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Algebra 2
EXPONENT RULES
11/10/08
Product of Powers
X3 ● X5 = X3+5 = X8
34 ● 35 = 34+5 = 39
Quotients of Powers
X8 = X8-3 = X3
X3
49 = 49-3 = 46
43
Power of a Power
(X4) 6 = X4●6 = X24
(53) 4 = 53●4 = 512 = 244140625
(XY) 6 = X6 Y6
(3X) 7 = 37 X7
Power of a Quotient
b
b8
7
( )8 =
( )4
a
a8
3
Rational Exponents
1
n
4
C  C
n
1
X7 7 X4
Simplify: get the terms down to less complicated.
Simplify: 3X²Y²(-2X³Y4)
(3●-2)(X²●X³)(Y²●Y4) = -6X5Y6
 y 
7 4
2 z 13 y 3
=
 y 7*4
 y 28

2 4 z 13*4 y 3*4 16 z 52 y 12
Evaluate: solve, get the answer.
a0 =1 20 = 1
Any number to the zero power is equal to 1.
a1 = a
Negative exponents
X-3 = 1/X3
3-4 = 1/34
21 = 2
1
11/14/08
Operations with Functions
Function: Relationship between two variables.
Relation: For every X you will have a Y.
Domain (X) Range (Y)
Are these functions?
Domain Range
Domain Range
X
Y
X
Y
2
3
3
2
3
4
5
3
4
8
6
3
5
10
6
5
6
12
7
6
Function
Not a function b/c X repeats.
Vertical Line Test : Use to determine if you have a function or not.
If you draw a vertical line that crosses the graph at
two points then you do not have a function.
This isn’t a function.
This IS a function.
2
FUNCTION NOTATION: Helps you know the value of your function.
Y = 3X + 5
Y = 7X + 10
f(x) = 3X +5
g(X) = 7X + 10
f(6) = 3(6) +5 g(6) = 7(6) + 10
f(6) = 18 + 5 = 23
11/14/08
g)6) = 42 + 10 = 52
Addition & Subtraction
f(X) + g(X)
f(X) = 5X² -2X +3
Add vertically
g(X) = 4X²+ 7X -5
5X² -2X +3
+ 4X²+ 7X -5
9X² +2X -2
Subtract vertically
Note: By subtracting the – changes all signs.
5X² -2X +3
5X² -2X +3
-(4X²+ 7X -5) becomes
-4X² -7X + 5
X² -9X + 8
Example 1:
Add:
f(X) = 5X³ + 6X² -7X +12
g(X) = 7X³
+ 5
12X³ + 6X² -7X + 17
Subtract:
5X³ + 6X² -7X +12
- 7X³
- 5
- 2X³ +6X² -7X +7
Multiplication/Division
Example 1:
f(X) = 5X²
g(X) = 3X -1
(5X²)(3X-1) = (5●3)(X²●X1)(5X²● -1) = 15X³- 5X²
Example 2:
f(X)= 6X4 -3X³-2X -4
3
g(X) = 5X²+7X +8
(5X²+7X +8)( 6X4 -3X³-2X -4)
Take each term in the first function and
multiply everything in the second function.
5X²(6X4 )+5X²(-3X³)+5X²(-2X)+ 5X²(-4)= 30X6-15X5
7X(6X4 )+7X(-3X³)+7X(-2X)+7X(-4)
=
-10X³-20X²
42X5-21X4
8(6X4)+8(-3X³)+8(-2X)+8(-4) =
Answer:
-14X²-28X
48X4-24X³
-16X -32
30X +27X +27X4 -34X³-34X2 - 44X -32
6
5
Division
f ( x) 3 x 2  3x 3  2 x  4

 .5  .25  .4  .90  .25  .65
g ( x)
6 x 2  16 x  10
f ( x) 6 x 4  3x 3  2 x  4

 1.2 x 2  .43x 2  2 x  .5
2
g ( x)
5x  7 x  8
11/18/08
Solving Systems of Equations by Graphing or Substitution
y  x  3

 y  3 x  5
Y = mx + b
Y = dependent
X = independent
Remember m = slope
b = y-intercept
Y-intercept is where the line crosses the Y axis.
For the first equation, Y=x-3, the line should look like this:
0
-0.5 0
1
2
3
The line crosses the y-axis
4
at -3.
-1
-1.5
-2
Series1
-2.5
-3
-3.5
4
11/21/08
The solution of a system is where the two graphs
of the equations intersect.
This graph has only one solution.
It is CONSISTENT and INDEPENDENT.
These are parallel lines. THEY DO NOT
INTERSECT. No solution.
It is INCONSISTENT.
Two equations with the same line.
Multiple solutions.
It is CONSISTENT and DEPENDENT.
5
Example 1: Graph the system
x  y  5

 x  5 y  7
Make a table for both equations.
X+y = 5
x-5y = 7
X Y
X Y
5
2
3
2
1.4
4
4
1
4
2.2
3
X+Y=5
2
X-5Y=-7
1
0
The graph has one solution.
1
2
3
4
(3,2) It is consistent and independent.
Solving with substitution
x  y  5

