11 ALGEBRA

CHAPTER REVIEW TEST 2B

1. Solve 9x+2 = 240 + 9x.

A)	1	B)	1	C) 1	D)	3	E)	5
	4		2			2		2

			(	1	3	27	2
2. Simplify				3)				.
				( 9) 3
A)	1			B) –27				C) –9	D) 27	E) 9
	27

	3x +4y	= 793
3.	Given	,	find 5x – y.
	3x 4y = 665
	A) 625ñ5		B) 125	C) 625
	D) 3125		E) 3125ñ5

4. Given 253x+1 = a, find 1252x+1 in terms of a.

A) a B) a2 C) a5 D) 4a E) 5a

5. Simplify	22n+3 +( 2)2n+3				.
	2n	( 2	2n	)
	( 2)
A) –8	B) –22n		C) –2			D) 2	E) 22n

6. Evaluate	23 +27	+29	.
	22 +26	+28
A) 24	B) 23		C) 22	D) 2	E) 1

7. Evaluate	4+ 12 + 7+4 3.
A) ñ3	B) 2ñ3	C) 1 + ñ3
D) 3 + 2ñ3		E) 4 + 4ñ3

8. Evaluate	( 4)2	3 ( 3)3 +	25.
A) –10	B) –2	C) 10	D) 12	E) 14

Chapter Review Test 2B

109

9. Evaluate 2430.2 + 640.666... – 1250.333... – (–11)0.

A) –10 B) –6 C) 0 D) 10 E) 13

	10. Evaluate	1				2			.
			3	2
							2
	A) ñ2	B) ñ3			C)			3 2		D)	2 3	E) 0
							2				3

11. Evaluate		20 45 – 15		32 +12	75	.
	15		20 +10	27 – 6	50
A) 1	B) 2		C) ñ2		D) ñ3		E) ñ5

12. Evaluate	2+ 2	2 – 2.
A) 2	B) 3	C) ñ2	D) ñ3	E) ñ6

13.a = ñ5 + ñ3, b = ñ6 + ñ2 and c = ñ7 + 1 are given. Order a, b and c.

A) a > b > c B) b > a > c C) a > c > b D) b > c > a E) c > b > a

14.	Evaluate	6	.
	3 2+		2
	A) 3 2+ 2		B) 3 2	C) 3 3 8 4 2
	D) 12		E) 33 2 – 2 2+ 2

15. Evaluate 0.01 4 0.21 5 .

0.0001

A) 1018 B) 1019 C) 1020 D) 1021 E) 1022

16. Find the sum of the roots of the equation

x2 4x+4 = 3.

A) 0

B) 2

C) 3

D) 4

E) 5

110	Algebra 11

CHAPTER REVIEW TEST 2C

1. Find x if x	x x = 4 45 .
A) 47 64	B) 7 64	C) 10 47
D) 4		E) 64

2. Evaluate (	7 + 5)( 4 7 + 4 5)( 4 7 4 5).
A) 12	B) 2ñ7 – 2ñ5	C) 2ñ7 + 2ñ5
	D) 2	E) 1

3. Simplify (0.5a0.25 + a0.75)2 – a0.5(0.25 + a0.5).

A)	a6	B)	a5	C)	a4	a3	E)	a
					D)

4.Find the sum of the roots of the equation 32x+2 – 3x+3 – 3x + 3 = 0.

A)	5	B) –2	C)	3	D) –1	E)	1
	2			2			2

5. Solve 7y 492 – y = 343ñ7.
A)	1	B) 2	C)	1	D) ñ2	E) –2
	2			2

6.	Simplify			2a				1 a.
				a+1
						1 a
	A)	1 a				B) 1 a			C) a2 1
		1+ a2
		D) 1 a2						E)	1 1+ a2
									a

7. Given A =

5 +1

, find

5 1

in terms of A.

2 +1

2 1

A)

2

B)

3

C)

4

D)

5

E)

6

A

A

A

A

A

8. Evaluate			2	5		.
		2	5	6 15
A) 1 3			B)		1 3		C)		2 1
2					2		3
	D) 1		2			E)	3 1
	3						2

Chapter Review Test 2C

111

9. Find the smallest integer value of x which satisfies x x 1 141 .

A) 48

B) 49

C) 50

D) 51

E) 52

a = 9x +5

10. Given , find a in terms of b.

b = 3 3x

A) 3 – b	B) b2 – 3b	C) b2 + 4
D) b2 – 6b + 7		E) b2 – 6b + 14

11. Evaluate	6+ 6+	6+...	+ 6 6	6 ... .
A) 2	B) 3	C) 5	D) 6	E) 12

12. a = x 1 and b = x2 + x + 1 are given.

