COLLEGE ALGEBRA II (MATH 010)
SPRING 2012
PRACTICE FOR EXAM II
SOLUTIONS
1. Expand the following logarithms.
x2 −x−2
(x+4)2
1
3
, x > 2 (b) ln(xex )
(a) ln
Solution:
1
x2 −x−2
= 13 ln [(x − 2)(x + 1)] − 13 ln(x + 4)2
(a) 3 ln (x+4)2
= 13 ln(x − 2) + 13 ln(x + 1) − 23 ln(x + 4)
(b) ln x + ln ex =ln x + x ln e
= ln x + x
2. Write each as a single logarithm.
x
2
)+ln( x+1
(a) 3 log5 u+4 log5 v (b) ln( x−1
x )−ln(x −1)
Solution:
(a) logh5 u3 v 4
i
x
x+1
(b) ln ( x−1
)( x+1
)
− ln(x2 − 1) = ln "x−1
− ln [(x
− 1)(x + 1)]
x
#
x+1
( x−1
)
(x−1)(x+1)
1
ln (x−1)2
= ln
=
= −2 ln(x − 1)
3. Evaluate using the change-of-base formula.
(a) log3 21 (b) log 31 71
Solution:
21
71
71
(a) log
(b) log
= − log
log 3
log 3
log 1
3
1
4. Solve each exponential equation.
(a) ( 35 )x = 71−x (b) 2x−5 = 8
Solution:
(a)
ln( 35 )x = ln 71−x
x ln( 53 ) = (1 − x) ln 7
x ln( 35 ) + x ln 7 = ln 7
7
x = ln( 3ln+ln
≈ 1.356
7)
5
(b)
log2 2x−5 = log2 23
x−5 =3
x =8
5. Solve each logarithmic equation. (a) log2 (5x) = 4
(b) log2 (x + 7) + log2 (x + 8) = 1
Solution:
(a) 5x = 24
x = 16
5
(b) log2 [(x + 7)(x + 8)] = 1
(x + 7)(x + 8) = 21
x2 + 15x + 54 = 0
(x + 6)(x + 9) = 0
Solution (b) is x = −6 since x = −9 is extraneous.
2
6. The half-life of radium-226 is 1600 years. Suppose
we have a 22 mg sample.
(a) Find a function m(t) = m0 e−rt that models the
mass remaining after t years. (r = lnh2 )
(b) After how long will only 18 mg of the sample
remain?
Solution:
ln 2
t
− 1600
(a) m(t) = 22e
= 22e−.000433t
(b)
18 = 22e−.000433t
ln 9 = ln 11 − .000433t ln e
1
9
t = − .000433
ln( 11
)
≈ 463 yrs
7. A bowl of soup begins to cool according to Newton’s
Law of Cooling. Its temperature at time t is given
by
T (t) = 65 + 145e−0.05t
where t is measured in minutes and T is measured
in o F.
(a) What is the initial temperature of the soup?
(b) After how long will the temperature be 100o F?
Solution:
(a) T (0) = 65 + 145e−0
= 210
(b)
100
35
−.05t
t
= 65 + 145e−.05t
= 145e−.05t
35
= ln( 145
)
7
= −20 ln( 29
)
≈ 28
3
8. Solve each system of equations.
(a)



x + y =8
x − y =4
(b)



2x + y = 1
4x + 2y = 3
Solution:
(a)
x
y+4+y
2y
y
x
=y+4
=8
=4
=2
=2+4=6
Solution:(6, 2)
(b)
y = 1 − 2x
4x + 2(1 − 2x) = 3
2 =3
Inconsistent
(c)
x
2(4 − 2y) + 4y
8
y
x
= 4 − 2y
=8
=8
=t
= 4 − 2t
Solution: (4 − 2t, t)
4
(c)



x + 2y = 4
2x + 4y = 8
9. Solve the following system of equations.









x − 2y + 3z
=7
2x + y + z
=4
−3x + 2y − 2z = −10
Solution:


















x − 2y + 3z
=7
5y − 5z = −10
− 4y + 7z
= 11
x − 2y + 3z
=7
y − z = −2
3z
=3
z
=1
y
= 1 − 2 = −1
x = 7 + 2(−1) − 3(1) = 2
Solution:(2, −1, 1)
5
10. Solve each system of equations.
(a)









x − y − z
=1
−x + 2y − 3z = −4
3x − 2y − 7z
=0
(b)









Solution:
(a)


















x − y − z
=1
y − 4z = −3
y − 4z = −3
x − y − z
=1
y − 4z = −3
0 =0









z =t
y = 4t − 3
x = 5t − 2
(b)


















