rational exponents and more word problems. when multiplying powers with the same base, we keep the...

Rational Exponents and More Word Problems

When multiplying powers with the same base, we keep the base and add the exponents.

When dividing powers with the same base, we keep the base and subtract the exponents.

The value of any power with exponent 0 is 1.

To change the sign of an exponent, change the base to its reciprocal.(To change the base to its reciprocal, change the sign of the exponent)

When we have a power of a power, we keep the base and multiply the exponents.

Exponent Laws Review

bccb aa

cc

a

b

b

a

10 a

cbc

b

aa

a

cbcb aaa

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

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

5x1

So the problem is actually solved like this…

b

a

The numbers π and are examples of irrational numbers.

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.

2

252 x

??2 25x

We raise both sides to some 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)(?) = 1The missing exponent must be 1/2 !

2

1

2

12 25x

RATIONAL EXPONENTS

Ex. Solve for x.

51 x

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

So what does an exponent of 1/3 mean?

Ex. Solve for x:81/3 = x We could cube both sides.

3

3

31

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.

In fact all the rational exponents are like this.

Ex. Evaluate: a) b) c)

a) 2 (26 = 64) b) 3 (34 = 81) c. 5 (53 = 125)

RATIONAL EXPONENTS

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?

6

1

64 4

1

81 3

1

125

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.

51

5 4949

31

3 6464

41

4 10001000

Let’s step it up a notch…in all of these examples the bases are also powers…We can write them as so:

51

25 25 7749

31

23 23 8864

41

34 34 10101000

RATIONAL EXPONENTS

2552

5 2 777

233

23 2 888

344

34 3 101010

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

RATIONAL EXPONENTS

51

25 25 7749

31

23 23 8864

41

34 34 10101000

In each case we now have a power of a power, so we can multiply the exponents.

3

2

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

31

2125 The exponent can be viewed as a 2 and a 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

31

125

We know that 1251/3 is 5! 25125So 32

RATIONAL EXPONENTS

25

52

Evaluate the following without a calculator.Show one intermediate step.

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

The exponent can be viewed as a 3 and a ½ (hmm…)OrThe exponent can be viewed as a ½ and a 3. (hmm.. this is very helpful, since we know the square roots of 25 and 4!)

RATIONAL EXPONENTS

2

3

4

25

2

13

4

25

3

21

4

25

8

125

2

53

Evaluate the following without a calculator.Show one intermediate step.

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

a) 45/2 b) 81−3/4 c) 10−2.5 d) 82/3

a) b) c) d)

32

2

4

5

5

21

27

1

3

81

3

3

41

100000

110

1

10

100

100

5

5

5

21

25

RATIONAL EXPONENTS

4

64

8

8

31

31

2

32

4

2

8

8

2

2

31

32

e) f) g)3 11251

49

4

3 58

5

1

5

125

125

1

13

3 1

2

7

7

2

49

4

1

1

32

12

1

2

8

8

5

5

53

3 5

RATIONAL EXPONENTS

e) f) g)

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 3

5

43

2

12

x

Now consolidate each exponent, remembering that a square root is an exponent of ½ and the cubed-root is the exponent 1/3

31

5

4

21

3

2

12

x

31

5

4

23

22

x

35

2

123

22

x

Since the bases are now equal, so are the exponents…then cross-multiply and solve 3

5

2

123

x

Remember: An exponential equation is one where the unknown (x) is in the exponent.

SOLVING EXPONENTIAL EQUATIONS (RATIONAL EXPONENTS)

521233 x

9

46

469

10369

x

x

x

Ex. Solve:

Try this one. Solve for x.

