1

Intermediate Algebra Diagnostic Test (IAD) Selected Solutions to the Sample Test Questions found in the CSUS Mathematics and Statistics Diagnostic Test Study Guide Note: à This Solution Manual is designed to help you understand how the answers to the IAD Sample Test in the above Study Guide were found. à This booklet is not designed to stand alone but to be used AFTER you have tried the questions in the Sample Test. à There is often more than one way to get to an answer and no attempt has been made to show all possible ways to a solution. Topic I: Elementary Operations Scientific Notation is a way of writing a number so that only one non-zero digit is to the left of the decimal point (that is, the one’s place), the rest of the digits are to the right of the decimal, and the number is times a power of 10. Scientific notation is useful for very large or very small numbers. For example, 232000 is written as 2.32 x 105 (the decimal moved to the left so the exponent, 5, is positive) 0.00026 is written as 2.6 x 10 – 4 (the decimal moved to the right so the exponent, 4, is negative) Remember when multiplying numbers to add the exponents and when dividing to subtract. The list of exponent rules is at the beginning of Topic III. 1. Express in Scientific Notation:

5.7 x 106( ) 2.3 x 10 −3( )

Separate the numbers from the powers of 10. Then multiply the numbers and then the powers of 10 (using the exponent rules).

5.7 x 2.3( ) x 10 6+ −3( )

This yields 13.11 x 103 which is not in scientific notation. Move the decimal one place to the left and add one to the exponent 3. The answer is: 1.311 x 104

2. Express in Scientific Notation:

3.81 x 10 −5

1.05 x 10 −7

Again, separate the numbers from the powers of 10.

3.811.05

⎛⎝⎜

⎞⎠⎟

x 10−5

10−7. Divide the

numbers and subtract the powers of 10 to get: 3.63 x 10 −5− −7( ) = 3.63 x 102 Notice we rounded the answer to two decimal places and that this answer is in scientific notation.

2

3. Simplify:

4x⋅x2y 3

24⋅

9y 4

To simplify fractions look for factors common to both the numerator and the denominator of the fractions.

4 ⋅ x ⋅ x ⋅ y ⋅ y ⋅ y ⋅ 3 ⋅3

x ⋅ 4 ⋅ 3 ⋅2 ⋅ y ⋅ y ⋅ y ⋅ y = 3x

2y

4. Add and simplify:

2 y − x( ) − 5⎡⎣ ⎤⎦ − 3 4 − x − y( )⎡⎣ ⎤⎦

Follow the order of operations carefully! Remember PEMDAS?

2 y − x( ) − 5⎡⎣ ⎤⎦ − 3 4 − x − y( )⎡⎣ ⎤⎦ =

2 y − x − 5⎡⎣ ⎤⎦ − 3 4 − x + y⎡⎣ ⎤⎦ =

2y − 2x −10 −12 + 3x − 3y =2y − 3y( ) + −2x + 3x( ) + −10 −12( ) =−y + x − 22 then rearrange to get x − y − 22

To evaluate expressions for specific values of the variables, substitute the given number for each variable in the expression. Hint: put any negative values in parentheses before substituting so that you don’t mess up the signs.

5. If x = ( – 1) and a = 2, then

x3( )−1⋅a4

a2x2=

Before you substitute here you can simplify the expression. Remember that

m−1 =

1m

so x−3 =1x3

. Now the fraction looks like this and can be simplified to

a4

x3a2x2=

a2

x5

Now substitute in the values for x and a.

22

−1( )5 = 4

−1 = − 4

3

6. If a = – 3, evaluate 1− a − 2 a + 2 +1

Absolute Value represents the distance from zero, so it is always positive.

1− −3( ) − 2 −3( ) + 2 +1 =

1+ 3 − 2 −1 + 1 =

4 − 2 1( ) + 1 = 3

Topic II: Rational Expressions To add or subtract rational expressions (fractions) you must have common denominators first. Then you can add and subtract the numerators.

