PM12 - Logarithms Practice Exam - Answers - Mrs. Baldwin's Online Classroom

[Pages:12]Logarithms Practice Exam - ANSWERS

Answers

1. C 2. A 3. D 4. C 5. B 6. D NR 1. 15 7. B 8. C NR 2. 4.00 9. A

10. D 11. C 12. C NR 3. 100 13. B 14. C NR 4. 5.40 15. C 16. C 17. B 18. D

19. A 20. B 21. B 22. C 23. B NR 5. 39.8 24. C 25. B 26. C 27. D 28. C

29. D 30. B 31. B 32. B

Each multiple choice & numeric response is worth 1 mark (Total = 40 marks) Each written response is worth 6 marks (Total = 18 marks)

To determine your score on the exam, add your marks and divide by 58.

Principles of Math 12: Logarithms Practice Exam - ANSWERS

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1) Graphical Solution: The question tells you that b > 0, so you could graph

y = 2x

and

y

=

1 2

x

to see what each case would look like. These graphs are symmetrical with respect to the y ? axis (x = 0).

( ) Algebraic

solution:

1 b

x

=

b-1

x = b-x . Compared to the original of bx, it's reflected in the y-axis.

The answer is C

2) To graph y = log3 x , change of base is required since the calculator only accepts base 10 logs.

In your calculator, you would use

y1 =

log x log 3

and

y2

= x-6.

The answer is A.

3)

log 1 5

1 x

1

= log x

log

1 5

Change of Base

log1-logx =

log1-log5

Division Law

-logx =

-log5

Since log1=0

logx =

log5

Cancel out the negatives

= log5x

Change of Base in reverse

The answer is D.

4)

P = 1 - w-0.246t

0.83 = 1 - w-0.246(43)

0.83 = 1 - w-10.578 -0.17 = -w-10.578

0.17 = w-10.578

( ) ( ) 1

1

0.17 -10.578 = w-10.578 -10.578

w = 1.18

Plug in P = 0.83 and t = 43 Subtract 1 on both sides Cancel out the negatives

Isolate w by raising each side to the reciprocal exponent

The answer is C. (You can also solve this equation by graphing y1 = 0.83 and y2 = 1- w-0.246(43) , then find the x-value of the point of intersection.)

Principles of Math 12: Logarithms Practice Exam - ANSWERS

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5)

( ) ( ) logx y3z - logx yz2

y3z = log x yz2

y2 = log x z

The answer is B.

6)

7 = (3 + b)4

1

[7]4

=

1

(3

+

b)4

4

47 =3+b

b = 47 -3

The answer is D.

Raise to reciprocal exponents Convert fractional exponent to a radical

NR #1)

5 log2 x + 5 log2 y

( ) = 5 log2 x + log2 y

= 5log2xy = 5log2 8 = 5(3)

= 15

The answer is 15.

Factor out the 5 Multiplication Law We know xy=8 Evaluate log28 using change of base

7) Graph y1 = 23x and y2 = 5-x-1 in your calculator and find the x-value of the point of intersection.

Remember to keep your exponent in brackets! The answer is B.

Principles of Math 12: Logarithms Practice Exam - ANSWERS

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8) The initial amount A0 is 32. The final amount A is 8. The length of time t is 21 hours. The growth b is ?.

We want to solve for P. t

A = A0 (b)P

21

8

=

32

1 2

P

21

8 32

=

1 2

P

21

1 4

=

1 2

P

1 2

2

=

1 2

21

P

2 = 21 P

2P = 21

P = 10.5 The half life is 10.5 hours, which is 630 minutes. The answer is C

NR #2)

( ) a5x = logc ca 3x+8

a5x = (a logc c)3x+8 a5x = (a)3x+8

5x = 3x + 8

2x = 8

x=4 The answer is 4.00.

Power Law

l ogc c = 1 Common Base

9) The point (0, a) can be transformed to the point (a, 0) by drawing the inverse graph.

The answer is A.

Principles of Math 12: Logarithms Practice Exam - ANSWERS

4



t

10) Use the formula A = A0 (b)P

A = the future score S A0 = the initial score 50000 t = elapsed days d b = rate. Decreasing percentage; subtract this decimal from 1. (0.973) P = the percentage loss is per day, so the period is 1.

Plug these into the formula to get S = 50000 (0.973)d

The answer is D.

11)

( ) dB = 10 log 1012 ? I

( ) dB = log 1012 ? I

10

dB

10 10 = 1012 ? I

dB

10 10 I=

1012

dB -12

I = 10 10

dB -120

I = 10 10 The answer is C

dB

dB 120 dB - 120

Common denominator: - 12 = - =

10

10 10

10

12)

dB-120 I = 10 10

150-120 I = 10 10

30 I = 1010

I = 103

I = 1000 The answer is C.

NR #3) 1

logb b-100

( ) logb b100

100 logb b = 100 The answer is 100

Principles of Math 12: Logarithms Practice Exam - ANSWERS

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