx

e

e

pi

f (x) = ex

logex lnx

y = logex y = lnx x = ey

y = lnx

y = logex

ey = x

Solve

Property Base e

Inverse Properties

Product Properties

Quotient Properties

Power Properties

�ln 1 = 0

�ln Mp = p ⋅ ln M

�lnMN

= ln M − ln N

�ln(M ⋅ N ) = ln M + ln N

�ln ex = x

�eln x = x

�ln e = 1

lnx = 3 lnx = 7 lnx = 9

ln e2x = 4 ln e3x = 6 ln e4x = 4

1. Get the term by itself

2. Take Natural Log of both sides

3. Use Power Property Rule

4.

5. Solve for x

x = .079

� � �

3ex+2 = 24

ex+2 = 8 e

ln ex+2 = ln8

(x + 2)ln e = ln8

x + 2 = ln8 lne = 1

x = ln8 − 2

2ex−2 = 18 5e2x = 25

ex

e3= e2x ex

ex= e2 ex2

ex= e6

DAY 6 - G R A P H I N G E X P O N E N T I A L F U N C T I O N S

Linear Quadratic Exponential

In an exponential equation, the variable is in the exponent position. “a” >0 since negative numbers raised to an even root do not produce real solutions.

Graph the following functions on the same coordinate plane.

y = mx + b y = ax2 + bx + c y = bx

y = 2x y = 3x

Graph

Each Graph will Contain the Following Points

For Parent Function , when b > 1 EXPONENTIAL GROWTH

Domain: ALL REAL NUMBERS

Range: ALL POSITIVE NUMBERS

x-intercept: NONE

y-intercept:

Contains: and

Horizontal Asymptote: Vertical Asymptote: NONE

Increasing int: ALL REAL NUMBERS Decreasing int: NONE

y = 4x

y = 5x

(0, 1) (1, b) (1,1b )

y = bx

(0, 1)

(1, b) (1,1b )

y = 0

On the same coordinate plane, graph:

Graph

y = ( 12 )

x

y = ( 13 )

x

y = ( 14 )

x

y = ( 15 )

x

Each Graph will Contain the Following Points

For Parent Function when 0 < b < 1 EXPONENTIAL DECAY

Domain: ALL REAL NUMBERS

Range: ALL POSITIVE NUMBERS

x-intercept: NONE

y-intercept:

Contains: and

Horizontal Asymptote: Vertical Asymptote: NONE

Increasing int: NONE Decreasing int: ALL REAL NUMBERS

(0, 1) (1, b) (−1,1b )

y = bx

(0, 1)

(1, b) (−1,1b )

y = 0

On the same coordinate plane graph each set of functions:

� �

y = 2x y = 2x+1

y = 2x y = 2x−1

� �

Adding a number to the exponent shifts the graph to the left

Subtracting a number from the exponent shifts the graph to the right

One the same coordinate plane graph each set of functions:

y = 3x y = 3x+1

y = 3x

y = 3x − 2

Adding a number to the function shifts the graph up

Subtracting a number from the function shifts the graph down

f (x) = 3x

f (x) = 3x + 2

f (x) = 4x

f (x) = 4x − 2

Parent function: Standard Form:

h is a horizontal shift. + h shifts graph left - h shifts graph right

k is a vertical shift. +k shifts graph up -k shifts graph down

y = bx y = bx−h + k

DAY 7 - G R A P H I N G L O G A N D N AT U R A L L O G S

Graph

Generally, it is easier to graph if we change this to an exponential function first.

We may find it easier to choose a “y” value and find the “x” value with log functions.

log2x

y = log2x 2y = x

-2

-1

0

1

2

3

�y �2y = x �(x , y)

Graph

Each graph will contain the points

log3x log5x

(1, 0) (b, 1) ( 1b

, − 1)

For the Parent Function , when a > 1

Domain: ALL POSITIVE NUMBERS

Range: ALL REAL NUMBERS

x-intercept:

y-intercept: NONE

Contains: and

Horizontal Asymptote: NONE Vertical Asymptote: x=0

Increasing int: ALL REAL NUMBERS Decreasing int: NONE

Graph

y = logax

(1, 0)

(1, b) ( 1b

, − 1))

y = log 13x

Graph

For the Parent Function , when 0 < b < 1

Domain: ALL POSITIVE NUMBERS

Range: ALL REAL NUMBERS

x-intercept:

y-intercept: NONE

Contains: and

Horizontal Asymptote: NONE Vertical Asymptote: x=0

Increasing int: NONE Decreasing int: ALL REAL NUMBERS

y = log 12x y = log 1

4x

y = logbx

(1, 0)

(1, b) ( 1b

, − 1))

DAY 8 - A P P L I C AT I O N O F E X P O N E N T I A G R O W T H , D E C AY A N D C O M P O U N D I N T E R E S T

Compound Interest Formula

when compounded n times a year

when compounded continuously

Jermael’s parents put $10,000 in investments for his college expenses on his first birthday. They hope the investments will be worth $50,000 when he turns 18. If the investment compounds continuously, approximately what rate of growth will they need to achieve their goal?

Hector invests $10,000 at age 21. He hopes the investments will be worth $150,000 when he turns 50. If the interest is compounded continuously, approximately what rate of growth will he need to achieve his goal?

A = P (1 +rn )

nt

A = Pert

9.5 %

9.3 %

Rachel invest $15,000 at age 25. She hopes the investments will be worth $90,000 when she turns 40. If the interest is compounded continuously, approximately what rate of growth will she need to achieve her goal?

Growth and Decay

For an original amount, , that grows or decays at a rate, k, for a certain time, t, the final amount, A, is:

or using r for rate

Researchers recorded that a certain bacteria population grew from 100 to 300 in 3 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment?

First find the unknown rate, then use the rate to find the number of bacteria.

11.95 %

A0

A = A0ekt A = A0ert

rate = 36.62 % A = 656,035.502

Researchers recorded that a certain bacteria population grew from 100 to 500 in 6 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment?

Researchers recorded that a certain bacteria population declined from 700,000 to 400,000 in 5 hours after the administration of medication. At this rate of decay, how many bacteria will there be 24 hours from the start of the experiment?

The half-life of radium-226 is 1590 years. How much of a 100mg sample will be left in 500 years?

First find the decay rate, then use the ratite find the amount of the sample remaining.

r = .2682

A = 62440.55

r = − .1119

A = 0

r = − .000436

A = 80.415

The half-life of magnesium-27 is 9.45 minutes. How much of a 10mg sample will be left after 6 minutes?

The half-life of radioactive iodine is 60 days. How much of a 50mg sample will be left after 40 days?

r = − .0733

A = 6.44

r = − .01155

A = 31.498