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RATIONAL EXPONENTS AND LOGARITHMS

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RATIONAL EXPONENTS AND
MORE WORD PROBLEMS
REVIEW
Last class we saw that
(2x5)(3x3) = 6x8
• Powers with a common base can be
easily multiplied and divided
• Powers themselves can be raised to
another power
• Powers with a negative exponent
are equivalent to the reciprocal base
and positive exponent.
• Exponential equations can be solved
if the powers contain the same base.
28 / 25 = 23 = 8
(4d6)2 = 16d12
3e-3 = 3
e3
42x = 83
(22)2x = (23)3
24x = 2 9
4x = 9
x = 9/4
RATIONAL EXPONENTS
A rational number is one that can be expressed
as a fraction a .
The numbers ПЂ and в€љ2 are
b
examples of irrational numbers.
Ex. Solve for x.
x пЂЅ 25
2
We know to solve this we
must “undo” the square
by taking the square root
of both sides.
But what exactly is the
square root?
In order for the square root to undo the
exponent 2, it must also be an exponent.
Notice that in the final answer
the �x’ term has an exponent of 1.
x
2
пЂЅ
25
x 1 пЂЅ п‚±5
So the
problem is
actually
solved like
this…
RATIONAL EXPONENTS
Ex. Solve for x.
x пЂЅ 25
2
пЂЁx пЂ©
2 ?
пЂЅ 25
We raise both sides to an exponent.
But what is this exponent?
?
We see that one the left-hand side
we have a power of a power so we
will multiply the exponents to get 1.
(2)(?) = 1
The
missing
exponent
is 1/2 !!
1
x пЂЅ п‚±5
пЂЁx пЂ©
2
1
1
2
пЂЅ 25 2
x пЂЅ п‚±5
So the square root is
actually an
exponent of 1/2
RATIONAL EXPONENTS
This means that an
exponent of 1/3 is the
same as the cubedroot. It is like asking
what cubed gives me
this base?
So what does an exponent of 1/3 mean?
Ex. Solve
for x
81/3 = x
пѓ¦
пѓ§83
пѓ§
пѓЁ
1
3
пѓ¶
пѓ· пЂЅ x3
пѓ·
пѓё
8пЂЅ x
3
We could cube both
sides.
The left-hand side is a
power of a power so we
multiply the exponents.
We know that 23 is 8 so
x = 2.
In fact all the rational exponents are like this.
Ex. a) 641/6
b) 811/4
c. 1251/3
a) 2 (26 = 64)
b) 3 (34 = 81)
c. 5 (53 = 125)
RATIONAL EXPONENTS
Remember that this all started by recognizing that the
square root was an exponent.
It follows, then, that all rational
exponents can be represented by
root signs.
Ex.
1
1
5
49 пЂЅ 49
3
5
1
64 пЂЅ 64 3
4
100 пЂЅ 100
In all of these examples the bases
are perfect squares. We can write
them as so.
5
7
2
пЂЁ пЂ©
пЂЅ 7
2
5
пЂЁ пЂ©
1
1
3
8 пЂЅ 8
2
2
3
4
10
2
пЂЁ
пЂЅ 10
2
пЂ©
1
4
4
RATIONAL EXPONENTS
5
7
2
пЂЁ пЂ©
пЂЅ 7
2
пЂЁ пЂ©
1
1
3
5
8 пЂЅ 8
2
2
4
3
10
2
пЂЁ
пЂЅ 10
2
пЂ©
1
4
In each case we have a power of a power and
can multiply the exponents.
5
7
2
пЂЁ пЂ©
пЂЅ 7
2
2
1
5
пЂЅ7
5
4
3
8
2
пЂЁ пЂ©
пЂЅ 8
2
1
2
3
пЂЅ 83
So now we can have rational
exponents where the
numerator is not 1.
10
2
пЂЁ
пЂЅ 10
2
пЂ©
1
1
4
пЂЅ 10
2
RATIONAL EXPONENTS
Evaluate the following with a calculator. Show one
intermediate step.
2
125
пЂЁ125 пЂ©
2
3
1
3
When we need to evaluate an expression
with a rational exponent, we can “rip
the exponent apart”.
