Unit 4 Homework Problems MAT150 College Algebra
© Andrew McKintosh 2010
All problems should be done WITHOUT a calculator unless otherwise indicated!!!!
Day 44 Homework: Topics: Graphing Rational Functions: Vertical Asymptotes, Horizontal Asymptotes, Yintercept, X-intercept, crossing the horizontal Asymptote.
Supplemental Reading:
1. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut40_ratgraph.htm (skip content
on oblique/slant asymptotes and example 6.)
2. http://www.purplemath.com/modules/grphrtnl.htm
3. http://www.coolmath.com/algebra/23-graphing-rational-functions/index.html
x 2  5x  6  x  6  x  1
x2  5x  6
1. y 
Find
each
of
the
following…..
y


x2  4
x2  4
 x  2 x  2
A) The equation(s) for the vertical asymptote(s)
From the factored form of the function
We can see that the vertical asymptotes will occur at x  2 and
x2
B) The equation(s) for the horizontal asymptote(s)
From the original form of the function we can see that the degree of the numerator equals the degree of the
denominator, therefore the horizontal asymptote will occur at y = “the ratio of lead coefficients” so……
1
y   y 1
1
C) The y-intercept(s)
Let x = 0 in the original form
6 3
and you can see that y 

4 2
 3
so the y intercept is  0, 
 2
D) The x-intercept(s)
From the factored form of the function we can see that
the x intercepts will occur at .  6,0 and
E) Does this function cross it’s horizontal asymptote? If so find out where.
x2  5x  6
1
 1 x 2  4   x 2  5 x  6  x 2  4  x 2  5 x  6  4  5 x  6
x2  4
2
 2 
5 x  6  4  5 x  2  x  
so, this function crosses it ' s H . A. at   ,1
5
 5 
F) Sketch the graph of this function on the axes below.
 1,0
(Day 44 Homework continued) :
© Andrew McKintosh 2010
x 2  3x  4
Find each of the following…..
x3  2 x 2  4 x  8
 x  4  x  1
 x  4  x  1
 x  4  x  1
 x  4  x  1
x 2  3x  4
y 3
 2



2
2
x  2 x  4 x  8 x  x  2   4  x  2   x  2   x  4   x  2  x  2  x  2   x  2 2  x  2 
2. y 
A) The equation(s) for the vertical asymptote(s)
From the factored form of the function
we can see that the vertical asymptotes will occur at x  2 and
x  2
B) The equation(s) for the horizontal asymptote(s)
Since the degree of the numerator is less than the degree of the numerator we know that
Y = 0 must be our H.A.
C) The y-intercept(s)
Let x = 0 in the original form
4
1
and you can see that y 

8
2
1

so the y intercept is  0,  
2

D) The x-intercept(s)
From the factored form of the function we can see that
the x intercepts will occur at .  4,0 and
E) Does this function cross it’s horizontal asymptote? If so find out where.
Yes, we already found this in part D)
F) Sketch the graph of this function on the axes below.
Note: We need to plot more points in order to sketch in the “outside” portions of this graph.
1 4  4

 25 1 25
 7  2  14
x 3 y 

1 5 5
x  3  y 
1,0
Day 45 Homework: Topics: Graphing Rational Functions:
© Andrew McKintosh 2010
Vertical Asymptotes, Horizontal Asymptotes,
Y-intercept, X-intercept, crossing the horizontal Asymptote.
Supplemental Reading: (The same as yesterday!)
1. y 
 x  3 x  3
x2  9

2 x  3x  5  2 x  5 x  1
2
A) The equation(s) for the vertical asymptote(s)
From the factored form of the function
we can see that the vertical asymptotes will occur at x  
5
2
and
x 1
B) The equation(s) for the horizontal asymptote(s)
From the original form of the function we can see that the degree of the numerator equals the degree of the
denominator, therefore the horizontal asymptote will occur at y = “the ratio of lead coefficients” so……
1
y
2
C) The y-intercept(s)
D) The x-intercept(s)
Let x = 0 in the original form
and you can see that y 
From the factored form of the function we can see that
9 9