7. Add and simplify:

xy 2 − x2yxy + y 2

+ 2x

First write 2x as 2x1

and factor the first expression:

xy y − x( )y x + y( ) +

2x1

Now simplify by canceling the y from numerator and denominator. Then make common denominators and add the numerators.

x y y − x( )y x + y( ) +

2x1

⋅x + y( )x + y( ) =

xy − x2 + 2x2 + 2xyx + y

=

3xy + x2

x + y

You could factor out an x from the numerator but it will not cancel with anything in the denominator (because of the + )

8. Add and simplify:

xy 3

y − x+

y 5 + xy 4

xy + y

xy 3

y − x+

y 5 + xy 4

xy + y=

xy 3

y − x+

y 4 y + x( )y x +1( ) =

4

After the y cancels, the common denominator is y − x( ) x +1( )

xy 3

y − x( ) ⋅x +1( )x +1( ) +

y 3 y + x( )x +1( ) ⋅

y − x( )y − x( ) =

x2y 3 + xy 3 + y 5 − x2y 3

y − x( ) x +1( ) =

y 5 + xy 3

y − x( ) x +1( )

9. Simplify:

y − x2 + 2x2

⋅ 4x2

x2 − y 2

Since we are multiplying (and not adding) rational expressions, we do not need common denominators!

y − x2 + 2x2

⋅ 4x2

x2 − y 2=

y − x( )4x2

2 1+ x2( ) x + y( ) x − y( ) Factor everything you can first.

Notice the two factors: y − x( ) in the numerator and

x − y( ) in the denominator.

These are called opposites (the same way that 2 and – 2 are opposites. You can factor a (– 1) from

y − x( ) . Then it becomes

y − x( ) = −1( ) x − y( ) . Now we can cancel…

−1( ) x − y( ) 4 2x2

2 1+ x2( ) x + y( ) x − y( )=

−2x2

1+ x2( ) x + y( )

10. Simplify:

1x + 2

− x + 3x2 + 3x + 2

+ 3x −1x2 −1

First factor the denominators to find the LCD (Least Common Denominator).

1x + 2

− x + 3x + 2( ) x +1( ) + 3x −1

x +1( ) x −1( ) The LCD is x + 2( ) x +1( ) x −1( ) (ie. the most

of any factor in one fraction). Now make equivalent fractions.

5

1x + 2

⋅ x +1( ) x −1( )x +1( ) x −1( ) − x + 3

x + 2( ) x +1( ) ⋅ x −1( )x −1( ) + 3x −1

x +1( ) x −1( ) ⋅ x + 2( )x + 2( ) =

x2 −1− x2 + 2x − 3( ) + 3x2 + 5x − 2

x + 2( ) x +1( ) x −1( ) =

x2 −1− x2 − 2x + 3 + 3x2 + 5x − 2x + 2( ) x +1( ) x −1( ) =

3x2 + 3xx + 2( ) x +1( ) x −1( ) =

3x x +1( )x + 2( ) x +1( ) x −1( ) =

3xx + 2( ) x −1( ) or 3x

x2 + x − 2

11. Simplify:

3y+ 1

4⎛⎝⎜

⎞⎠⎟

y − 14y

⎛⎝⎜

⎞⎠⎟

First separate the numerator and denominator and find an LCD for each.

The common denominator for

3y+

14

is 4y so,

3y

⋅ 44

+ 14

⋅ yy

= 12 + y4y

.

The common denominator for y −

14y

is 4y so,

y1

⋅ 4y4y

− 1

4y =

4y 2 −14y

which factors to 2y +1( ) 2y −1( )

4y. Now

3y+ 1

4⎛⎝⎜

⎞⎠⎟

y − 14y

⎛⎝⎜

⎞⎠⎟

is now

12 + y4y

2y +1( ) 2y −1( )4y

which can be multiplied by

4y4y

to yield

12 + y4y

2y +1( ) 2y −1( )4y

⋅

4y1

4y1

=12 + y

2y +1( ) 2y −1( ) or 12 + y4y 2 −1

6

12. Simplify:

−1x +1

+ x2

1x +1

− 3x

Separate the numerator and denominator and find the LCD of each. The numerator’s LCD is 2(x + 1) and the denominator’s LCD is x(x = 1). Now make equivalent

fractions and combine to get

−1x +1

22

⎛⎝⎜

⎞⎠⎟+ x

2 x +1

x +1⎛⎝⎜

⎞⎠⎟

1x +1

xx

⎛⎝⎜

⎞⎠⎟

− 3x

x +1x +1

⎛⎝⎜

⎞⎠⎟

=

x2 + x − 22 x +1( )−2x − 3x x +1( )

Writing this horizontally and dividing fractions yields

x + 2( ) x −1( )2 x +1( ) ÷

−2x − 3x x +1( ) becomes

x + 2( ) x −1( )2 x +1( ) ⋅

x x +1( )−2x − 3

. This simplifies to

x + 2( ) x −1( )2 x +1( )

⋅ x x +1( )− 2x + 3( ) =

−x x + 2( ) x −1( )2 2x + 3( ) .