The exponent can be
viewed as a square and
an exponent of 1/3
However, we are not responsible to know
1252. Is there another way?
пѓ¦
пѓ§ 125
пѓ§
пѓЁ
1
3
пѓ¶
пѓ·
пѓ·
пѓё
2
We know that
1251/3 is 5!
пѓ¦
пѓ§ 125
пѓ§
пѓЁ
1
3
2
пѓ¶
пѓ· пЂЅ 52
пѓ·
пѓё
Since the new
expression is a power
of a power, we
multiply the
exponents. It does not
matter what order
there are in. So let’s
switch them.
2
125
3
пЂЅ 25
RATIONAL EXPONENTS
Evaluate the following with a calculator. Show one
intermediate step.
3
пѓ¦ 9 пѓ¶2
пѓ§ пѓ·
пѓЁ4пѓё
1
пѓ¦ пѓ¦ 9 пѓ¶3 пѓ¶ 2
пѓ§пѓ§ пѓ· пѓ·
пѓ§пѓЁ 4 пѓё пѓ·
пѓЁ
пѓё
пѓ¦
пѓ¶
2
9
пѓ§пѓ¦ пѓ¶ пѓ·
пѓ§пѓ§ 4 пѓ· пѓ·
пѓ§пѓЁ пѓё пѓ·
пѓЁ
пѓё
1
3
When we need to evaluate an expression
with a rational exponent, we can “rip
the exponent apart”.
The exponent can be
viewed as a cube and an
exponent of 1/2
The exponent can be
viewed as an exponent of
ВЅ and a cube
OR…
And we know
the square root
of 9 and 4.
3
27
пѓ¦3пѓ¶
пѓ§ пѓ· пЂЅ
8
пѓЁ2пѓё
RATIONAL EXPONENTS
Try these:
1. Evaluate each without a calculator by showing an intermediate step.
a) 45/2
b) 81-3/4
a) пѓ¦ 1
пѓ§42
пѓ§
пѓЁ
c) 100-2.5
5
пѓ¶
пѓ· пЂЅ 2 5 пЂЅ 32
пѓ·
пѓё
c)
пЂ­5
100
2
пѓ¦
пЂЅ пѓ§ 100
пѓ§
пѓЁ
1
2
пѓ¶
пѓ·
пѓ·
пѓё
b) пѓ¦ 1
пѓ§ 81 4
пѓ§
пѓЁ
пѓ¶
пѓ·
пѓ·
пѓё
пЂ­5
пЂЅ
пЂ­3
пЂЅ3
пЂ­3
пЂЅ
1
27
пЂ­5
пЂЅ 10
1
10
5
пЂЅ
1
100000
RATIONAL EXPONENTS
Try these:
1. Evaluate each without a calculator by showing an intermediate step.
d)
3
d)
125
3
пЂЁ
3
пѓ¦
e) пѓ§
пѓ§
пѓЁ
пЂ­1
125
пЂ­1
пЂ©
пЂ­1
125
пЂЁ5 пЂ©
e) пѓ¦
пѓ§
пѓ§
пѓЁ
4 пѓ¶
пѓ·
49 пѓ·пѓё
4 пѓ¶
пѓ·
49 пѓ·пѓё
пЂ­1
пЂ­1
пЂ­1
f)
пЂЁ
8
f)
пЂЁ
8
3
3
пЂ©
пЂ­5
пЂ©
пЂЁ 8пЂ©
пЂ­5
3
1
пѓ¦2пѓ¶
пѓ§ пѓ·
пѓЁ7пѓё
5
7
2
2
1
пЂ­1
пЂ­5
2
пЂ­5
1
5
32
RATIONAL EXPONENTS
Try these:
2. Solve the exponential equation
пЂЁ 8пЂ©
xпЂ«4
пЂЅ
Notice that we do not know
the cubed-root of 32 or the
square root of 8.
1
3
32
But we can get COMMON BASES
пЂЁ2пЂ©
xпЂ«4
3
пѓ¦ 3
пѓ§ 2
пѓЁ
пЂЁ пЂ©
пѓ¦
пѓ§22
пѓ§
пѓЁ
3
1
2
пЂЅ
3
1
2
пѓ¶
пѓ·
пѓё
пѓ¶
пѓ·
пѓ·
пѓё
xпЂ«4
xпЂ«4
Now simplify each side, remembering
that a square root is an exponent of ВЅ
and the cubed-root is the exponent 1/3
5
1
Power of a power….