5 5
the x intercepts will occur at .  3,0 and
 9
so the y intercept is  0, 
 5
E) Does this function cross it’s horizontal asymptote? If so find out where.
1
x2  9
 2
 1 2 x 2  3x  5   2  x 2  9   2 x 2  3x  5  2 x 2  18
2 2 x  3x  5
13
 13 1 
3x  5  18  3x  13  x  
so, this function crosses it ' s H . A. at   , 
3
 3 2
F) Sketch the graph of this function on the axes below.
3,0
(Day 45 Homework continued) :
2. y 
© Andrew McKintosh 2010
 x  3 x  2 
 x  3 x  2 
 x  3 x  2 
 x  2
x2  5x  6
 2



3
2
2
x  3x  x  3 x  x  3  1 x  3  x  3  x  1  x  3 x  1 x  1  x  1 x  1
A) The equation(s) for the vertical asymptote(s)
From the factored form of the function
we can see that the vertical asymptotes will occur at x  1 and
ALSO note that we have a “hole” when x  3 y 
3  2
1

3  13 1 8
x 1
 1
 3, 
 8
B) The equation(s) for the horizontal asymptote(s)
Since the degree of the numerator is less than the degree of the numerator we know that
Y = 0 must be our H.A.
C) The y-intercept(s)
Let x = 0 in the original form
6
and you can see that y   2
3
so the y intercept is  0, 2
D) The x-intercept(s)
From the factored form of the function we can see that
the only x intercept will occur at .  2,0
E) Does this function cross it’s horizontal asymptote? If so find out where.
Yes, we already found this in part D)
F) Sketch the graph of this function on the axes below.
HOLE
Note: We need to plot one more point in order to sketch in the far left “outside” portions of this graph.
x  2
y
 4
4

 1 3 3
4

 2,  
3

x3
Day 46 Homework: Topics: Exponential Functions and their graphs,
© Andrew McKintosh 2010
compound interest, Logarithmic Functions and their graphs,
the definition of logarithm.
Supplemental Reading:
1. http://www.coolmath.com/algebra/17-exponentials-logarithms/index.html (check out lessons 1, 3, 4, 7 and 8)
2. http://www.purplemath.com/modules/expofcns.htm
3. http://www.purplemath.com/modules/graphexp.htm
4. http://www.purplemath.com/modules/graphlog.htm
5. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut42_expfun.htm
6. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut43_logfun.htm
1. Use your calculator and evaluate each of the following. Round your answer to three decimal places.
A) f  x   3.4x
3.4
1.3
B) f  x   10x
x  1.3
at
10
 4.908
C) f  x   ex
at
.24
e3.2  24.533
1
 
3
3  2
x  .24
 .004
1
D) f  x    
3
x  3.2
at
x2
at
x  3
1
1
  3
3
2. Graph the following two functions on the same set of axes… (Fill out the tables!!!)
f  x   3x
x
0
1
2
-1
-2
x
0
1
2
-1
-2
1
g  x   
3
x
f (x)
1
3
9
1/3
1/9
g (x)
1
1/3
1/9
3
9
3. $20,000 is placed in a savings account that earns 8% interest. Find the amount of money after 5 years IF…
A) interest is compounded quarterly
 .08 
A  20000 1 

4 

B) interest is compounded monthly
12 5
4  5
 .08 
A  20000 1 

 12 
 $29,718.95
C) interest is compounded continuously
.08 5
A  20000e
 $29,836.49
 $29,796.91
(Day 46 Homework continued) :
4. Repeat problem 3. Is the interest rate is 10%.
A)
© Andrew McKintosh 2010
B)
 .10 
A  20000 1 