Topic III: Exponents and Radicals Recall some properties of exponents:

x−n =1xn

example: 2−3 =123

ab

⎛⎝⎜

⎞⎠⎟

n

=an

bn

23

⎛⎝⎜

⎞⎠⎟

4

=24

34

ab( )n = anbn 3x( )2= 32 x2

am( )x= amx 23( )5 = 215

aman = a m+n( ) x3 ⋅ x4 = x7

am

an= am−n

x5

x2= x3

xn = x1n 53 = 5

13

13. Evaluate: 8−2

3 8−2

3 = 23( )−23 = 2

31 ⋅ −2

3( ) = 2−2 =122

=14

14. Simplify: 10 ⋅ 20 10 ⋅ 20 = 10 ⋅10 ⋅2 = 10 2

7

15. Simplify: 16x2 + 24y 2 + 36z2

16x2 + 24y 2 + 36z2 = 4 4x2 + 6y 2 + 9z2( ) = 2 4x2 + 6y 2 + 9z2( )

16. Simplify:

3ab12⎛

⎝⎜⎞⎠⎟

4a2b( )−2

ab3

2

3 ⋅ a ⋅b12 ⋅4−2 ⋅a−4 ⋅b−2

a ⋅b3

2=

3 ⋅b12+ −2( )

42 ⋅a4 ⋅b3

2=

3

16a4b3

2 ⋅b3

2=

316a4b3

17. Simplify:

x−3y 2( )2x−6y 2( )−2

x−6y 4x12y −4 = x −6+12( )y 4+− 4( ) = x6y 0 = x6

18. Simplify for a > 0: 8a3 + a 32a + 16a33 22 ⋅2 ⋅a2 ⋅a + a 42 ⋅2 ⋅a + 23 ⋅2 ⋅a33 = 2a 2a + 4a 2a + 2a 23 = 6a 2a + 2a 23

19. Simplify:

43 x3

2y4

3

16x−1

2y 2

⎛

⎝⎜⎜

⎞

⎠⎟⎟

−12

42 ⋅4 ⋅ x3

2 ⋅ x12

42 ⋅ y 2 ⋅ y−4

3

⎛

⎝⎜⎜

⎞

⎠⎟⎟

−12

=4 ⋅ x

32+ 1

2( )

y2

1 +− 43

⎛⎝⎜

⎞⎠⎟

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟

−12

=4

−12 x2( )−1

2

y2

3⎛⎝

⎞⎠

−12

=y

13

2x

20. Simplify: a3x+2

a4− x

a3x+2

a4− x= a3x+2− 4− x( ) = a3x+2−4+ x = a4x−2

21. Simplify:

z2x−4y 3

y −2zx−2

z2x2y 3y 2

x4z=

z ⋅ z ⋅ x2 ⋅ y 5

x2 ⋅ x2 ⋅ z=

y 5zx2

22. Express without radical signs: x23 ⋅ x5 x2

3 ⋅ x15 = x

23 + 1

5( ) = x10

15+3

15( ) = x13

15

23. Express without radical signs: z73

z7

3 = z7

3⎛⎝

⎞⎠

12= z

76

24. Solve for y: 2x =

5 − 3y4

Multiply both sides by 4 to clear the fraction.