пѓ¦ 1 пѓ¶3
пЂЅпѓ§ 5пѓ·
пѓЁ2 пѓё
пЂЁ пЂ©
пЂЅ 2
пЂ­5
1
3
пѓ¦
пѓ§2
пѓ§
пѓЁ
3 x пЂ« 12
2
пЂ­5
пѓ¶
пѓ· пЂЅ пЂЁ2 пЂ© 3
пѓ·
пѓё
Since the bases are
equal so are the
3 x пЂ« 12
exponents.
2
пЂЅ
пЂ­5
3
RATIONAL EXPONENTS
3 x пЂ« 12
пЂЅ
пЂ­5
2
We can (finally) solve for x.
3
3 x пЂЅ пЂ­ 3 . 33 пЂ­ 12
пѓ¦пЂ­5пѓ¶
3 x пЂ« 12 пЂЅ 2 пѓ§
пѓ·
пѓЁ 3 пѓё
3x пЂЅ
пЂ­ 10
3 x пЂЅ пЂ­ 15 . 33
x пЂЅ пЂ­ 5 . 11
пЂ­ 12
3
2
4пЂ­ x
1
3. Solve for x.
9 3 пЂЅ 27
пЂЁ3 пЂ©
2
4пЂ­ x
3
9 пЂЅ 27
1
3
33 пЂЅ 3
2
3
пЂЅ
пЂ©
4пЂ­ x
6
12 пЂ­ 3 x
6
6
2*6
12 пЂ­ 3 x
2
6
3
12 пЂ­ 3 x
3
6
пЂЅ пЂЁ3
пЂЅ
6
пЂЅ 12 пЂ­ 3 x
3
4 пЂЅ 12 пЂ­ 3 x
пЂ­ 8 пЂЅ пЂ­3x
xпЂЅ
8
3
BACK TO SOME WORD PROBLEMS
We have seen that patterns with a common ratio can be
described with an exponential equation.
Ex. 120,60,30,15,7.5…
This can be written as
t n пЂЅ 120 ( 0 . 5 )
n пЂ­1
We have also seen that when a data table is
given we can use this general equation:
x
y пЂЅ a (r )
period
Where �a’ is the initial amount (at the
beginning) and period is when the
value of y increases by r times
MORE WORD PROBLEMS
x
y пЂЅ a (r )
period
Where �a’ is the initial amount (at the
beginning) and period is when the
value of y increases by r times
Ex 1. A certain population has been seen to
triple every 12 years. In 1950, there was 2500
individuals. How many was there in 2011?
The equation is:
x
y пЂЅ 2500 ( 3 ) 12
61
y пЂЅ 2500 ( 3 ) 12
y пЂЅ 2500 ( 3 )
We need to
solve for y
knowing that
x = 2011-1950
5 . 0833
y пЂЅ 2500 ( 266 . 2873 )
y пЂЅ 665718 . 28
In 2011 there was
665718
individuals.
MORE WORD PROBLEMS
x
y пЂЅ a (r )
period
Where �a’ is the initial amount (at the
beginning) and period is when the
value of y increases by r times
Ex 2. A certain population has been seen to triple
every 12 years. In 1950, there was 2500 individuals.
When will there be 67500 individuals?
The equation is:
We need to
solve for x
knowing that
y = 67500
x
y пЂЅ 2500 ( 3 ) 12
Isolate the
power!!
Common
bases
anyone?
x
67500 пЂЅ 2500 ( 3 ) 12
27 пЂЅ ( 3 )
x
x
12
3 пЂЅ ( 3 ) 12
3
3пЂЅ
In 36 years (1986)
there would be
67500 individuals.
x
12
x пЂЅ 36
MORE WORD PROBLEMS
Ex 3. A teacher’s salary increases by 6% every 15 years. In
2000, a starting teacher’s salary was $35000. What will it be in
2020?
We can start this by looking at a table of values.
x (years since 2000)
0
Starting salary (in thousands of 35
dollars)
15
37.1
30
45
39.33 41.69
I’m rich!
To calculate the salary
6% of 35
after 15 years, we can
2.1 + 35 = 37.1
= 0.06(35)
take 6% of the present
= 2.1
salary and add it to the
present salary.
