RATIONAL EXPONENTS AND MORE WORD PROBLEMS

REVIEW

Last class we saw that

• Powers with a common base can be easily multiplied and divided

(2x5)(3x3) = 6x8

28 / 25 = 23 = 8

• Powers themselves can be raised to another power

(4d6)2 = 16d12

• Powers with a negative exponent are equivalent to the reciprocal base and positive exponent.

3e-3 = 3 e3

• Exponential equations can be solved if the powers contain the same base.

42x = 83

(22)2x = (23)3

24x = 2 9

4x = 9x = 9/4

RATIONAL EXPONENTS

A rational number is one that can be expressed as a fraction a .

bThe numbers π and √2 are examples of irrational numbers.

Ex. Solve for x.

252 xWe know to solve this we must “undo” the square by taking the square root of both sides.

252 x

But what exactly is the square root?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.

5x1

So the problem is actually solved like this…

RATIONAL EXPONENTS Ex. Solve for x.

252 x

??2 25x

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.

5x1

(2)(?) = 1

The missing exponent is 1/2 !!

5

252

1

2

12

x

x So the square root is actually an exponent of 1/2

RATIONAL EXPONENTS

So what does an exponent of 1/3 mean?

Ex. Solve for x81/3 = x

We could cube both sides.

3

3

3

1

8 x

The left-hand side is a power of a power so we multiply the exponents.

38 x We know that 23 is 8 so x = 2.

This means that an exponent of 1/3 is the same as the cubed-root. It is like asking what cubed gives me this base?

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. 5

15 4949 3

13 6464 4

14 100100

In all of these examples the bases are perfect squares. We can write them as so.

5

125 2 77 3

123 2 88 4

124 2 1010

RATIONAL EXPONENTS

In each case we have a power of a power and can multiply the exponents. In each case we have a power of a power and can multiply the exponents.

5

125 2 77 3

123 2 88 4

124 2 1010

5

2

5

125 2 777

3

2

3

123 2 888

2

1

4

124 2 101010

So now we can have rational exponents where the numerator is not 1.

RATIONAL EXPONENTS

Evaluate the following with a calculator. Show one intermediate step.

3

2

125When we need to evaluate an expression with a rational exponent, we can “rip the exponent apart”.

3

12125

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?

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

3

1

125

We know that 1251/3 is 5!

2

2

3

1

5125

251253

2

RATIONAL EXPONENTS

Evaluate the following with a calculator. Show one intermediate step.

2

3

4

9

When we need to evaluate an

expression with a rational exponent, we can “rip the exponent apart”.

2

13

4

9

The exponent can be

viewed as a cube and an exponent of 1/2

OR…

The exponent can be viewed as an exponent of ½ and a cube

3

2

1

4

9

And we know the square root of 9 and 4.3

2

3

8

27

RATIONAL EXPONENTS

Try these:1. Evaluate each without a calculator by showing an intermediate step.

a) 45/2 b) 81-3/4 c) 100-2.5

a) b) 3224 5

5

2

1

27

1381 3

3

4

1

100000

1

10

110100100

55

5

2

1

2

5

c)

RATIONAL EXPONENTS

Try these:1. Evaluate each without a calculator by showing an intermediate step.

d) e) f)3 1125

1

49

4

3 58

d)

5

1

5

125

125

1

13

3 1

2

7

7

2

49

4

1

1

e) f)

32

12

1

2

8

8

5

5

53

3 5

RATIONAL EXPONENTS

Try these:2. Solve the exponential equation

34

32

18

x Notice that we do not know the cubed-root of 32 or the square root of 8.

But we can get COMMON BASES

35

43

2

12

xNow simplify each side, remembering that a square root is an exponent of ½ and the cubed-root is the exponent 1/3

3

1

5

4

2

13

2

12

x

Power of a power….

3

15

4

2

3

22

x

3

52

123

22

xSince the bases are equal so are the exponents.

3

5

2

123

x

RATIONAL EXPONENTS

3

5

2

123

x We can (finally) solve for x.