4 

4  5
C)
12 5
 $32,772.33
 .10 
A  20000 1 

 12 
 $32,906.18
.10  5
A  20000e
 $32,974.43
5. Graph y  log 2 x  x  2 y
x
y
0
1
2
-1
-2
1
2
4
1/2
1/4
6. Use the definition of logarithm to convert the following logarithmic equations into their equivalent exponential
form.
A) 3  log 2 8
8  23
 1 
B)  2  log5  
 25 
1
 52
25
C ) log10 1  0
1  100
D) ln e  1
e  e1
7. Use the definition of logarithm to convert the following exponential equations into their equivalent logarithmic
form.
A) 53 
1
125
 1 
3  log5  
 25 
B) 64  43
C ) 104  10, 000
3  log 4 64
4  log10 10000
D ) 50  1
0  log5 1
Day 47 Homework: Topics: Properties of logarithms.
© Andrew McKintosh 2010
Supplemental Reading:
1. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut44_logprop.htm
2. http://www.coolmath.com/algebra/17-exponentials-logarithms/10-inverses-tricks-01.htm
3. http://www.purplemath.com/modules/logs.htm
4. http://www.purplemath.com/modules/logrules.htm
1. Evaluate the following WITHOUT using a calculator.
A) log 3 81
B) log 6 1
log 3 34
0
4
 1 
C ) log 4  
 64 
1
log 4 3
4
log 4 4 3
D) log 2 213
13
E ) ln e13
F ) log100
log e e13
log10 102
13
2
3
2. Use the change of base formula, and your calculator, to evaluate the following. Round your answer to three
decimal places in A and B.
 1 
B) log3  
 75 
1
log
75
log 3
A) log 2 28
ln 28
ln 2
C ) log100 1000
log10 1000
log10 100
 4.807
 1.5
 3.930
3. Find he exact value of log3 5 9 WITHOUT using a calculator.
1
2
1
log3 5 9  log3 9 5  log3  32  5  log3 35 
4. Expand logb
4
b3 x 2
y6
1
log b
4
2
5
3
1
3 1
3
3
1
3
 b3 x 2  4
b3 x 2
b4 x2
3 1
3
4 2
2
4
2
2

log

log

log
b
x

log
y

log
b

log
x

log
y
  log b x  log b y


b
b
b
b
b
b
b
6
6
3
4 2
2
y
 y 
y2
5. Condense 3log x  2log y  5log z
3log x  2 log y  5log z  log x 3  log y 2  log z 5  log
x3
x3 5
x3 z 5
5

log
z

log

z

log
y2
y2
y2
Day 48 Homework: Topics: More with properties of logarithms.
Supplemental Reading: Same as yesterday!
© Andrew McKintosh 2010
1. Find the value for each of the following without using calculator.
C) log4  16
B) ln 3 e2
A) log100
Does not exist!
(You cannot feed
a negative number to
a logarithm!)
2
log10 100
ln e 3
log10 10 2
2
log e e 3
2
D) log5 75  log5 3
log 5
2
3
E) log6 2  log6 18
F) 3ln e6  2ln e5
log 6  2 18 
ln  e 6   ln  e5 
log 6 36  log 6 62  2
ln e 18  ln e10
75
3
log 5 25  log 5 5  2
2
3
ln  e 18  e10 
2. Expand ln
x5 y 2
z4
ln  e 18 10 
1
 52 
 x5 y 2  2
x5 y 2
x y
ln
 ln  4   ln  2 
4
z
 z 
 z 


5
ln e 8  8
5
 ln x 2 y  ln z 2  ln x 2  ln y  ln z 2

5
ln x  ln y  2 ln z
2
3. Condense 5log 2 y  log 2 z  3log 2 x
5log 2 y  log 2 z  3log 2 x  log 2 y 5  log 2 z  log 2 x3  log 2
4. Write 4 3 
3  log 4
1
in logarithmic form.
64
1
64
5. Write 7  log 2 128 in exponential form.
128  27
y5
y5
 log 2 x3  log 2 3
z
xz
2
Day 49 Homework: Topics: Solving Exponential Equations
© Andrew McKintosh 2010
Supplemental Reading:
1. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut45_expeq.htm
2. http://www.purplemath.com/modules/solvexpo.htm
3. http://www.coolmath.com/algebra/17-exponentials-logarithms/11-solving-exponential-equations-01.htm
1. Solve the following exponential equations by “making the bases the same”.
A. 32 x 1  27
B. 2  4x1  8
32 x 1  33
2 x  2
2  4 x 1 8

