Download 3/3/06

Square roots
Square root the top and the bottom
separately.
Leave answer as a fraction – not a decimal.
Yes, you can do this in your head!
0.4  0.4  0.16 is like knowing 4  4  16 and
moving the decimal two places left.
25
25 5


36
36 6
0.16  0.4
Simplify nth roots

3
320  3 64  5  3 64  3 5  4 3 5


Another example:
3
375  3 125  3  3 125  3 3  5 3 3


Find a number in the X3 column on your
yellow sheet that goes into 320 evenly (in
this case 64).
64 is a “perfect cube” because it can be
written as a product of whole numbers (in
this case 64=4x4x4).
64 gets to come “outside” of the radical
only because it’s “perfect”. When it comes
outside, you write the root (in this case 4).
Any numbers that aren’t “perfect” have to
stay “inside” the radical.
x 3 and y 3 are both perfect cubes because
3


3
x11 y 5  3 x3 x3 x3 x 2 y 3 y 2  xxxy 3 x 2 y 2  x 3 y 3 x 2 y 2

x3  x and
3
y3  y .
These “perfect” cubes get to come
“outside” of the radical.
Break the x11 into multiples of x3 and then
any remainder. The remainders are stuck
“inside” the radical because they aren’t
“perfect”.
Note: on the multiple choice problems,
make sure the answer you choose has a
root that matches the one in the problem.
In other words x 2 y 3 x 2 y1 not x 2 y x 2 y1
3
x8 y 4
8  3  2 with a remainder of 2
4  3 = 1 with a remainder of 1
x 2 y 3 x 2 y1
There’s a shortcut if you want to try it. Take
the power on the x (in this case 8) and divide
it by the root (in this case 3). 3 goes into 8
2 whole times with a remainder of 2. This
means that the x on the outside has a power
of 3 and the x on the inside has a power of 2.
Similarly, the exponent on the y is 5 and 3
goes into it 1 whole time with 2 left over.
Simplify expressions with rational exponents
x
1 4
x
4 5
x
5 20
x
16 20
x
21 20
1

x 21 20
Another example:
x 1 2  x 2 3  x 3 6  x 4 6  x 7 6 
16
 x 1 3 
 6 
 x 
x
  x 1 3( 6) 
16

1 318 3 1 6
  x17 3 
16
Alternatively:
16
 x 1 3 
 6 
 x 
  x 1 36) 
16

 x17 18
x 

1 3 1 6
x 
6
1
x7 6
16
1
Convert both fractions to a denominator of 20
and then add them. The answer here had a
negative exponent and you shouldn’t write
answers with negative exponents so use the
rule where you put the expression in the
denominator and make the exponent positive.
1
am  m
a
Remember to do the ( ) first.
When you divide numbers with like bases,
subtract the exponents.
am
 a mn
n
a
In this case the exponents on x are
1
6
and  6 or
3
1
Remember that when you subtract a negative,
change it to adding a positive.
The common denominator in this case is 3 so
18
multiply both the 6 and the 1 by 3 which is
.
3
Add the exponents keeping the denominator
of 3. Then use the power to a power rule and
x 1 18
 1 
x
x 1 18( 1)  x 1 1818 18  x17 18
multiply the exponents.
a 
m n
 a mn
Write square roots as exponential expressions
Express using fractional exponents:
g
g  2 g1  g1 2
Another example:
Write
12z in exponential form.
Remember that when we write a square root,
we usually don’t write the little 2 but it’s there.
Similarly, the g has a power of 1 even though
we don’t usually write it. The power (1) is the
top of the fraction and the root (2) is the
bottom of the fraction.
12 z  2 12 z   12 z 
1
12
Express using fractional exponents
6
6
j
5
The power (5) is the top of the fraction and the
root (6) is the bottom of the fraction.
j5  j5 6
Page 1
Simplify expressions with fractional exponents
x
m
Simplify:  8