Now take 6% of 37.1
and add it to 37.1
Now take 6% of 39.33
and add it to 39.33
MORE WORD PROBLEMS
Ex 3. A teacher’s salary increases by 6% every 15 years. In
2000, a starting teacher’s salary was $35000. What will it be in
2020?
We can start this by looking at a table of values.
x (years since 2000)
0
15
Starting salary (in thousands of 35
dollars)
37.1
1.06
Now we can find the equation.
So let’s find the common ratio.
30
45
39.33 41.69
1.06
1.06
6% of x = 0.06x
0.06x + x = 1.06x
Each term is
1.06 times as
big as the last
Of course! We got each
successive value by taking 6% of
the last value and adding it to
the last value.
MORE WORD PROBLEMS
Ex 3. A teacher’s salary increases by 6% every 15 years. In
2000, a starting teacher’s salary was $35000. What will it be in
2020?
We can start this by looking at a table of values.
x (years since 2000)
0
Starting salary (in thousands of 35
dollars)
When a value increases
or appreciates by a
certain rate, the
common ratio will be
1 + rate/100
15
37.1
30
45
39.33 41.69
When a value decreases
or depreciates by a
certain rate, the
common ratio will be
1 - rate/100
MORE WORD PROBLEMS
Ex 3. A teacher’s salary increases by 6% every 15 years. In
2000, a starting teacher’s salary was $35000. What will it be in
2020?
We can start this by looking at a table of values.
x (years since 2000)
0
15
Starting salary (in thousands of 35
dollars)
37.1
30
45
39.33 41.69
So the equation for this data is
x
20
y пЂЅ 35 (1 . 06 ) 15
y пЂЅ 35 (1 . 06 ) 15
Plug in 20 for x (years
since 2000) to solve
for y
y пЂЅ 37 . 83
In 2020 a starting
teacher will make
$37 830
LI
OR
FE
Another exponential situation involves radioactive material.
Certain isotopes of the elements that make up the world
around us are unstable. Each nucleus decays randomly but
overall there is a pattern.
The half life of a
We model this with HALF-LIFE
substance is the time for
half of the material to
decay into something else.
Ex. A certain radioactive
After two half-lives, Вј of
substance has a half of 8
the original amount
days. It initially contained
remains.
90 mg. When will there be
5.625 mg left?
All half-life questions have
a common ratio (and base)
of ВЅ.
In 32 days, there will be 5.625
mg left.
x
x
пѓ¦ 1 пѓ¶8
y пЂЅ 90 пѓ§ пѓ·
пѓЁ2пѓё
пѓ¦ 1 пѓ¶8
5 . 625 пЂЅ 90 пѓ§ пѓ·
пѓЁ2пѓё
x
пѓ¦ 1 пѓ¶8
пЂЅпѓ§ пѓ·
16 пѓЁ 2 пѓё
1
4
x
пѓ¦1пѓ¶
пѓ¦ 1 пѓ¶8
пѓ§ пѓ· пЂЅпѓ§ пѓ·
пѓЁ2пѓё
пѓЁ2пѓё
4пЂЅ
x
8
x пЂЅ 32
LI OR FE
Ex 2. A certain substance has a half-life of 65 minutes. When will there be
1
th of the original amount?
64
However, we can still solve
the problem. The equation
would be
x
Notice that we were not
given the original amount.
пѓ¦ 1 пѓ¶ 65
y пЂЅ aпѓ§ пѓ·
пѓЁ2пѓё
x
пѓ¦ 1 пѓ¶ 65
a пЂЅ aпѓ§ пѓ·
64
пѓЁ2пѓё
1
The y value is 1/64 times
�a’ or 1 ( a )
64
x
пѓ¦1пѓ¶
пЂЅпѓ§ пѓ·
64 пѓЁ 2 пѓё
1
6
65
The next step would be to
divide both sides by �a’.
x
пѓ¦1пѓ¶
пѓ¦ 1 пѓ¶ 65
пѓ§ пѓ· пЂЅпѓ§ пѓ·
пѓЁ2пѓё
пѓЁ2пѓё
6пЂЅ
x
65
x пЂЅ 390
INTEREST
INTEREST
Another type of exponential equation is the compound interest
equation.
This is different than simple
interest which simply gives
you a set amount each time.
With compound interest
you’re getting interest
on the interest.
Compound interest gives
you a set percentage of the
amount in your account.