64

3 279x

6

312

3

2

33

33

279

6312

32

64

331

2

64

31

x

x

x

x

3

89

24

924

93612

312326

x

x

x

x

x

SOLVING EXPONENTIAL EQUATIONS (RATIONAL EXPONENTS)

Applications (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…

Using the exponential pattern formulawe get 1)5.0(120 n

nt

period

x

rAy )(0

We have also seen that for most applications involving time, we use a more general version:

Where:x measures time since the starty is the amount at time x,r is the common ratio,A0 is the original amount (ie, the

amount at time x = 0), andperiod is the amount by which the x

values increase.(Note, sometimes x is replaced by t to emphasize that it measures time.)

11 )( n

n rtt

Ex 1. A certain population has been seen to triple every 12 years. In 1950, there was 2500 individuals.a) What is the population in 2011?

12)3(2500x

y

We know:A0 = 2500 (initial population…)r = 3 (…which triples…)period = 12 (…every 12 years)

We want to find the population (y) in 2011(when x = 2011 – 1950 = 61years)

28.665718

)2873.266(2500

)3(2500

)3(25000833.5

1261

y

y

y

y

periodx

rAy )(0

In 2011, the population was 665 718

Applications (word problems)

12)3(2500x

y

12)3(250067500x

12)3(27x

123 )3(3x

123

x

36xIsolate the power!! Common bases anyone?

Ex 1. A certain population has been seen to triple every 12 years. In 1950, there was 2500 individuals.b) When was the population 67 500?

We know:A0 = 2500 (initial population…)r = 3 (…which triples…)period = 12 (…every 12 years)

We want to find x (number of years since 1950) when y = 67 500

Applications (word problems)

In 36 years, that is in 1986, the population was 67 500

x (years since 2000) 0 15 30 45

y (salary, in thousands of dollars) 35

Ex 2. Bob’s salary increases by 6% every 15 years. In 2000, his salary was $35 000. What will it be in 2020?

We could start with the equation, but sometimes starting with a table is easier:

37.1 39.326 41.686

To calculate the salary after 15 years, we can take 6% of the present salary and add it to the present salary…OrTake 106% of the present salary

6% of 35= 0.06(35)= 2.1

2.1 + 35 = 37.1

106% of 35= 1.06(35)= 37.1

106% of 37.1= 1.06(37.1)= 39.326

106% of 39.33= 1.06(39.326)= 41.686

Applications (% increase)

Let’s find the equation of this pattern. We’ll need the common ratio.Each term is 1.06 times as big as the last, CR = 1.06… but of course!

×1.06 ×1.06 ×1.06


x (years since 2000) 0 15 30 45

y (salary, in thousands of dollars) 35 37.1 39.326 41.686


So the equation for this data is:

Plug in 20 for x (notice we don’t plug in 2020) to solve for y.

15)06.1(35x

y

83.37

)06.1(35

)06.1(35

1520

15

y

y

yx

In 2020 Bob’s salary will be $37 830

periodx

rAy )(0


x (years since 2000) 0 15 30 45

y (salary, in thousands of dollars) 35 37.1 39.326 41.686


Shortcut for “% increase” (appreciation)and “% decrease” (depreciation) questions

When a value increases or appreciates by a certain percentage, the common ratio will be:

r = 1 + (percent increase in decimal form)

When a value decreases or depreciates by a certain rate, the common ratio will be:

r = 1 − (percent decrease in decimal form)

Notice that even when something depreciates in value, r is still positive.

if r > 1, the values of y will be increasingif 0 < r < 1, the values of y will be decreasingif r = 1, the values of y will not change(if r < 0, the values of y will alternate between + and −)

Ex 3. In a certain rural town, the population is decreasing by 25% every 6 years. If there are 2300 people in the town this year, how long will it take before there are only 5 people left?

Since this is a % decrease question, we can calculate r as follows:r = 1 – 0.25 = 0.75This means that every 6 years, only 75% of the previous population remains.

period = 6 yrsA0 = 2300y = 5x = ?