2
2 3


1
8
23

 x
n
m
3
 8
253 2  2 253  2 15625  125 but I prefer to take
the root first and then use the exponent.
First rewrite it with a positive exponent, then
find the cube root of 8 (which is 2) and then
take that answer to the power of 2.
  5  5  5  5  125
3
25
 m x n or
The bottom of the fraction is the root.
2  2  2  8 so 3 8 equals 2
The bottom of the fraction is the root and the
top is the power. You could also write it as
81 3  3 8  2
253 2 
n
1
3
2

1
 2
2

1
4
Solve equations containing radicals
Solve:
x  5  6
Solve:
x  2  3  8
There is no way for the square root of a
negative number to equal a negative number
so the answer to this problem is “no solution”.
This is similar to the previous problem, but
you need to add 3 to both sides first and then
realize that you can’t square root any number
to get a negative.
x  2  3  8
+3
x  2  5
4 x  2  3x  4
Solve:

+3
4x  2
 
2
3x  4
4 x  2  3x  4
 3x
-3x

2
1. Square both sides.
2. Put the x ’s on the left side by adding 3x to
both sides.
3. Put the numbers on the right by adding 2
to both sides.
1x  2  4
+2
Solve:

3
3
+2
x  2
x  6  5
x6

3
  5 
3
x  6  125
+6
+6
This is different from the square root problems
because you CAN take the cube root of a
number and get a negative answer.
1. Cube both sides
2. Add 6 to both sides
x  119
Page 2
Solve exponential equations
1
 273 x 4
9
2
3  33(3 x  4)
Solve:
2  9 x  12
12
Get a common base on both sides of the
equation, in this case use 3. Once you have
this, you can just use the exponents because
they have to be equal if the bases are equal.
Then SOLV4X!
+12
10  9 x
10
9
Solve: 5.87 x  24
x log 5.87  log 24
x
log 24
log 5.87
x  1.796
Find the solution: 3x  4 x8
3 x  4 x  48
x log 3  x log 4  8log 4
x log 3  x log 4  8log 4
x(log 3  log 4)  8log 4
8log 4
x
log 3  log 4
x  38.5507
x
Take the log of both sides and then treat the
log5.87 as a constant (which it is) and divide
both sides by it. The answer rounds to 1.796
or 1.8. The  sign means “is approximately
equal to” and should be used to show that you
rounded an answer.
You can’t get a common base using 3 and 4.
I can’t find an easier way to do this problem
although there are other ways to do it. You
can evaluate the logs sooner but then you
might have a rounding error that you don’t get
if you wait until the last minute to use your
calculator.
Page 3
WP: Roots (word problems)
A certain gas will escape from a storage tank
according to the formula e  130 p1 2 , where e
represents the amount escaping in gallons per
minute, and p represents the pressure in
pounds per square inch. How much gas is
escaping when the pressure is 1600 pounds
per square inch? Round your answer to the
nearest whole number.
e  130(1600)1 2
You are given the pressure ( p  1600 ), so
substitute 1600 for p . Take the square root of
1600 to get 40 and then multiply 40 by 130.
The answer is expressed in gallons per
minute.
e  130  40
e  5200 gal/min
The volume of a sphere can be given by the
formula V  4.18879r 3 . You have to design a
spherical container that will hold a volume of
65 cubic inches. What should the radius of
your container be?
65  4.18879r 3
65
 r3
4.18879
You are given the volume needed ( V  65 ), so
substitute 65 for V . To get r 3 by itself, divide
both sides by 4.18879 and then cube root
both sides to get just r . You can simplify the
left side using you calculator in one step:
3
(65 / 4.18879) or 3 x (65 / 4.18879) . Both the
3
m.
65
r
4.18879
r  2.49 inches
3
Page 4
and the
can be found by pressing
Find Inverses of functions
Find the inverse of f ( x)  7 x 2 . Determine if
the inverse is a function.
x  7 y2
x
 y2
7
x
y
7
f 1 ( x )  
f ( x) is the same as y, so the equation is
really y  7 x 2 . To find the inverse, switch the
x and y and solve for y.
x
7
Write the equation for the inverse f ( x) 
y 5
8
8x  y  5
x 5
8
x
To determine whether or not it’s a function
you can graph it. Remember to graph both
x
x
and y  
. This will make a
y
7
7
parabola that opens to the right and if you
apply the vertical line test, it proves that it is
not a function.
f ( x) is the same as y, so the equation is
y 5
really x 
. To find the inverse, switch
8
the x and y and solve for y.
This is a linear equation so it will be a
function.
y  8x  5
f 1 ( x )  8 x  5
Exponential equations to log form
b x  a can be written as logb a  x
2
27 is the base, 2
Write the equation 27 3  9 in logarithmic
form.
log 27 9  2
3
3
is the exponent and 9 is
the answer.
Log equations to exponential form
logb a  x can be written as b x  a
Write the equation log 243 9 
form.
2
in exponential
5
243 is the base, 2
the answer.
2
243 5  9
Page 5
5
is the exponent and 9 is
Find logs by converting to exponential form
b x  a can be written as logb a  x
baseexponent  answer can be written as logbase answer  exponent
Evaluate log10 10 5
log10 105  x
10 x  105
answer: -5
Evaluate log5 25
log 5 25  x
5x  25
5 x  52
answer: 2
 1 
Evaluate log 3 