2x − 2y + 3z
=6
y − 4z = −12
y − 4z
=7
2x − 2y + 3z
=6
y − 4z = −12
0 = 19
Inconsistent
6
2x − 2y + 3z = 6
4x − 3y + 2z = 0
−2x + 3y − 7z = 1
11. Solve each system using matrices.
(a)



x + y =8
x − y =4
(b)


x + 2y = 4
2x + 4y = 8

Solution:
(a)


1 1 | 8
1 −1 | 4




 R2−R1−>R2

→
1 1 | 8
0 1 | 2
1 1 | 8
0 −2 | −4


R1−R2−>R1


→

(− 21 )R2

1 0 | 6
0 1 | 2

1 2 | 4
0 0 | 0


x = 6, y = 2
(b)


1 2 | 4
2 4 | 8


 R2−2R1−>R2

→
y = t, x = 4 − 2t
7

→
12. Solve the following system using a matrix.









x − y =6
2x − 3z = 16
2y + z = 4
Solution:





1 −1 0 | 6
2 0 −3 | 16
0 2 1 | 4












 R2−2R1−>R2 


→


1 −1 0 | 6
0 2 −3 | 4
0 0 4 | 0
1 −1 0 | 6
0 1 − 32 | 2
0 0 1 | 0








1
2 R2
→




1 −1 0 | 6
0 1 − 32 | 2
0 0 4 | 0


 R2+ 3 R3−>R2

2

→










1 0 0 | 8
0 1 0 | 2
0 0 1 | 0







 R3−R2−>R3

→






1
4 R3
→
1 −1 0 | 6
0 1 0 | 2
0 0 1 | 0
x = 8, y = 2, z = 0
8
1 −1 0 | 6
0 2 −3 | 4
0 2 1 | 4


 R1+R2−>R1

→

13. Solve each system using a matrix.
(a)









2x − 2y − 2z = 2
2x + 3y + z = 2
3x + 2y = 0
(b)









−x + y + z = −1
−x + 2y − 3z = −4
3x − 2y − 7z
=0
Solution:
(a)










2 −2 −2 | 2
2 3 1 | 2
3 2 0 | 0

1 −1 −1 | 1
0 5 3 | 0
3 2 0 | 0











1
2 R1
→




1 −1 −1 | 1
2 3 1 | 2
3 2 0 | 0



 R3−3R1−>R3 


→


1 −1 −1 | 1
0 5 3 | 0
0 0 0 | −3


 R2−2R1−>R2

→

1 −1 −1 | 1
0 5 3 | 0
0 5 3 | −3


 R3−R2−>R3

→




 Inconsistent

(b)










−1 1 1 | −1
−1 2 −3 | −4
3 −2 −7 | 0
1 −1 −1 | 1
0 1 −4 | −3
3 −2 −7 | 0









 −R1 
 → 



1 −1 −1 | 1
−1 2 −3 | −4
3 −2 −7 | 0



 R3−3R1−>R3 


→


1 −1 −1 | 1
0 1 −4 | −3
0 0 0 | 0

 R2+R1−>R2

→

1 −1 −1 | 1
0 1 −4 | −3
0 1 −4 | −3





z = t, y = 4t − 3, x = 5t − 2
9



 R3−R2−>R3

→

14. Use the following matrices to compute the given expression, if possible.


A=
0 3 −5
1 2 6
(a) A + B
(f) CA
Solution:



B=
(b) B + C

4 1 0
−2 3 −2


(c) −2A (d) AB

(a) A + B = 
4 4 −5
−1 5 4


(b) B + C = undefined

(c) − 2A = 
0 −6 10
−2 −4 −12
(d) AB = undefined

(e) BC = 

(f) CA =




10
22 6
14 −2


1 14 −14
2 22 −18
3 0 28








C=

4 1
6 2
−2 3
(e) BC





15. Perform the following operations using the given matrices, if possible.

D=




3
2
1






(a) DE (b) EF
Solution:

(a) DE =




E=
−2 −1
F =
4
−4







3x
2x
2y
4z
−6 −3
−4 −2
−2 −1






(b) EF = −4 (c) F E = 
− y + z
+ w =6
− 3z
+ 2w = 1
+ z − 5w = 4
− 4w = 5
Solution:









(c) F E
16. Write the following system in matrix notation.









3 −1 1 1
2 0 −3 2
0 2 1 −5
0 0 4 −4
11








x
y
z
w
















=
6
1
4
5








−8 −4
8 4