2
2
4 x 1  4
x  1
x 1  1
2 x  1  3
x0
2
D.  
3
C. 3x  9 x  27
2
3x   32   33
2 x2  x  3  0
3x  32 x  33
 2 x  3 x  1  0
x2
2
3x  2 x  33
2
x
3
2
x 1
x 1

9
4
2
 
3
x 1
3
 
2
2
 
3
x 1
  2  1 
   
 3  


x 1
x  2x2  3
2
2
   
3
3
x  1  2
E. 2  32 x1   1  17
2
2
2
x  1
2  32 x 1   18
32 x 1  9
32 x 1  32
2x  1  2
x
1
2
2. Solve the following exponential equations. Give an exact answer AND an approximate answer rounded to three
decimal places.
A. 2x1  3
B. 2  32 x1   1  15
ln 2 x 1  ln 3
2  32 x 1   16
 x  1 ln 2  ln 3
x 1 
ln 3
ln 3
 x  1
 2.585
ln 2
ln 2
ln 8
ln 3
ln 8
2 x  1 
ln 3
1 ln 8
x 
2 2 ln 3
x  .446
2x 1 
32 x 1  8
ln 32 x 1  ln 8
 2 x  1 ln 3  ln 8
C. 32 x 1  5x 2
ln 32 x 1  ln 5x  2
 2 x  1 ln 3   x  2  ln 5
2 x ln 3  ln 3  x ln 5  2 ln 5
ln 3  2 ln 5  x ln 5  2 x ln 3
D. e2 x  3ex  4  0
ln 3  2 ln 5  x  ln 5  2 ln 3
ln 3  2 ln 5
x
ln 5  2 ln 3
 7.345  x
 e   3e
 e  4  e
x 2
x
4  0
x
x
 1  0
ex  4  0
ex  4
x  ln 4
x  1.386
ex  1  0
e x  1
no solution
Day 50 Homework: Topics: Solving Logarithmic Equations
© Andrew McKintosh 2010
Supplemental Reading:
1. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut46_logeq.htm
2. http://www.purplemath.com/modules/solvelog.htm
3. http://www.coolmath.com/algebra/17-exponentials-logarithms/15-solving-logarithmic-equations-01.htm
4. http://www.coolmath.com/algebra/17-exponentials-logarithms/14-tricks-to-help-with-calculus-01.htm
1. Solve the following logarithmic equations.
A. ln x  ln 4  0
B. ln  x  3  1  2
ln x  ln 4
ln  x  3   1
x4
x  3  e1
x  3 e
C. log  3  x   2
D. 2log3 1  x   3  11
3  x  10
2 log 3 1  x   8
2
log 3 1  x   4
3  x  100
 x  97
1  x  34
1  x  81
 x  80
x  97
x  80
E. log4  x   log4  x  12  3
F. log3  x   log3  x  6  3
log 4 x  x  12   3
log 3 x  x  6   3
x  x  12   4
x  x  6   33
3
x 2  12 x  64
x 2  6 x  27
x 2  12 x  64  0
x 2  6 x  27  0
 x  16  x  4   0
 x  9  x  3  0
x  16 x  4
x  9 x  3
Extraneous
Extraneous
G. log2  x   log2  x  6  2
H. log3  x  5  log3  x  3  2
x
 22
x6
x
4
x6
x  4  x  6
x5
 2
x3
x5
 32
x3
x5 1