 729 
 1 
log 3 
x
 729 
1
3x 
729
1
3x  6
3
x
3  36
answer: -6
Convert to exponential form and SOLV4X.
Since both sides have a base of 10, then
x  5 .
A shortcut would be to just look at the
exponent on the 10. Don’t make it harder
than needed! 
Convert to exponential form and SOLV4X.
The exponent that you have to put on 5 to get
25 is 2 so the answer is x  2 .
Convert to exponential form and SOLV4X.
The exponent that you have to put on 3 to get
729 is 6, but it’s in the bottom of a fraction so
the exponent has to be -6.
Page 6
Find natural logs of numbers
ln b is read “natural log of b” and is a simplified way of writing loge b .
Do these types of problems by using e as the base.
There is an e function on your calculator above the / button.
Evaluate ln 34
ln 34  3.526
log e 34  3.526
Calculator: L 34 e.
Convert to exponential form.
e3.526  34
 1 
Evaluate ln  8 
 e
 1 
ln  1 
e 8 
 
log  e   x
ln e
1
Convert from radical form to a fractional
exponent.
You can skip the last few steps if you
recognize that you are looking for the
exponent on e .
8
1
8
   x just look at the
When you get to log e e
e
e e
x
1
x  1
exponent on the e .
8
8
Page 7
1
8
Properties of logs, simplify expressions
log a xy  log a x  log a y
log a
Express in terms of logarithms of x, y
and z:
3 xy 4
log a 2
z
log a 3 xy 4  log z 2
x
log a x  log a y
y
log a x m  m log a x
Use the rules for logarithms of products, quotients
and powers to write as the sum or difference of
logarithms:
log b
log a 3  log a x  log a y 4  log z 2
9
log b x
log a 3  log a x  4 log a y  2 log z
Given logc 2  0.315 , log c 3  0.500 and
5
logc 5  0.732 , find log c .
12
5
log c 2
2 3
log c 5   2 log 2  log 3
 x7 y3 
x7 y3

log
b
8 
z8
 z 
7
9
 log b y
1
3
 log b z
1
9
7
 log b
x 9y
z
8
9

8
3
9

9
7
1
8
log b x  log b y  log z
9
3
9
Write as a single logarithm: 4logb x  8logb y
log b x 4  log b y 8
log b x 4
log b y 8
0.732   2  0.315  0.500 
.398
Properties of logs, find equivalent logs
log a xy  log a x  log a y
log a
x
log a x  log a y
y
 64 
Find an expression equivalent to log a 1  3 
x 

 x 3 64 
log a  3  3 
x 
x
log a x m  m log a x
Find an expression equivalent to log a
 x 3  64 
log a 

