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
1
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
2
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
3
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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