unit 5 - logarithm and exponential functions - teacher notes 5 - logarithm... · use the product...
TRANSCRIPT
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