6

0

)75.0(2300

)(x

period

x

y

rAy

6

6

6

)75.0(00217.0

)75.0(2300

5

)75.0(23005

x

x

x

We can’t write these with common bases, so we can’t solve this yet. Stay tuned for logarithms which will help us determine that the answer is 127.9 years

Applications (% decrease)

Another exponential situation involves the half-life of radioactive materials. Certain chemicals are unstable, and decay at a predictable rate.We model this with HALF-LIFE

The half life of a substance is the time it takes for half of the original material to decay. After two half-lives for example, only ¼ of the original amount remains.

Applications (half-life)

All half-life questions have:r = ½, andperiod = half life

periodx

rAy 0

lifehalfx

Ay

2

10

period = half-life = 8 daysr = 1/2A0 = 90 mgy = 5.625 mgx = ?

Ex 4. A certain radioactive substance has a half of 8 days. It initially contained 90 mg. When will there be 5.625 mg left?

8

2

190625.5

x

8

2

10625.0

x

84

2

1

2

1x

328

4

x

xIn 32 days, there will be 5.625 mg left.


periodx

rAy 0

Ex 5. A certain substance has a half-life of 65 minutes. When will there be of the original amount? th

64

1

lifehalfx

Ay

2

10

65

00

2

1

64

x

AA

65

2

1

64

1x

656

2

1

2

1x

39065

6

x

x

period = half-life = 8 daysr = 1/2A0 = ?x = ?y = um

Notice that we were not given A0, the original amount, but y value is 1/64th of A0, we use y =

After 6 and a half hours, there will remain but 1/64th of the original amount.


periodx

rAy 0

640A

640A

Divide both sides by A0

Another application of exponential growth is compound interest.

This is different than simple interest which simply gives you a set amount period.

Compound interest gives you a set percentage of the amount in your account each compounding period. After the first period, this amount includes some interest already earned.With compound interest you’regetting interest on the interest.

Applications (compound interest)

Compounding periods can be:daily (1/365 of a year)weekly (1/52 of a year)quarterly (1/4 of a year)semi-annually (1/2 of a year)monthly (1/12 of a year)etc.

Careful!In these types of questions:

x measures the number of years,period is the fraction of a year per

compounding period

You get 6% (that is 0.06) in the year but you get it spread over 4 compounding periods (quarterly).Each compounding period you’re getting:6% ÷ 4 = 1.5%, (that is 0.015)Recall that this is a 1.5% increase, so r = 1 + 0.015 = 1.015

Ex 6. You invest $100 in an account paying 6% annual interest, compounded quarterly. How much will you have in 10 years?


period = 1/4r = 1 + 0.015 = 1.015A0 = 100x = 10y = ?

periodx

rAy 0

So the equation would look like this:

41015.1100x

y

Now we can plug in x =10

Applications (compound interest)Ex 6. You invest $100 in an account paying 6% annual interest, compounded quarterly. How much will you have in 10 years?

41

015.1100 xy

104015.1100y

40.181$y

14

015.1100 xy

xy 4015.1100

Here is a shortcut formula for compound interest:

nt

n

iPA

1

Where A is the $ amount in the account at time t (years)P is the principle (initial) $ amount (when t = 0)i is the decimal value of the annual interest raten is how many times per year the interest is compoundedt is the number of years

Look for terms like:daily (n = 365),weekly (n = 52)quarterly (n = 4)semi-annually (n = 2) monthly (n = 12)


Ex 7. An bank account earns interest compounded monthly. The investment doubles in 9.27 years. Calculate the annual interest rate.

nt

n

iPA

1

)27.9(12

1212

iPP

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

25.111

1212

i

Raise both sides to the

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

25.111

1

25.1111

25.111

25.1111

1212

i

1212 25.111

1 i

i

1212 25.111

1

%5.7

075.0

i

i


Ex 8. 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

1 12

05.01100

A

)10(365

2 365

05.01100

A

70.1641 A 87.1642 A

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.

nt

n

iPA

1


rational exponents and more word problems. when multiplying powers with the same base, we keep the...

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