Math 103: College Algebra
Instructor: Julio C. Herrera
Exam 3
Name:
October 18, 2016
Exam Score:
Instructions: This exam covers the material from chapter six. Please read each question
carefully before you attempt to solve it. Remember to show all of your work clearly to get
credit. The exam is closed book and calculators are not allowed. Good luck!
Problem 1: Suppose an investment account is opened with an initial deposit of $12, 000
earning 7.2% interest compounded continuously. How much will the account be worth after
30 years?
Solution:
We are told that the interest is compounded continuously, indicating that we should use
the formula A(t) = P ert , where P is the initial deposit.
A(t) = 12, 000e0.072(30)
= $104, 053.65
The account will be worth $104, 053.65 after 30 years.
Problem 2: Graph the function f (x) = 2x−2 ; give the horizontal asymptote, the domain,
and the range.
Solution:
We know that all basic exponential functions are asymptotic to the x-axis unless there is a
vertical shift then that horizontal asymptote shifts accordingly. So our horizontal asymptote
is y = 0. Although this graph shifted to the right 2 units from its parent graph, it does not
affect the domain or range. There are no restrictions on our x-values, therefore the Domain
is (−∞, ∞) and the graph has not been shifted up or down. The Range is (0, ∞) and the
graph is sketched below.
Page 1 of 5
Math 103: College Algebra
Instructor: Julio C. Herrera
Exam 3
Problem 3: Is the following true:
October 18, 2016
log3 (27)
= −1? Verify the result.
log4 (1/64)
Solution:
If log3 (27) = x ⇒ 3x = 27 ⇒ x = 3
If log4 (1/64) = y ⇒ 4y =
⇒
1
⇒ y = −3
64
log3 (27)
3
=
= −1
log4 (1/64)
−3
Thus, the assertion
log3 (27)
= −1 is true.
log4 (1/64)
Problem 4: For the function f (x) = log(5x + 10) + 3, state the domain, range, vertical
asymptote, and x- and y-intercepts.
Solution:
Recall that log x is defined for x > 0. For the domain of f (x), we must keep
5x + 10 > 0
5x > −10
x > −2
The domain and range of f (x) are (−2, ∞) and (−∞, ∞), respectively. This graph shifts
to the left 2 units, therefore the vertical asymptote, which normally occurs at x = 0, will
be at x = −2. To find the x-intercept we set y = 0 and solve for x:
0 = log(5x + 10) + 3
⇒ −3 = log(5x + 10)
⇒ 10−3 = 5x + 10
x = −1.9998
The x- intercept is (−1.9998, 0). To find the y-intercept we set x = 0.
f (0) = log(5(0) + 10) + 3
= log(10) + 3
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Math 103: College Algebra
Instructor: Julio C. Herrera
Exam 3
October 18, 2016
=1+3
=4
Therefore, the y-intercept is (0, 4). In summary,
Domain: (−2, ∞)
Range: (−∞, ∞)
Vertical Asymptote: x = −2
x-intercept: (−1.9998, 0)
y-intercept: (0, 4)
r
y
Problem 5: Use the properties of logarithms to expand ln y
. Rewrite each ex1−y
pression as a sum, difference, or product of logs.
Solution:
r
y
ln y
1−y
y 1/2
= ln y
(1 − y)1/2
y 3/2
= ln
(1 − y)1/2
= ln y 3/2 − ln(1 − y)1/2
= 3/2 ln y − 1/2 ln(1 − y)
Problem 6: Prove that logb (n) =
1
for any positive integers b > 1 and n > 1.
logn (b)
Solution:
Let b and n be positive integers such that b > 1 and n > 1. By the change-of-base formula,
logb (n) = x
=
logn (n)
logn (b)
Page 3 of 5
Math 103: College Algebra
Instructor: Julio C. Herrera
=
Exam 3
October 18, 2016
1
logn (b)
This proves the assertion.
Problem 7: Use logarithms to solve e2x − ex − 6 = 0
Solution:
e2x − ex − 6 = 0
⇒ (ex − 3)(ex + 2) = 0
⇒ (ex − 3) = 0 or (ex + 2) = 0
⇒ ex = 3 or ex = −2
Notice that ex = −2 makes no sense as it implies that x = ln(−2). Hence, ex = 3 ⇒ x =
ln(3). Thus, the solution is x = ln(3).
r kt
. Use the definition of
Problem 8: Recall the compound interest formula A = a 1 +
k
a logarithm along with properties of logarithms to solve the formula for time t.
Solution:
r kt
A=a 1+
k
A
r kt
⇒ = 1+
a
k
r kt
⇒ ln(A/a) = ln 1 +
k
r
⇒ ln(A/a) = kt ln 1 +
k
⇒t=
ln A − ln a
r
k ln 1 +
k
Problem 9: Recall
the formula for calculating the magnitude of an earthquake,
2
S
M = log
. One earthquake has magnitude 3.9 on the MMS scale. If a second
3
S0
earthquake has 750 times as much energy as the first, find the magnitude of the second
quake. Round to the nearest hundredth.
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Math 103: College Algebra
Instructor: Julio C. Herrera
Exam 3
October 18, 2016
Solution:
The first earth quacke can be described by the equation
2
M1 = log
3
S1
S0
The second earth quake has 750 times as much energy as the first:
S2 = 750S1
2
S2
⇒ M2 = log
3
S0
750S1
2
= log
3
S0
2
2
S1
= log 750 + log
3
3
S0
=
2
log 750 + M1
3
We are told that the first earthquake M1 = 3.9. Hence,
=
2
log 750 + 3.9
3
= 1.9167 + 3.9
≈ 5.82
The second earthquake had a magnitude of 5.82.
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