Find a3b1/2.
A) x – 1	B) (x 1) x3 1	C) x+1
D) (x3 1) x 1		E) x3 – 1

13. Evaluate	42+	42 ...	.
	3 16 3 16 ...
A) 3	B) 7	C) 4		D)	9	E) 5
	2				2

14. Order the numbers a =	3, b = 3 4 and c = 4 6.
A) a < b < c	B) c < b < a
C) c < a < b	D) b < c < a

E) b < a < c

15. Given 2x – y + 2y – x = 11, find 4x – y + 4y – x.
A) 119	B) 121	C) 123	D) 22	E) 44

16. Given	3 x 5		x 1, find the sum of all possible
values of x.
A) 3	B) 7		C) 12	D) 16	E) 20

112	Algebra 11

CHAPTER 3

A. EXPONENTS

Up to now your study of math has included the study of linear, quadratic and trigonometric functions. In this book we will look at two new types of function: exponential functions and logarithmic functions. As we shall see, these functions are inverses of each other.

Let us begin by recalling what you already know about exponents and exponential expressions. In the exponential expression ap, a is called the base and p is called the exponent. We read this expression as ‘the pth power of a’, or ‘p to the power a’. We can also read a2 as ‘a squared’ and a3 as ‘a cubed’.

1. Integer Exponents

An exponential expression is a short-hand way of writing a multiplication operation in which the factors are all equal. In other words, for a natural number n we can write

an = a a a … a where a appears as a factor n times.

Remember that the first power of any number is the number itself, and any non-zero number to the power zero has value 1:

a1 = a, a0 = 1 where a 0, and 00 is undefined.

We can also define the negative power of a non-zero base as

	a– n =	1	where a \ {0} and n .
		an

Finally, remember that in an expression such as 2a5, 5 is the exponent of a, not the exponent of 2a. Similarly, –24 and (–2)4 are different: –24 = –16 while (–2)4 = 16.

1 Evaluate the expressions.

EXAMPLE

a. 50

b. (–7)0

c. 41

d. 2–2

Solution We can use the properties given above.

1

1

a. 5

0

0

1

–2

= 1

b. (–7)

= 1

c. 4

= 4

d. 2

=

=

22

4

Using the properties above, we can easily derive the following additional properties for any integers m and n and any non-zero real numbers a and b:

1. am an = am+n

2.

am

= am – n

3. (am)n = am n

an

am

m

m

m

a

m

4.

(a b) = a

b

5.

(

)

=

.

b

bm

114	Algebra 11

EXAMPLE 2

Solution

Be careful!

76 74 724

96 92

93

(2x)3 2x3

a2 – b2 = (a – b) (a + b)

Simplify each expression, given that all variables represent positive real numbers. Leave each answer as a single number or fraction with positive exponents.

a. 76 74

b.

96

c. (2x)3

d.

x 2 y 2

93

x 1 y 1

6

a. 76 74 = 76+4 = 710

b.

93 = 96 3 = 93

c. (2x)3 = 23x3 = 8x3

9

1

1

y2 x2

x

2

y

2

x

2

2

x

2

2

2

x

2

x y

d.

=

y

=

y

=

y

x 1 y 1

y x

1 1

y x

x2 y2

x

y

x y

=

( y x )(y+ x)

x y

(x y)2

y x

=x+ y

xy

index

n a

radical radicand sign

Check Yourself 1

Simplify each expression, given that all variables are positive. Leave each answer as a single number or fraction with positive exponents.

	a.	33 34	b.	78	c.	3(x2y)2	d.	x–1 + y–1
				72				x–2 – y–2
	Answers
	a.	37	b.	76	c.	3x4y2	d.	xy
								y – x

2. Roots and Ratical Expressions

Let a and b be real numbers, and let n be an integer greater than or equal to 2. Then an nth root of a is a number which, when raised to the power n, is equal to a. In other words, b is an nth root of a if and only if bn = a.

For example, 4 is a third root of 64 because 43 = 64, and –2 is a third root of –8 because (–2)3 = – 8. Moreover, both 3 and –3 are second roots of 9 because 32 = (–3)2 = 9. Note that a second root is usually called a square root and a third root is called a cube root.

We write n a to mean the principal nth root of a. In this notation, a is the radicand, n is the index, and the symbol ñ is called the radical sign. We do not write the index when n = 2.

Exponential and Logarithmic Functions

115

The principal nth root n a identifies

a.the positive root of a when n is even and a is positive.

b.the unique root which has the same sign as a when n is odd.

When n is even and a is negative, the principal nth root of a is undefined in the set of real numbers.

By the definition above we can write 3 64 4,

3 –8 2,

9

3 and

5 32 2.

We can also write parts a and b of the definition above as follows:

n

n

a if n is odd.

For n 2,

n , and a ,

a

=

|a| if n is even.

Any expression in the form

n a is called a radical expression, or simply a radical. We can

write radical expressions in different ways:

3 64 3 43

4, 3

–8 3 (–2)3

2,

9 (–3)2

32 3

and 5 32 5 25 2.

3 Evaluate each radical expression.

EXAMPLE

a.