This amount may include
some interest already
earned.
Ex. You invest $100 in an account
paying 6% interest a year
compounded quarterly. How much
will you have in 10 years?
You get 6%
in the year
but you get it
spread over
4 times
(quarterly).
Each time
you get an
interest
payment
you’re
getting
6%/4 =
1.5%
INTEREST
INTEREST
Ex. You invest $100 in an account
paying 6% interest a year
compounded quarterly. How much
will you have in 10 years?
So the equation would
look like this:
Simplifying
the exponent
gives
0 . 06 пѓ¶
пѓ¦
y пЂЅ 100 пѓ§ 1 пЂ«
пѓ·
4 пѓё
пѓЁ
Now we can plug
in x =10
x
0 . 06 пѓ¶ 14
пѓ¦
y пЂЅ 100 пѓ§ 1 пЂ«
пѓ·
4 пѓё
пѓЁ
4x
0 . 06 пѓ¶
пѓ¦
y пЂЅ 100 пѓ§ 1 пЂ«
пѓ·
4 пѓё
пѓЁ
The amount of
interest earned
each period
4 (10 )
y = $181.40
x is time
in years.
The period
is Вј of a
year
INTEREST
INTEREST
In general, the compound interest is
rпѓ¶
пѓ¦
A пЂЅ P пѓ§1 пЂ« пѓ·
nпѓё
пѓЁ
nt
Where
пѓ�A is the amount in the account at
time, t
пѓ�P is the principle (initial) amount
пѓ� r is the decimal value of the interest
rate
пѓ� n is how many times per year the
interest is compounded.
Look for terms like: daily (n =365),
semi-annually (n = 2), weekly (n = 52)
and monthly (n =12)
INTEREST
INTEREST
Ex 2. An bank account earns interest compounded
monthly. The investment doubles in 9.27 years. Calculate
the annual interest rate.
r пѓ¶
пѓ¦
A пЂЅ P пѓ§1 пЂ«
пѓ·
12 пѓё
пѓЁ
r пѓ¶
пѓ¦
2 P пЂЅ P пѓ§1 пЂ«
пѓ·
12 пѓё
пѓЁ
r пѓ¶
пѓ¦
2 пЂЅ пѓ§1 пЂ«
пѓ·
12 пѓё
пѓЁ
12 t
When the money
doubles there will be
2P in the account.
So A = 2P
12 ( 9 . 27 )
111 . 25
We are now
solving for the
base. We must
“undo” the
exponent.
Raise
both
sides to
the
1/111.25.
INTEREST
INTEREST
Ex 2. An bank account earns interest compounded
monthly. The investment doubles in 9.27 years. Calculate
the annual interest rate.
r пѓ¶
пѓ¦
2 пЂЅ пѓ§1 пЂ«
пѓ·
12 пѓё
пѓЁ
111 . 25
1
1
1
2 111 . 25
111 . 25 111 . 25
пѓ©пѓ¦
пѓ№
r пѓ¶
пЂЅ пѓЄпѓ§ 1 пЂ«
пѓ·
пѓє
12 пѓё
пѓЄпѓ« пѓЁ
пѓєпѓ»
1
2
111 . 25
пЂЅ 1пЂ«
r
12
2
111 . 25
пЂ­1 пЂЅ
r
12
1
пѓ¦
пѓ¶
111
.
25
12 пѓ§ 2
пЂ­ 1пѓ· пЂЅ r
пѓ§
пѓ·
пѓЁ
пѓё
r пЂЅ 0 . 075
r пЂЅ 7 .5 %
INTEREST
INTEREST
Ex 3. Which is better: 5% interest per year compounded
monthly or 5% per year compounded daily?
Let’s assume an initial investment of $100 and a term
of 10 years.
0 . 05 пѓ¶
пѓ¦
A пЂЅ 100 пѓ§ 1 пЂ«
пѓ·
12
пѓЁ
пѓё
A пЂЅ 164 . 70
12 (10 )
0 . 05 пѓ¶
пѓ¦
A пЂЅ 100 пѓ§ 1 пЂ«
пѓ·
365 пѓё
пѓЁ
365 (10 )
A пЂЅ 164 . 87
Because the interest is compounded more often (even
though each time it is a smaller percentage) the account
paying the daily compounded interest is better.
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