Logarithms and Log Properties
Definition
y = log b x is equivalent to x = b y
Algebra Cheat Sheet
Basic Properties & Facts
Arithmetic Operations
Properties of Inequalities
If a < b then a + c < b + c and a  c < b  c
 b  ab
a  =
c c
ab + ac = a ( b + c )
a
  a
b =
c
bc
a b
<
c c
a b
If a < b and c < 0 then ac > bc and >
c c
If a < b and c > 0 then ac < bc and
a
ac
=
b b
 
c
a c ad + bc
+ =
b d
bd
a c ad  bc
 =
b d
bd
ab ba
=
cd d c
a+b a b
= +
c
c c
a
  ad
b =
 c  bc
 
d 
ab + ac
= b + c, a  0
a
Properties of Absolute Value
if a  0
a
a =
if a < 0
a
a 0
a = a
a +b  a + b
Exponent Properties
an
1
= a n m = m n
am
a
a na m = a n+ m
(a )
n m
( ab )
a
n
n
= a nm
a 0 = 1, a  0
n
=a b
1
= n
a
a
 
b
n
n
bn
b
=  = n
a
 a
n
m
n
a =a
m n
1
n
a = nm a
( ) = (a )
a = a
Properties of Radicals
2
Complex Numbers
i = 1
1
m
n
n
1
m
i = 1
a = i a, a  0
( a + bi ) + ( c + di ) = a + c + ( b + d ) i
( a + bi )  ( c + di ) = a  c + ( b  d ) i
( a + bi )( c + di ) = ac  bd + ( ad + bc ) i
( a + bi )( a  bi ) = a 2 + b2
n
ab = a b
a + bi = a + b
n
a na
=
b nb
( a + bi ) = a  bi Complex Conjugate
2
( a + bi )( a + bi ) = a + bi
n
n
a n = a , if n is odd
n
a n = a , if n is even
n
2
For a complete set of online Algebra notes visit http://tutorial.math.lamar.edu.
2
) = r log
b
x
x
log b   = log b x  logb y
 y
The domain of log b x is x > 0
Quadratic Formula
Solve ax 2 + bx + c = 0 , a  0
x 2 + 2ax + a 2 = ( x + a )
2
x 2  2 ax + a 2 = ( x  a )
2
b ± b 2  4 ac
2a
If b 2  4ac > 0 - Two real unequal solns.
If b 2  4ac = 0 - Repeated real solution.
If b 2  4ac < 0 - Two complex solutions.
x=
x 2 + ( a + b ) x + ab = ( x + a )( x + b )
x3  a 3 = ( x  a ) ( x2 + ax + a 2 )
2
b logb x = x
Factoring Formulas
x 2  a 2 = ( x + a )( x  a )
d ( P1 , P2 ) =
+ ( y2  y1 )
r
log b ( xy ) = log b x + logb y
log x = log 10 x common log
where e = 2.718281828K
x3 + 3ax 2 + 3a 2 x + a 3 = ( x + a )
2
log b ( x
Special Logarithms
ln x = log e x
natural log
Distance Formula
If P1 = ( x1, y1 ) and P2 = ( x2 , y2 ) are two
points the distance between them is
n
a
a
  = n
b
b
1
n
=a
an
n n
Triangle Inequality
( x2  x1 )
log b b x = x
Example
log 5 125 = 3 because 53 = 125
Factoring and Solving
a
a
=
b
b
ab = a b
Logarithm Properties
log b b = 1
log b 1 = 0
Complex Modulus
x3  3ax2 + 3a 2 x  a 3 = ( x  a )
3
3
Square Root Property
If x2 = p then x = ± p
x3 + a 3 = ( x + a ) ( x2  ax + a 2 )
Absolute Value Equations/Inequalities
If b is a positive number
p =b

p =  b or p = b
x 2 n  a 2 n = ( x n  a n )( x n + a n )
If n is odd then,
x n  a n = ( x  a ) ( x n1 + ax n 2 + L + a n 1 )
xn + an
= ( x + a)( x
n 1
 ax
n 2
2 n 3
+a x
L + a
n 1

b < p < b
p >b

p <  b or
p>b
)
Completing the Square
(4) Factor the left side
Solve 2 x 2  6 x  10 = 0
2
2
(1) Divide by the coefficient of the x
x 2  3x  5 = 0
(2) Move the constant to the other side.
x 2  3x = 5
(3) Take half the coefficient of x, square
it and add it to both sides
2
2
9 29
 3
 3
x 2  3x +    = 5 +    = 5 + =
4 4
 2
 2
© 2005 Paul Dawkins
p <b
3
29