3 125

b.

3 64

c. ( 5)2

d.

4 16

Solution a.

3 125 3 53 5

b. 3 64

3 ( 4)3 4

c.

( 5)2 | 5| 5

d.Since the radicand (–16) is negative and the index is even, the radical 4 16 is undefined in the set of real numbers.

We can use the following laws to evaluate and simplify radical expressions:

				1.	n a b n a n b	2.			n a						n a				3.		n m a m n a.
															n b
										b
	4 Simplify the expressions.
EXAMPLE
							8
	a. 3		11		b.	3													c. 5 8a2 5 4a3
						27
Solution a.							b.			8						3	8		3		23		2
		3	11		3 2 11 6 11				3
										27						3	27				33		3
																			3
	c.	5	8a2 5		4a3 5 (8a2 )(4a3 ) 5 32a5								5 25 a5						5 (2a)5					2a

116	Algebra 11

Check Yourself 2

1.	Evaluate each expression.
	a.	36	b.	100	c.	3 729	d.	3	343		e.	5		1
	f.	6 64	g.	6 729	h.	25	i.	3		8	j.	3			343
						49								1000
										125
Answers
1.	a. 6		b. –10		c.	9	d. –7				e	–1
	f. 2		g.	undefined in	h.	5	i.	–	2		j.			7
						7			5			10

am an = am+n

am bm = (a b)m

3. Rational Exponets

					1		where n and n 2 indicates a root with index n:		1
	An exponent of the form								a	n	= n a	.
					n
	1						1
	For example, 32 =			3 and			53	= 3 5.
	1
	The expression a	n		is equivalent to the principal nth root of a, and so it satisfies the properties
								1

of principal roots. For example, (–4)2 is undefined in the set of real numbers because

1

(–4)2 4, which is not a real number.

Taking into account the definition of the nth root of a number we can also write

1

an b if and only if bn = a.

EXAMPLE 5

Solution

Evaluate each expression.

	1	1						1					1				1			(32)	1
a.	1002	b. 814					c.	646			d. ( 32)4				e. ( 32)5				f.		5
a.	1	ó100 = 10										b.	1
	1002 =												814 = 4 81= 3
c.	1											d.	1
	646 = 6 64 = 2												( 32)4 is undefined in
	1												1	1			1		1
e.	( 32)5 = 5 –32 = –2											f.	(32) 5 =		=			=
														1		5	32		2
														325
					1														m
Now that we have defined					a	n	, we can use the properties of exponents to define a n													for any
rational number			m	:
			n
							m	1		1			m
						a n =(am )			n	=(a	n	)m or a n = n am ( n a)m						.

Exponential and Logarithmic Functions

117

EXAMPLE 6

Solution

1

33 = 27 273 = 3

1

25 = 32 325 = 2

1

43 = 64 3 = 4

1

52 = 25 252 = 5

EXAMPLE 7

Solution

1

23 =8 83 =2

1

24 =16 164 =2

1

53 =125 1253 =5

EXAMPLE 8

Solution

EXAMPLE 9

Evaluate each expression.

2	3	c. 64	2	3
a. 273	b. 325		3	d. 252

21

a.273 =(273 )2 = 32 = 9

31

b.325 =(325 )3 = 23 = 8

c.	64	2	=	1		=	1			1	=	1
		3
					2		1			2		16
				64	3		(643 )2			4

31

d.252 =(252 )3 = 53 =125

Evaluate the expressions.

a.	4		1	8		2	+16		3				b.								1				1
			2			3			4						(0.0016)4 +(0.125) 3
a.			1	8		2		3		1		1			1					1		1			1					1			1			1					4 2+1									3
	4	2			3		+16		4 =					+				=						+				=						+					=									=
										1			2				3			2		1			1					2			2				3							8						8
										42		83			164							(83 )2			(164 )3								2			2
							1			1								1				1			1								1
b.														16				4			125 3				164							1253					2					5				7
	(0.0016)						4 +(0.125)3				=								+					=					+						=					+					=
													10000															1						1			10					10				10
																					1000				10000			4		1000				3

Write each expression as a quotient with positive integer exponents, given that all variables represent positive real numbers.

	a. (x		y )	3				b. (x			4				1	)(x			1	y			5	)
												y
		4	12	4							3				4				3				4
						y12 3			3						3			y9
				3				(y12 )4					y12 4
	a. (x 4 y12 )4				(		)4
						x4			3				x	4	3			x3
								(x4 )4							4
		4	1	1	5			4	1				1		5						4		1		1		5		y
	b. (x 3 y 4 )(x3 y4 ) x 3 x3 y												4 y4 x									3 y						x 1 y
																					3				4		4
																													x

Factorize each expression using the given common factor, given that x represents a positive real number.

5	1	1	3	21x	1	2
a. 3x4	x4 ; x4		b. 9x5		5 ; 3x–	5

118	Algebra 11