8x = 5 − 3y 8x − 5 = −3y y =

8x − 5− 3

y =5 − 8x

3

25. Simplify: 10x

2x

10x

2x=

102

⎛⎝⎜

⎞⎠⎟

x

= 5x

8

26. Rationalize the denominator:

23 − 7

Multiply the numerator and

denominator by the conjugate:

3 + 7( )

23 − 7

⋅ 3 + 73 + 7

=2 3 + 7( )

3 − 7=

2 3 + 7( )−2 ⋅ 2

=− 3 − 7

2

Topic IV: Linear Equations and Inequalities There are 2 methods of solving a system of linear equations (ie. 2 equations with 2 variables). Either method can be used, so chooses the one that seems easier for the problem. You will get the same answer with either method. Substitution: solve one equation for one of the variables and substitute that expression into the other equation. It is a good idea to put parentheses ( ) around the expression being substituted. This method is best when one equation can be solved for one variable without getting fractions. Elimination: multiply each equation by a number so that when you add the equations together, one variable is eliminated. This method is best when solving for any variable creates fractions. REMEMBER: You must solve for both variables. So after you find one answer, solve for the other (by either method).

27. Solve:

12

x + 3y = −5

5x − 2y = 14

⎧⎨⎪

⎩⎪ First, multiply the first equation by 2 to clear the

fractions. Now the problem looks like this:

x + 6y = −105x − 2y = 14

⎧⎨⎩

You can now use either

method to solve it. Using the Elimination method, multiply the first equation by ( – 5).

−5x − 30y = 50 5x − 2y = 14

−32y = 64 y = − 2

Now we can use substitution to find x.

x + 6 −2( ) = −10

x −12 = −10x = 2

So, the answer is x = 2, y = – 2 or the point: 2,− 2( )

28. Solve:

4x = 5y2x + 3y = 22

⎧⎨⎩

First, rewrite so it looks like

4x − 5y = 02x + 3y = 22

⎧⎨⎩

Now, use Elimination again. Multiply the second equation by – 2.

9

4x − 5y = 0−4x − 6y = −44

−11y = −44 y = 4

Substitute y = 4 into the first equation to get…

4x = 5 4( ) x = 5 So the solution is x = 5, y = 4 or the point: 5,4( )

For #29, remember that when solving problems with square roots, be careful to square both sides of the equation and always plug your answers back into the original equation to check for extraneous solutions. Some answers will not work!! 29. Solve for x: 2 − 1− x = −3 You could square both sides now but it is messier. It’s better to add – 2 to both sides first.

2 − 1− x = −3

− 1− x = −5

1− x = 51− x = 25x = −24

Then check the answer.

30. Solve for m:

17 −19m > −4 m + 5( ) Remember that when multiplying or dividing by a negative number, the inequality sign changes direction. (Adding or subtraction negative does not change the direction of the inequality sign)

17 −19m > −4m − 20−15m > −37

m <3715

31. Solve for x:

3 5 − 4x( ) − 3 − 7x( ) = − 2 + 9x( )

15 −12x − 3 + 7x = −2 − 9x12 − 5x = −2 − 9x4x = −14

x =−144

x =−72

Topic V: Quadratic Polynomials, Equations, Inequalities 32. Multiply:

x2 − 5x( ) 4x2 − 3x + 2( ) You can use the distributive property or rectangle multiplication.

10

4x2 – 3 x +2 x2 – 5x Now combine like terms to get: 4x4 − 23x3 +17x2 −10x 33. Factor completely: 4a2 − 9a2b4 First look for factors in common for each term. Here it is a

2 .

a2 4 − 9b4( ) This is a difference of squares and yields: a2 2 − 3b2( ) 2 + 3b2( )

34. Solve for x: x2 +

52

x =32

First multiply both sides by 2 to clear the fractions.

2x2 + 5x − 3 = 02x −1( ) x + 3( ) = 0

2x −1= 0 or x + 3 = 0

x =12

or x = −3

35. Solve for x: x4 + x2 − 2 = 0 Let m = x2 . Then the question becomes how to factor m2 + m − 2 = 0 . This is much easier.

m + 2( ) m −1( ) = 0 which gives us m = – 2 or m = 1.

Now substitute x2 for m and solve for x.

x2 = −2 or x2 = 1

x = ± i 2 or x = ±1

36. If you solve this equation by completing the square, what do you add to each side of the equation? x2 + 3x = 4 To complete the square means to take the x2-term and the x-term and form a perfect square by adding something. This “something” can be calculated by dividing the x-term coefficient by 2 and squaring the result.

The coefficient of the x-term is 3. So, divide 3 by 2 and square it.

32

⎛⎝⎜

⎞⎠⎟

2

=94

So the answer to this question is to add 94

to complete the square. This is how it

would solve from there.