x3 9
9  x  5   1 x  3
x  4 x  24
9 x  45  x  3
3x  24
8 x  48
x8
x6
log 2
x
2
x6
log 3
Day 51 Homework: Topics: Applications involving exponential equations
© Andrew McKintosh 2010
Supplemental Reading:
1. http://www.coolmath.com/algebra/17-exponentials-logarithms/12-solving-exponential-equations-rate-time01.htm
2. http://www.coolmath.com/algebra/17-exponentials-logarithms/13-radioactive-decay-decibel-levels-01.htm
3. http://www.purplemath.com/modules/expoprob2.htm
4. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut47_growth.htm
1. Complete the table for a savings account in which interest is compounded continuously.
Initial Investment
$10,000
$5000
$165,298.89
A)
B)
C)
Annual % Rate
4%
6.9%
12%
A) T2 : 20, 000  10, 000e.04t  2  e.04t  ln 2  .04t  t 
Time to Double
17.33 yrs
10 yrs
5.78 yrs
Amount after 15 yrs
$18,221.19
$14,075.53
$1,000,000
ln 2
 17.33 years
.04
A 15  : A 15   10, 000e.0415  $18, 221.19
B) % : 10, 000  5000e
r 10 
 2  e10 r  ln 2  10r  r 
ln 2
 .069  6.9%
10
A 15  : A 15   5000e.06915  $14, 075.53
C ) T2 : 2 P  Pe.12t  2  e.12t  ln 2  .12t  t 
.12 15
P : 1, 000, 000  Pe
P
1, 000, 000
.12 15 
e
ln 2
 5.78 years
.12
 $165, 298.89
2. Determine how long it would take for $1000 to double if it is invested at an interest rate of 7% compounded…
Round your answer to the nearest hundredth of a year.
A) Annually
B) Monthly
C) Continuously
12 t
1t
12
t
 .07 
t
 .07 
2000  1000e.07t  2  e.07t
2000  1000 1 
  2  1.07  2000  1000 1  12   2  1.0583
1




ln 2  ln e.07t  ln 2  .07t
12 t
t
ln 2  ln 1.07   ln 2  t ln 1.07 
ln 2  ln 1.0583  ln 2  t ln 1.0583
ln 2
 t  t  9.90 yrs
ln 2
ln
2
.07
t
 10.24 yrs
t


9.93
yrs
ln 1.07 
ln 1.0583



3. The half life of radioactive radium

226





Ra  is 1620 years. What percent of a present amount of radioactive
radium will remain after 100 years? (Round your answer to the nearest tenth of a percent)
1
ln
1
1
1
k 1620
1620 k
A0  A0 e
 e
 ln  1620k  k  2
2
2
2
1620
A  A0 e
1
ln
2 100 
1620
 A  .958115578 A0
So about 95.8% will remain after 100 years.
4. The population P of a city (in thousands) is P  548ekt where t = 0 represents the year 2001. In 1969 the
population was 238,000. Find the value of k and use the result to predict the population in the year 2025.
Round your answer to the nearest thousand.
(note: if 2001 = year zero then 1969 = year # -32)
238
ln
238
238
k  32 
32 k
548
238  548e

e
 ln
 32k  k 
548
548
32
238
548  24 
32
ln
so the amount in 2025  year # 24  A  548e
So the population in 2025 will be about 1,024,000
 1024.314356
Day 52 Homework: Topics: Solving Linear systems of equations in two variables
© Andrew McKintosh 2010
By graphing and substitution.
Supplemental Reading:
1. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut49_systwo.htm (Read
everything up through example 6)
2. http://www.purplemath.com/modules/systlin1.htm (read the first four pages only)
3. http://www.coolmath.com/algebra/12-2x2-systems-of-equations/index.html (read items 1, 2 and 4 only)
1. Solve each of the following systems of equations by Graphing.
3

y   x  2
B) 
2
 2 x  3 y  7
 x y  3
A) 
2 x  3 y  1
 x  3y  1
C) 
 2 x  3 y  1
 2,1
 2,1
 2, 1
2. Solve each of the following systems of equations by the Substitution method.
3

 x y  3
y   x  2
A) 
B) 
2
2 x  3 y  1
 2 x  3 y  7
x  y  3  x  3 y
 3

2x  3  x  2   7
2 x  3 y  1  2  3  y   3 y  1
 2

9
 6  2 y  3 y  1
2x  x  6  7
2
 6  5 y  1
4
x

9
x  12  14
5y  5  y  1
x  3  y  x  3 1  x  2
 2,1
 3x  6 y  7
D) 
 2 x  4 y  2
2 x  4 y  2   2 x  4 y  2
x  2 y 1
3 x  6 y  7  3  2 y  1  6 y  7
6y  3 6y  7
13 x  26  x  2
3
3
y   x  2  y    2   2  1
2
2
2,