3
 x

 ( x  4)( x 2  4 x  16) 
log a 

x3


2
log a ( x  4)  log a ( x  4 x  16)  3log a x
This is a really hard problem because you
have to get a common denominator so you
can subtract the fractions and then know how
Page 8
1
 61  2
log a  
 64 
1
 log a 61  log a 64 
2
61
64
to factor x3  64 into ( x  4)( x 2  4 x  16) .
Solve log equations
Solve: log2  x  8  2
Convert to exponential form.
2 2  x  8
1
 x 8
4
8  8
1 32

4 4
31
x
4
Solve: log3  x  6  log3 x  4
SOLV4X
Subtract 8 from both sides.
Convert -8 into a fraction.
Subtract the fractions.
x
log 3
 x  6  4
1x
 1x
x
 x  6
34 
x
81x  x  6
Combine the logs.
Convert to exponential form.
SOLV4X
Multiply both sides by x .
Subtract 1x from both sides.
Divide both sides by 80.
Reduce the fraction.
80 x  6
6
80
3
x
40
x
Solve equations involving natural logs
Find x if e9.2 x  6 and you are given
ln 6  1.7918
log e 6  9.2 x
Convert to log form.
Because loge x  ln x , you can substitute
1.7918 for ln 6 and divide by 9.2.
ln 6  9.2 x
ln 6
9.2
x  .1948
x
Page 9
Graph exponential and log functions
Graph f ( x)  2 x  5
Graph y  log 2 x  3
Graph f ( x)  2 x on your calculator. The
equation is shifted 5 units down from this
parent equation.
You can’t graph y  log 2 x on your calculator
because it only has log and ln functions.
Instead, use this conversion:
log x
logb x 
log b
In the case of this example, that would be
y   log x / log 2  3 . What do you think the -3
does to the graph?
Graph y   log( x 1)  / log 2 on your calculator.
Graph y  log 2 ( x  1)
What is the difference between this problem
and the previous one. Hint: the -3 was outside the
( ) in the last one and the -1 is inside the ( ) on this one.
This makes a difference in how the graph is shifted.
Page 10
WP: Interest Problems
The compound amount with continuous
compounding is given by A  Pe rt , where P is
the principal, r the rate and t the time in
years. If the rate is 9.5%, find how long it
takes for the money to double, that is for
A  2 P .  ln 2  0.69315
2 P  Pe.095t
Substitute 2P for A .
Divide both sides by P .
Convert to ln form.
Divide both sides by .095.
Substitute 0.69315 for ln2.
It will take 7.3 years for the money to double.
2  e.095t
ln 2  .095t
ln 2
.095
0.69315
t
.095
t  7.3 years
t
If $10,000 is invested at 6% compounded
monthly for 7 years the compounded amount
84
is given by A  10, 000 1.005  . Given that
log1.005  0.00217 , find log A . (Note that
Note that you are asked to find
log A not just A .
Convert to log form and simplify.
10, 000  104 )
log A  log104  84 log1.005
log A  4  84  0.00217
log A  4.18228
Page 11
WP: Growth & decay problems
A radioactive substance decays so that the
amount A present at time t (years) is
A  A0 e 2.5t . Find the half-life (time for half to
decay) of this substance.
.5 A0  A0 e
 ln.5  0.69315
A0 is the “initial amount” of the substance.
Substitute 0.5A for A0 because you want the
half-life of the initial amount of the substance.
2.5t
Divide both sides by A0 .
.5  e 2.5t
ln .5  2.5t
Convert to ln form.
ln 0.5
2.5
0.69315
t
2.5
t  0.277 years
The number of a certain product that will be
sold t years after the product is introduced is
given by 7300ln  5t  2 . How many of the
t
product will be sold 7 years after the product
is introduced? Round the answer to the
nearest whole number.
7300  ln  5t  2 
Solve for t .
Substitute 7 for t.
Simplify using a calculator.
Round 26,359.7 to 26.360.
7300  ln  5  7  2 
7300  ln 37
26,360
Page 12