Mathematics Numbers: Exponents Science and Mathematics Education Research Group Supported by UBC...

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Mathematics Numbers: Exponents Science and Mathematics Education Research Group Supported by UBC Teaching and Learning Enhancement Fund 2012-2014 Department of Curriculum and Pedagogy FACULTY OF EDUCATION a place of mind

Transcript of Mathematics Numbers: Exponents Science and Mathematics Education Research Group Supported by UBC...

MathematicsNumbers: Exponents

Science and Mathematics Education Research Group

Supported by UBC Teaching and Learning Enhancement Fund 2012-2014

Department of Curriculum and Pedagogy

FACULTY OF EDUCATIONa place of mind

Properties of Exponents

Review of Exponent Laws

Let a and b be positive real numbers. Let x and y be real numbers.

10 a

yxyx aaa

yxy

x

aa

a

xyyx aa

xx

aa

1

x

x

x

b

a

b

a

xx

a

b

b

a

yxaa yx ifonly and if

Exponent Laws I

How can the following expression be simplified using exponents?

33333

B or A EitherE.D.

C.

B.

A.

33

3

23

5

23

)23(

33

3

Press for hint

yxyx aaa

Solution

Answer: B

Justification: Group the expression into two separate powers of 3:

This gives the final expression 33 + 32. This cannot be simplified any further unless the powers of three are calculated.

Answer A (35) is incorrect since 33 + 32 ≠ 35. Notice that we only add exponents when two powers with the same base are multiplied together, not added:

33333 33

yxyx aaa

23

yxyx aaa

3692733333

3582723D.

1255)23(C.

3692733B.

2433A.

33

33

23

5

Solution Continued

The values we are comparing in this question are small, so we can calculate the final values of each expression:

Only answer B matches the value of the expression in the question. The other answers give a different final value, so the expressions are not equivalent.

Exponent Laws II

Simplify the following expression:

2

25

7

77

simplified be CannotE.D.

C.

B.

A.

23

5

5

3

77

7

17

17

Press for hint

2

2

2

5

2

25

7

7

7

7

7

77

Solution

Answer: A

Justification: This expression can be simplified in several ways:

17

77

7

)77(7

7

77

3

03

2

032

2

25

17

77

77

7

7

7

7

7

77

3

03

2225

2

2

2

5

2

25

Factor out 72 from the numerator:

Split the fraction into the sum of two fractions:

Exponent Laws III

Simplify the following expression:1210

10

43

12

16

112

14

1

12

16

E.

D.

C.

B.A.

Press for hint

xxx baab

Solution

Answer: E

Justification: Write 1210 as a product of powers with base 3 and 4:

Alternatively, you can write the denominator as a power with base 12:

16

1

4

1

43

43

43

)43(

43

1210121210

1010

1210

10

1210

10

16

1

4

1

412

12

4)43(

12

443

12

43

122210

10

210

10

21010

10

1210

10

Exponent Laws IV

Simplify the following expression:

98

2

22

4

simplified be CannotE.

D.

C.

B.

A.

9

94

54

76

2

122

122

122

1

Solution

Answer: B

Justification: The power of 4 can be rewritten as a power with a base of 2:

Factor out a power of 2 from the denominator to cancel with the numerator:

98

4

98

22

98

22

98

2

22

2

22

2

22

)2(

22

4

54544

4

494484

4

98

4

22

1

)22(2

2

2222

2

22

2

Exponent Laws V

Write the following as a single power with base 4:

0170

17

444

4

4of power single a as written be CannotE.D.

C.

B.

A.

1

0

2

1

17

4

4

4

4

Solution

Answer: E

Justification: This expression cannot be simplified any further.

14

4

44

417

17

017

17

0017

17

017

17

44

4

44

4

017

17

017

17

404

4

44

4

1. Incorrectly adding exponents

2. 40 = 1, not 0

0

17

17

17

017

17

4

4

4

4

44

4

3. Splitting the denominator

Common errors include:

Exponent Laws VI

Write the following as a single power of 2:

545

25.05.0

2

2of power single a as written be CannotE.D.

C.

B.

A.