1


 x  3y  1
C) 
 2 x  3 y  1
x  3y  1  x  1 3y
2 x  3 y  1
2 1  3 y   3 y  1
2  6 y  3 y  1
3 y  3  y  1
x  1  3 y  x  1  3 1
x  2
 2,1
 5 x  15 y  10
E) 
 3 x  9 y  6
3 x  9 y  6  3 x  9 y  6  x  3 y  2
5 x  15 y  10  5  3 y  2   15 y  10
 15 y  10  15 y  10
 10  10
3  7
No Solution !
An infinite # of solutions of the form  3 y  2, y 
Day 53 Homework: Topics: Solving Linear systems of equations in two variables
© Andrew McKintosh 2010
By elimination.
Supplemental Reading:
1. http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut49_systwo.htm (start reading
after example 6)
2. http://www.purplemath.com/modules/systlin5.htm (read this page only)
3. http://www.coolmath.com/algebra/12-2x2-systems-of-equations/index.html (read item 3 only)
1. Solve each of the following systems of equations using the Elimination method!!!
A)  x  y  3

2 x  3 y  1
*2  
2x  2 y  6
 2 x  3 y  1
5y  5  y  1
x y 3
x 1  3  x  2
B) 
3
y   x  2
2


2 x  3 y  7
*2   2 y  3x  4  3x  2 y  4 *3  9 x  6 y  12


2 x  3 y  7

2 x  3 y  7 *2   4 x  6 y  14
13 x
3
3
y   x  2  y    2   2  y  3  2  y   1
2
2
C)  x  3 y  1



 2 x  3 y  1
 2,1
 26  x  2
 2, 1
*  2   2 x  6 y  2
 2 x  3 y  1
 3 y  3  y  1
x  3 y  1  x  3 1  1  x  3  1  x  2
D)  3x  6 y  7



2 x  4 y  2
 2,1
*2   6 x  12 y  14
*3  6 x  12 y  6
0  20  No Solution !
E)  5 x  15 y  10


3 x  9 y  6
*3  15 x  45 y  30
*5   15 x  45 y  30
0  0  find infinite generating set
5 x  15 y  10  5 x  15 y  10  x  3 y  2
So the infinite generating set will be
 3 y  2, y 
Day 54 Homework: Topics: Solving Linear systems of equations in two variables
By elimination.
Supplemental Reading: SAME AS YESTERDAY!!!!
1. Solve each of the following systems of equations using the Elimination method!!!
*2  
A) 
 x  2y  3

2 x  4 y  1

2x  4 y  6
 2 x  4 y  1
0  7 No Solution!
B) 2 x  y  1


4 x  2 y  2
*  2   4 x  2 y  2
 4x  2 y  2
0 = 0 So we must find the infinite generating set!
 x, 2 x  1
2 x  y  1   y  2 x  1  y  2 x  1
C)  3 x  2 y  5


2 x  5 y  7
*2   6 x  4 y  10
*3  6 x  15 y  21
19 y  11  y  
 3x  2 y  5

2 x  5 y  7
*5   15 x  10 y  25
*  2   4 x  10 y  14
19 x
D)  1
2
 2 x  3 y  4

  x  3 y  7

5
*6  
 39  x 
3 x  4 y  24
11
19
 39 11 
 , 
 19 19 
39
19
*3 
9 x  12 y 
*5  5 x  3 y  35 *4   20 x  12 y  140
 11x
3 x  4 y  24

5 x  3 y  35
*5   15 x  20 y  120
*3  15 x  9 y  105
 11y  15  y  
 68 15 
 , 
 11 11 
72
15
11
 68  x 
68
11
© Andrew McKintosh 2010