3

52123x

123

103

x

33.153

1233.33

x

x

11.5x

3. Solve for x.

6

43 279

x

6

312

3

2

33

33

279

6

312

3

2

6

433

12

6

4

3

1

x

x

x

x

3

8

38

3124

3123

6*26

312

3

2

x

x

x

x

x

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

1)5.0(120 nnt

We have also seen that when a data table is given we can use this general equation:

period

x

ray )(

Where ‘a’ is the initial amount (at the beginning) and period is when the value of y increases by r times

MORE WORD PROBLEMS

period

x

ray )( 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:

12)3(2500x

y We need to solve for y knowing that x = 2011-1950

28.665718

)2873.266(2500

)3(2500

)3(25000833.5

12

61

y

y

y

y

In 2011 there was 665718 individuals.

MORE WORD PROBLEMS

period

x

ray )( 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:

12)3(2500x

y

We need to solve for x knowing that y = 67500

12)3(250067500x

In 36 years (1986) there would be 67500 individuals.

Isolate the power!!

12)3(27x

Common bases anyone?

123 )3(3x

123

x 36x

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 30 45

Starting salary (in thousands of dollars)

35

To calculate the salary after 15 years, we can take 6% of the present salary and add it to the present salary.

6% of 35= 0.06(35)= 2.1

2.1 + 35 = 37.1

37.1

Now take 6% of 37.1 and add it to 37.1

39.33

Now take 6% of 39.33 and add it to 39.33

41.69

I’m rich!

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 30 45

Starting salary (in thousands of dollars)

35

37.1

39.33

41.69

Now we can find the equation. So let’s find the common ratio.

1.06 1.06 1.06Of course! We got each

successive value by taking 6% of the last value and adding it to the last value.

6% of x = 0.06x0.06x + x = 1.06xEach term is 1.06 times as big as the last

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 30 45

Starting salary (in thousands of dollars)

35

37.1

39.33

41.69

When a value increases or appreciates by a certain rate, the common ratio will be 1 + rate/100

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 30 45

Starting salary (in thousands of dollars)

35

37.1

39.33

41.69

So the equation for this data is

15)06.1(35x

y

Plug in 20 for x (years since 2000) to solve for y

83.37

)06.1(35 15

20

y

y

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.

We model this with HALF-LIFEWe model this with HALF-LIFE

Ex. A certain radioactive substance has a half of 8 days. It initially contained 90 mg. When will there be 5.625 mg left?

The half life of a substance is the time for half of the material to decay into something else. After two half-lives, ¼ of the original amount remains.

All half-life questions have a common ratio (and base) of ½.

8

2

190

x

y

8

2

190625.5

x

8

2

1

16

1x

84

2

1

2

1x

328

4

x

x

In 32 days, there will be 5.625 mg left.

LI OR FE

Ex 2. A certain substance has a half-life of 65 minutes. When will there be

of the original amount? th

64

1

Notice that we were not given the original amount.

Notice that we were not given the original amount.

However, we can still solve the problem. The equation would be

65

2

1x

ay

The next step would be to divide both sides by ‘a’.

65

2

1

64

1x

aa

The y value is 1/64

times ‘a’ or )(64

1a

65

2

1

64

1x

656

2

1

2

1x

39065

6

x

x

INTERESTINTEREST

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.

Compound interest gives you a set percentage of the amount in your account. This amount may include some interest already earned.

With compound interest you’re getting interest on the interest.

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%

INTERESTINTEREST

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: 4

1

4

06.01100

x

y

The amount of interest earned each period

x is time in years.

The period is ¼ of a yearx

y4

4

06.01100

Simplifying the exponent gives

Now we can plug in x =10

)10(4

4

06.01100

y y =

$181.40

INTERESTINTEREST

In general, the compound interest is nt

n

rPA

1

Where A is the amount in the account at time, tP is the principle (initial) amount r is the decimal value of the interest rate 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)

INTERESTINTEREST

Ex 2. An bank account earns interest compounded monthly. The investment doubles in 9.27 years. Calculate the annual interest rate.

tr

PA12

121

)27.9(12

1212

rPP

When the money doubles there will be 2P in the account.So A = 2P

25.111

1212

rWe are now solving for the base. We must “undo” the exponent.

Raise both sides to the 1/111.25.

INTERESTINTEREST

Ex 2. An bank account earns interest compounded monthly. The investment doubles in 9.27 years. Calculate the annual interest rate.

25.111

1212

r

25.111

125.111

25.111

1

1212

r

1212 25.111

1 r

1212 25.111

1 r

r

1212 25.111

1

%5.7

075.0

r

r

INTERESTINTEREST

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.

)10(12

12

05.01100

A

)10(365

365

05.01100

A

70.164A 87.164A

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.