1

0

1

2

2

2

2

2

Solution

Answer: B

Justification: This expression can be simplified in many ways because all the terms can be expressed as a power of 2. Two possible solutions are shown below.

Cancel terms where possible and collect like terms:

Express all terms as a power of 2:

15455545554

5545

222222)2(2

1)22(25.0

5.0

2

1455

45

554

5

25.0

1

2

15.0

5.0

225.0

5.0

2

Exponent Laws VII

Which of the following powers is the largest?

34

5

2

3

4

5

6E.

5D.

4C.

3B.

2A.

4352 25 56

Solution

Answer: B

Justification: There are generally no rules when comparing powers with different bases.

7776

1

6

1

255

644

813

322

55

2

3

4

5

6

43525 34256

Note: Large exponents tend to have more impact on the size of a number than large bases.

2100 1002

Exponent Laws VIII

How many of the following terms are less than 0?

23

4E.

3D.

2C.

1B.

0A.

232)3( 23

Solution

Answer: C

Justification: Simplify each term separately.

Be careful when dealing with negatives on exponents. Although every expression has a negative sign, only –32 and –3-2 are negative. Think about order of operations: brackets come before exponents, which come before multiplication.

93

9

13

9

13

9)3(

2

2

2

2

99

10

9

19

)3(3033 2222

Exponent Laws IX

How many of the following are less than 1?

5E.

4D.

3C.

2B.

1A.

22222 )5.0(,)5.0(,)5.0(,2,)5.0(

Solution

Answer: D

Justification: Simplify each term separately to find the terms less than 1.

42)5.0(

4

1

)2(

1)5.0(

4)2()5.0(

4

1

2

12

4

1

2

1)5.0(

22

22

22

22

22

Greater than 1

Less than 1

Less than 1

Less than 1

Less than 1

Exponent Laws X

Consider adding 3 until you obtain the value 333 as shown below:

How many terms are there in the summation?

33

3

3

33

13

11

32

33

33

n

n

n

n

n

E.D.

C.

B.

A.

3333...333

n terms

Solution

Answer: C

Justification: The summation is equal to 3n.

It becomes straightforward to solve for n after this step.

33

33

33

33...333

nn terms

32

1

33

33

3

3

3

33

n

n

n

Exponent Laws XI

Let p and q be positive integers. If , which of the following are always true?

qp

qp

qp

qp

qp

22

22

22

22

22

E.

D.

C.

B.

A.

qp

Solution

Answer: D

Justification: Rewrite the two expressions using positive exponents of p and q:

It is now much easier to compare the two expressions.

Remember that dividing by a larger denominator gives a smaller result.

qq

pp

2

12,

2

12

qp 2

1

2

1 qp 22 since

Exponent Laws XII

Let p and q be positive integers and p > q. If b > 0, find all values of b such that

is always true.

1

1

10

10

1

b

b

b

b

b

E.D.C.B.A.

qp bb

Solution

Answer: C

Justification: Rewrite the inequality using positive exponents of p and q:

The LHS is larger than the RHS only if bp < bq , since numbers with smaller denominators are larger. Therefore b must be between 0 and 1 to make bp < bq (since p > q). For example, 0.52 < 0.53.

In order to choose between answers B and C, consider when b = 1.

Since we have to include the equality case, the answer is C:

qp bb

11

qp 1

1

1

1 since 1 to any power is still 1

10 b

Exponent Laws XIII

Solve for c.

xc

c

c

c

c

2E.

8D.

4C.

2B.2

1A.

122 22

xx c

Solution

Answer: B

Justification:

2

2

2)2(

22

22

1

1

)2()2(

)2()2(

1

1

c

c

cxx

xx

c

c

xx

xx

since xyyx aa

since if and only if yx aa yx

divide both sides by (then subtract exponents)

x2

Exponent Laws XIV

Simplify the following:

2

1

2

199

199

1

199

199

E.

D.

C.B.A.

2

22

)99100(

99100

Press for hint

))((22 bababa Difference of squares:

Solution

Answer: D

Justification: The numerator is a difference of squares:

OR

))((22 bababa

1122

199199

1

)99100(

)99100(

)99100(

)99100)(99100(

1199199

1

)99100(

)99100(

)99100)(99100(

)99100)(99100(