4x4 – 3x3 +2x2

– 20x3 + 15x2 – 10x

11

x2 + 3x +94= 4 +

94

x +32

⎛⎝⎜

⎞⎠⎟

2

=254

x +32= ±

52

x =−3 2

+52

or x =−3 2

−52

x = 1 or x = − 4

37. Solve for x: 2x2 + 4x +1= 0 Solving this quadratic equation requires the Quadratic Formula. Any equation in the

form ax2 + bx + c = 0 can be solved by this formula: x =

−b ± b2 − 4ac2a

.

In this problem a = 2, b = 4, and c = 1.

x =− 4 ± 42 − 4 2( ) 1( )

2 2( )x =

− 4 ± 16 − 84

x =− 4 ± 8

4=

x =− 4 ± 2 2

4

x =2 − 2 ± 2( )

4

x =− 2 ± 2

2

38. Omitted 39. Omitted

40. Simplify:

i2 − 5i

Recall that i = −1 . This gives us i2 = −1, i3 = − i, and i 4 = 1. So any power of i can

be simplified to i, – 1, – i, or 1 by dividing the exponent by 4 and looking at the remainder. For example, i345 = i86 ⋅4 ⋅ i1 = 1⋅ i = i .

To simplify

i2 − 5i

we must multiply by the conjugate of the denominator (same ends,

different sign between). Here the conjugate is 2 + 5i.

i2 − 5i

⋅ 2 + 5i2 + 5i

= 2i + 5i 2

4 − 25i 2 but i 2 = – 1, so 2i + 5i 2

4 − 25i 2 = 2i − 54 + 25

= − 5 + 2i29

12

41. Solve for z: 3z2 − 4z + 2 = 0 Use the Quadratic Formula.

z =− −4( ) ± −42( ) − 4 3( ) 2( )

2 3( )z =

4 ± 16 − 246

z =4 ± − 8

6 which simplifies to

4 ± 2i 26

z =2 2 ± i 2( )

6 3 The answer is: z =

2 ± i 23

42. Solve for a: 3 −

11+ 2a

=1

a −1 Note that a cannot be 1 or

−

12

.

The LCD of the denominators is (1 + 2a)(a – 1) so multiply each term by that to clear the fractions.

3 1+ 2a( ) a −1( ) −1 a −1( ) = 1 1+ 2a( )3 2a2 − a −1( ) − a −1( ) = 1+ 2a

6a2 − 3a − 3 − a +1= 1+ 2a6a2 − 6a − 3 = 0

3 2a2 − 2a −1( )3

=03

2a2 − 2a −1= 0 Now use the Quadratic Formula to solve for a

a =− −2( ) ± −2( )2

− 4 2( ) −1( )2 2( )

a =2 ± 4 + 8

4 which simplifies to

2 ± 2 34

a =2 1± 3( )

4 So, a =

1± 32

43. For what values of m is

2m −1( ) m − 5( ) > 0?

First solve for m when 2m −1( ) m − 5( ) = 0 .

m =

12

, m = 5

Now find these numbers on a number line. These are called critical points. We need to see what happens on each side of these critical points. Choose test points 0, 1, and 6. ½ 5

0 1 6

13

Substitute the test numbers into the inequality to see whether the product is greater than 0 or less than 0.

2 ⋅0 −1( ) 0 − 5( ) = 5 which is greater than 0 5 > 0

2 ⋅1−1( ) 1− 5( ) = − 4 which is less than 0 − 4 < 0

2 ⋅6 −1( ) 6 − 5( ) = 11 which is greater than 0 11 > 0

So, every number greater than 5 and every number less than ½ will yield a product

that is greater than 0. Therefore, the answer is m <

12

or m > 5

Topic VI: Graphing and Geometry 44. Find the equation of the line which passes through the points (1, 3) and (7, – 2).