x  =
2
4

(5) Use Square Root Property
3
29
29
x =±
=±
2
4
2
(6) Solve for x
3
29
x= ±
2
2
For a complete set of online Algebra notes visit http://tutorial.math.lamar.edu.
© 2005 Paul Dawkins
Functions and Graphs
Constant Function
y = a or
f ( x) = a
Graph is a horizontal line passing
through the point ( 0, a ) .
The graph is a parabola that opens right
if a > 0 or left if a < 0 and has a vertex
  b 
b 
at  g    ,   .
  2a  2a 
Line/Linear Function
y = mx + b or f ( x ) = mx + b
Graph is a line with point ( 0,b ) and
slope m.
Circle
2
2
( x  h) + ( y  k ) = r2
Graph is a circle with radius r and center
( h, k ) .
Slope
Slope of the line containing the two
points ( x1 , y1 ) and ( x2 , y2 ) is
y2  y1 rise
=
x2  x1 run
Slope – intercept form
The equation of the line with slope m
and y-intercept ( 0,b ) is
y = mx + b
Point – Slope form
The equation of the line with slope m
and passing through the point ( x1 , y1 ) is
m=
y = y1 + m ( x  x1 )
2
( x  h)
2
f ( x) = a ( x  h) + k
2
The graph is a parabola that opens up if
a > 0 or down if a < 0 and has a vertex
at ( h, k ) .
Parabola/Quadratic Function
y = ax 2 + bx + c f ( x ) = ax 2 + bx + c
The graph is a parabola that opens up if
a > 0 or down if a < 0 and has a vertex
 b
 b 
at   , f     .
 2a  2a  
( y  k)
+
2
=1
a2
b2
Graph is an ellipse with center ( h, k )
with vertices a units right/left from the
center and vertices b units up/down from
the center.
( x  h)

(y k)
Division by zero is undefined!
32  9
 32 =  9 ,
(x
(x
)
2 3
 x5
a
a a
 +
b+c b c
1
 x 2 + x3
x2 + x3
( x + a)
2
x2 + a2  x + a
( x + a)
n
x+ a
 x n + a n and
( y  k)
=1
2
2 ( x + 1)  ( 2 x + 2 )
2
( 2x + 2)
2
2
 2 ( x + 1)
2
= 9 Watch parenthesis!
n
x+a 
n
x+n a
= ( x + a )( x + a ) = x 2 + 2 ax + a 2
2
5 = 25 = 3 2 + 4 2  3 2 + 4 2 = 3 + 4 = 7
See previous error.
More general versions of previous three
errors.
2 ( x + 1) = 2 ( x2 + 2 x + 1) = 2 x 2 + 4 x + 2
a
ab

b c
 
c
units up/down from the center and
asymptotes that pass through center with
b
slope ± .
a
a
  ac
b 
c
b
2
© 2005 Paul Dawkins
( 2x + 2)
2
= 4 x2 + 8x + 4
Square first then distribute!
See the previous example. You can not
factor out a constant if there is a power on
the parethesis!
1
 x2 + a2   x2 + a2
( x  h) = 1

b2
a2
Graph is a hyperbola that opens up and
down, has a center at ( h, k ) , vertices b
For a complete set of online Algebra notes visit http://tutorial.math.lamar.edu.
( 3)
= x2 x2 x 2 = x6
( x + a)
 x2 + a 2
x+a 
)
2 3
1
1
1 1
=
 += 2
2 1+ 1 1 1
A more complex version of the previous
error.
a + bx a bx
bx
= +
=1+
a
a a
a
Beware of incorrect canceling!
 a ( x  1) =  ax + a
Make sure you distribute the “-“!
a + bx
 1 + bx
a
2
2
a
b
Graph is a hyperbola that opens left and
right, has a center at ( h, k ) , vertices a
units left/right of center and asymptotes
b
that pass through center with slope ± .
a
Hyperbola
2
Reason/Correct/Justification/Example
2
2
 0 and  2
0
0
2
Hyperbola
2
Error
 a ( x  1)   ax  a
Ellipse
2
Parabola/Quadratic Function
y = a ( x  h) + k
Common Algebraic Errors
Parabola/Quadratic Function
x = ay2 + by + c g ( y ) = ay2 + by + c
 x2 + a2 = (  x2 + a2 )2
Now see the previous error.
a
 
a
1
 a   c  ac
=   =    =
 b   b   1  b  b
   
c c
a a
   
 b  =  b  =  a  1  = a
  
c
 c   b   c  bc
 
1
 
For a complete set of online Algebra notes visit http://tutorial.math.lamar.edu.
© 2005 Paul Dawkins