 x  5 y  7
1)
First take the first equation and solve for X
X + Y =5
-Y
-Y
X = 5-Y Replace this value for X in second equation and solve for Y.
2) X-5Y=-7
5-Y -5Y =-7
5 – 6Y = -7
-5
-5
-6Y = -12
-6 -6
Y = 2 Take your value for Y, 2 and replace it in the first equation
and solve for x.
3) X + Y = 5
X+2=5
-2 -2
6
X=3
Solution of the system is (3,2)
Example 2:
2X + Y = 3
3X – 2Y = 8
1) 2X + Y =3
-2X
-2x
Y = -2X +3
2) 3X – 2(-2X + 3) = 8
3X + 4X – 6 = 8
7X -6 = 8
+ 6 +6
7X = 14
7
7
X=2
3) 2(2) + Y = 3
4+Y=3
-4
-4
Y = -1
Solution (2,-1)
11/25/08
2x+5y=15
To solve this you want to be able to cancel out one of the
-4x+7y=-13
variables. Find which one you could cancel by multiplying one
of the equations.
2(2x+5y)=15(2)
Multiply both sides by 2, now you can cancel out
4x+10y = 30
the x.
-4x+7y = -13
17y = 17
17
17
Y=1
Now put your value for y into the first equation and solve for X.
2x+5(1) = 15
2x +5 = 15
-5 -5
2x
= 10
2
2
X=5
Solution for this system is (5,1)
7
Example 1:
3x -7y =8
5x -6y =10
To cancel the Xs out you need to multiply by -3and 5.
Then solve for Y.
(5)3x -7y = (8)(5)=
15x-35y=40
(-3)5x -6y =(10)(-3) = -15x+18y=-30
0 -17y = 10
-17 -17
Y = -.58 Now replace the value of Y in the first equation and solve for X
3x – 7(-.58) = 8
3x + 4.06 = 8.00
- 4.06 -4.06
3x
= 3.94
3
3
X = 1.31 Solution is (1.31, -.58)
Example 2:
2x + 5y = 15
2x - 5y = -13
4x
=2
4
4
X = ½ = .5
The Ys will cancel out.
2(.5) + 5y = 15
1 + 5y = 15
-1
-1
5y = 14
5
5
Y = 2.8 Solution (.5, 2.8)
8
12/1/08
EXPONENTIAL FUNCTION
Bacteria cells double (2) every hour. You start with 25 cells.
Hrs. Cells
1800
0
25
1
50
1400
2
100
1000
3
200
600
4
400
5
800
6
1600
1600
1200
Series1
800
400
200
0
0
The function is f(x) = 25(2)x
5
10
f(x) = ab x
a - is what you start with. b – is the multiplier
x = time ( secs., hrs. ,days, years etc.)
Example 1
Population in the US
1990 248,718,301
Population grows 8% per decade (every 10 years).
In 2025 what will the population be.
2025- 1990 = 35 years or 3.5 decades
Rules for % Multiplier
Growth: If you have a % 1st you change the percent to a decimal.
8% = .08 (Count 2 spaces to the left from end, add zeros if needed)
Next add 1: 1 + .08 = 1.08
f(x) = 248718301 (1.08)3.5
f(x) = 325,604,866
Decay: Change the percent then SUBTRACT from 1.
Example; 15% decay rate, 15% = .15 1-.15 = .85
9
12/5/08
Compounded Interest Formula
A = P (1 +
r nt
)
n
r= interest rate n = how it is compounded t = time
Compounding terms: Annually = 1
Semi-annually = 2
Quarterly = 4
Daily = 365
Weekly = 526
Example 1.
P = $250
A = 250 (1 +
4% compounded annually for years.
.04 1*6
)
1
A = 250(1.04)6
A = $316.33
A = 250(1.01)24
A = $317.43
Compounded quarterly
A = 250 (1 +
f(x) = bx
.04 4*6
)
4
b is the multiplier when 0<b<1 then you have a decay rate.
When b is greater than 0 and less than 1 – decay
12/8/08
Logarithms
Exponential
Logarithm
bx = y
x = logb y
7³ = 343
4 = log7 343
35 = 243
5 = log3 243
105 = 100000
5 = log10 100000
Exponent
b x =y
base
Exponent
x = logb y
base
10
One-to-One Property
Exponents are the same
bx= by
b3= b3
x=y
5 = log2 32
25 = 32
4 log 3V = 34=V
25 = 25
V = 81
12/9/08
Product Property
34 * 36  346  310
log( m * n)  log m  log n
.47 + .60 = 1.07
Log 36 = log(6●6) = log 6 + log 6
1.55 + 1.55 = 3.1
Quotient Property
m
 log m  log n
n
3
log  log 3  log 2
2
3
log  .47  .30  .17
2
log
5
 log 5  log 2
2
log 2.5  .69  .30  .39
log 2.5  log
Power Property
log P m  m log P
log 34  4 log 3
log 34  4(.47)  1.08
Exponential – Logarithmic Inverse Property
log b b x  X
b log  X
x
3log4  4  log 34  4
11
Change of Base
log x
log b X 
log b
log 4 .60
log 3 4 

 .27
log 3 .47
Solve log 3 ( x 2  7 x  5)  log 3 (6 x  1)
x²+7x-5= 6x+1
-6x
-6x
x²+1x-5= 1
-1 -1
x²+1x-6=0 Factor (x+3)(x-2)
x+3=0
x-2=0
-3 -3
+2 +2
X = -3
x=2
Use positive solution first and replace in equation
log 3 (2 2  7(2)  5  log 3 6(2)  1
log 3 (4  14  5)  log 3 (12  1)
log 3 13  log 3 13
True answer
If it wasn’t true you would then use the negative solution.
12
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