First we need to find the slope which is: m =

y2 − y1

x2 − x1

. So, in this case, m =

3 − −2( )1− 7

which simplifies to m =

−5 6

. Note: you can make either point x1, y1( ) , but be careful

not to mix them up. Now choose a point (either one will work) and substitute it and the slope into the formula: y = mx + b . Using the point (1, 3), we get

3 =

−5 6

⎛⎝⎜

⎞⎠⎟

1( ) + b 3 +56= b b =

236

So, the equation of the line is

y =

−5 6

x +236

. If you need the standard form of the line, multiply through by 6 and

move the x term to get: 5x + 6y = 23 45. Find the equation of the line with a slope of 3 which passes through the point (0, – 2). Using the information given and y = mx + b , we get

−2 = 3 0( ) + b b = −2 So the equation of the line is y = 3x − 2 . 46. Find the distance between points A and B on the graph. First find the coordinates of the points on the graph. A = ( – 3, 1) and B = (2, 2). Now use the distance formula to find the length of the line segment between the

points: d = x2 − x1( )2

+ y2 − y1( )2 In this case,

d = 2 − −3( )( )2

+ 2 −1( )2 d = 52 +12 d = 26

47. What is the midpoint between the points ( – 5, 3) and (7, – 2)? The midpoint is the average of the x’s and the average of the y’s.

x1 + x2

2,

y1 + y2

2⎛

⎝⎜⎞

⎠⎟ = −5 + 7

2,

3 + −2( )2

⎛

⎝⎜

⎞

⎠⎟ = 1, 1

2⎛⎝⎜

⎞⎠⎟

14

48. Find the quadratic function whose graph is shown in the figure. A quadratic function graphs as a parabola and has the form y = ax2 + bx + c . If the a is positive the graph opens upward. If the a is negative, the parabola opens downward. In this case the a is negative. When x = 0, then y = 2, so the point (0, 2) can be used to find the c. So, our function now looks like: y = −ax2 + bx + 2 . Using both points (2, 0) and ( – 2, 0), we get a system of equations:

0 = −a 22( ) + b 2( ) + 2

0 = −a −2( )2( ) + b −2( ) + 2

⎧⎨⎪

⎩⎪ or

0 = −4a + 2b + 20 = −4a − 2b + 2

⎧⎨⎩

Adding these two equations together, we get:

0 = −8a + 412= a Substituting

12

for a and solving for b we find that b = 0

So, our function is now y = −12

x2 + 2

49. Find the slope of the line, l, in the figure given. Locate two points on the line. The easiest to find are (0, – 3) and (2, 0). Using the

slope formula: m =

y2 − y1

x2 − x1

, we get m =

0 − −3( )2 − 0

=32

, which is the slope of the line, l.

50. Graph the line y = 4 This says for any x the y value is 4. This is a horizontal line, whose graph is in the answer key on page 27. 51. Graph the function: 3x – 4y – 5 = 0.

First solve the equation for y. This gives you y =

34

x −54

. Find the point

0, −5 4

⎛⎝⎜

⎞⎠⎟

on

the y – axis. Then move up 3 units and then 4 units to the right. Connect the two points to make the line. The graph is in the answer key on page 27. 52. Graph the equation: y = 2x2 −1 Note: this is NOT a straight line because the x is squared. Make a table with some positive and negative values for x and calculate the y values. Then plot the points you find. The graph is in the answer key on page 27.

x y = 2x2 −1 0 – 1 1 1 – 1 1 2 7 – 2 7

15

53. OMIT 54. Find x. This is a right triangle so you can use the Pythagorean Theorem to find x. The Pythagorean Theorem is a

2 + b2 = c2 when a and b are the legs of a right triangle and c is the hypotenuse (longest side). In this case, 72 + x2 = 102 . Solving, we get

49 + x2 = 100x2 = 51

x = 51

55 through 60. OMIT Topic VII: Logarithms and Functions 61. If

f x( ) = 3x2 − 4x + 5, then f u − 2( ) =

Substitute (u – 2) for each x in the function.

f u − 2( ) = 3 u − 2( )2− 4 u − 2( ) + 5

= 3 u2 − 4u + 4( ) − 4 u − 2( ) + 5

= 3u2 −12u +12 − 4u + 8 + 5 = 3u2 −16u + 25

62. If f h( ) = h +

3h+ 2 find f −3( )

Substitute (– 3) for each x in the function.

f −3( ) = −3 + 3−3

+ 2

= −3 −1+ 2 = −2

63. Write logb a = x in exponential notation.

This is the definition of a logarithm. logb a = x ⇔ a = bx 64. Solve for x: 43x = 16 First write both sides in the same base. Here you can just change the 16 to 42.

43x = 42 . This is only true when 3x = 2. That is, when x =

23

.

65. Solve for x: logx 125 = 3 First take this out of log notation and put it into exponential notation: 125 = x3 . 125 = 53 so 53 = x3 You can see that x must equal 5. 66. Evaluate: log64 16 First, set this equal to x then take it out of log notation.

log64 16 = x 16 = 64x 42 = 43( )x

2 = 3x so, x =23

16

67. Solve for x: log5 2x + 5( ) = 3 Take this out of log notation.

2x + 5( ) = 53

2x + 5 = 1252x = 120 x = 60

68. Solve for x:

log2 x + log2 x − 7( ) = 3

Remember the property of logs that says: loga + logb = log ab( )

So,

log2 x x − 7( )⎡⎣ ⎤⎦ = 3

x2 − 7x = 23

x2 − 7x − 8 = 0

x − 8( ) x +1( ) = 0 x = 8 or x = −1

But x = – 1 does not work since you cannot take the log of a negative number. so the answer is x = 8 is the only answer.

69. Simplify: log5 25 − log4

116

⎛⎝⎜

⎞⎠⎟− log7 7

We cannot combine these by the properties of logs because the bases are different. So we must evaluate each separately. Set each term to equal a different variable.

log5 25 = x log4

116

⎛⎝⎜

⎞⎠⎟= y log7 7 = z

25 = 5x 1

16= 4y 7 = 7z

52 = 5x 4−2 = 4y 71 = 7z

2 = x −2 = y 1= z

So, x – y – z = 2 – ( – 2 ) – 1

log5 25 − log4

116

⎛⎝⎜

⎞⎠⎟− log7 7 = 3

Topic VIII: Word Problems 70. You can exchange $2.00 for £1.7. How many pounds (£) can you get for $15?

Use ratios: 1.72

=x

15 Then cross multiply and solve for x.

2x = 1.7 15( ) x =

25.52

x = £12.75

71. The side of a square is doubled. How much does the area change? If the square has a length of x, then A = x2 . When we double x, it becomes 2x.

17

So, now the area is: A = 2x( )2

A = 4x2 So the area is four times larger. 72. If 73% of a number is 365, what is the number?

For percentages like this use this ratio:

partwhole

=percent

100 remembering that what

comes after the word “of” is the whole.

So, this problem becomes

365whole

=73

100 Then cross multiply.

73 ⋅whole = 36500 whole = 36500

73 The whole number is 500.

73. 10 is to 28 as 35 is to what number?

Use ratios again. 1028

=35x

10x = 28 ⋅35 x =28 ⋅35

10 x = 98

74. The product of two numbers is – 48. Their sum is 13. What are the numbers? Product is the answer to a multiplying problem. Sum is the answer to an adding problem. (Difference is the answer to a subtracting problem. Quotient is the answer to a dividing problem) Let x be one number and y be the other number. x ⋅ y = −48 and x + y = 13 Solve this system of equations by substitution. y = (13 – x )

x 13 − x( ) = −48

13x − x2 = −480 = x2 −13x − 480 = x −16( ) x + 3( )

If x = 16, then y = – 3, and if x = – 3, then y equals 16.

75. A rectangle has an area of 40 and a perimeter of 26. What are the dimensions of the rectangle? Let L be the length of the rectangle and W be the width. Area = LW and 2L + 2W = Perimeter 40 = LW and 2L + 2W = 26 Solve for W and use substitution to solve this system of equations.

W = 13 − L( ) So

40 = L 13 − L( )40 = 13L − L2

L2 −13L + 40 = 0L − 8( ) L − 5( ) = 0

L= 8 or L =5 If L = 8, then W = 5 and if L = 5, then W = 8 So the dimensions of the rectangle are 5 by 8. 76. Five times a positive number is equal to the difference of the square of that number and 36. What is the number?

18

Let n be the number. Then five times the number is 5n. The square of the number is n2 . The equation is then…

5n = n2 − 360 = n2 − 5n − 36

0 = n − 9( ) n + 4( )n = 9 or n = −4

But n is a positive number so the answer is only 9. I have tried to be clear and concise in these solutions. There is often more than one way to solve a problem, so you can get the right answer by a different method. Please contact me with comments or suggestions on ways to improve this solution manual. My email address is: morrisb@csus.edu Thank you! β