Simplify the expression.
112)
ln(e11 ln e)
112)
A)
11 ln(11)
B)
ln(11)
C)
11
D)
e11
Differentiate.
113)
y = ln 1 – x
(x +3)5
113)
A)
ln 6x –8
(x +3)6
B)
4x –8
(x +3)6
C)
4x –8
(x +3)(1 – x)
D)
(x +3)5
1 – x
Find the derivative of the function.
114)
y = ln (6+ x2)
114)
A)
2
x
B)
2x
x2+ 6
C)
12
x
D)
1
2x + 6
Differentiate.
115)
e3x
115)
A)
e3x
B)
1
3e3x
C)
3x
D)
3e3x
Simplify.
116)
32p ·9p·4p
116)
A)
186p
B)
300p
C)
182p
D)
cannot be simplified
117)
Find the first and second derivatives of f(x) =1 – 2x
ex.
117)
A)
f'(x) =3 – 2x
ex
f”(x) =2x – 5
ex
B)
f'(x) = 4xex– 2ex
f”(x) = 4xex+ 2ex
C)
f'(x) =2x – 3
ex
f”(x) =5 – 2x
ex
D)
f'(x) = 2xex– 3ex
f”(x) = 2xex–ex
Simplify.
118)
ln 8– ln 4
118)
A)
ln(2)
B)
ln 2
C)
ln 1
2
D)
ln(4)
Solve for x.
119)
e(x2+ 9) ·e(6x) = 1
119)
A)
x = ln 1
2
B)
x = 0
C)
x = ± 3
D)
x = – 3
22
Simplify the expression.
120)
e–2 ln 4
120)
A)
16
B)
e2
C)
2
D)
1
16
Solve the problem.
121)
Suppose that the amount in grams of a radioactive substance present at time t (in years) is given by
A(t) =380e–0.32t. Find the rate of change of the quantity present at the time when t =5.
121)
A)
24.6 grams per year
B)
4.9 grams per year
C)
–4.9 grams per year
D)
–24.6 grams per year
Solve the equation for x.
122)
6 ln x –8=5
122)
A)
x =1
6e13
B)
x = ln 13
6
C)
x =1
6 ln 13
D)
x =e13/6
123)
If 5t+5t+5t= 75, find t.
123)
A)
t = 2
B)
t = 5
C)
t = 25
D)
t cannot be determined
23
Differentiate.
124)
ln 3x
124)
A)
1
2x ln 3x
B)
1
6x
C)
1
3x ln 3x
D)
1
6x ln 3x
E)
none of these
125)
Find k such that 3–x/2 =ekx for all x.
125)
A)
ln 3
B)
1
2 ln 3
C)
–1
2 ln 3
D)
ln –3
2
Differentiate.
126)
ex2+ 2 ln(xe)
126)
A)
2ex+ 2 1
ln(xe)
·xe
B)
2xex2+ 2e 1
x
C)
x2ex2– 1 + 2e 1
xe·xe– 1
D)
none of these
Solve for x.
127)
2 – ln(x + 3) = ln 4
127)
A)
x =1
4e2– 3
B)
x = ln 4 – 1
C)
x = – 3
D)
x = 2e
24
128)
Which of the following functions y = f(x) satisfy y’= 32y, f (0) =1
2?
(I) y = 32e1/2x
(II) y =e16x
(III) y =1
2e32x
(IV) y =1
2x32
128)
A)
I only
B)
III only
C)
IV only
D)
I and II
E)
none of these
Differentiate.
129)
(6e2x – x)3
129)
A)
3(6e2x – x)2(12e2x – 1)
B)
3(12xe2x – 1 – 1)2
C)
3(12ex– 1)2
D)
3(6e2x – x)2(12e2x)
Solve the problem.
130)
A company begins an advertising campaign in a certain city to market a new product. The
percentage of the target market that buys the product is a function of the length of the advertising
campaign. The company estimates this percentage as 1 –e–0.03t where t = number of days of the
campaign. The target market is estimated to be 1,000,000 people and the price per unit is $0.60.
The cost of advertising is $3000 per day. Find the length of the advertising campaign that will
result in the maximum profit.
130)
A)
75 days
B)
50 days
C)
60 days
D)
54 days
B
Differentiate.
131)
y =ex
4x2+3
131)
A)
ex(4x2–8x +3)
(4x2+3)2
B)
ex+4x2–8x +3
(4x2+3)2
C)
ex–1(4x2–8x +3)
(4x2+3)2
D)
ex–1(4x2+3) –8x ex
(4x2+3)2
Simplify.
132)
ln e7/3
132)
A)
3
7
B)
7
3e
C)
7
3
D)
3
7e
Differentiate.
133)
f(x) =e3x
133)
A)
3e3x
B)
3ex
C)
1
3e3x
D)
e3x
134)
Find an equation of the tangent line to the graph of y =x3 ln(–2x) at x = – 1.
134)
A)
y + ln 2 = 4(x – 1)
B)
y = (1 + 3 ln 2)(x + 1) – ln 2
C)
y = (x + ln 2) – 1
D)
y – 1 = 4(x – ln 2)
Differentiate.
135)
f(x) =5–e–x
135)
A)
e–x
B)
5–e–x
C)
–e–x
D)
5+e–x
Solve the equation.
136)
2(12 – 2x) =64
136)
A)
3
B)
–3
C)
6
D)
32
Solve the problem.
137)
Suppose that the population of a certain type of insect in a region near the equator is given by
P(t) =10 ln (t + 10), where t represents the time in days. Find the rate of change of the population
when t =4.
137)
A)
1.0 insects
B)
1.4 insects
C)
2.5 insects
D)
0.7 insects
Find the derivative of the function.
138)
y = ln (9x3– x2)
138)
A)
27x – 2
9x2
B)
27x – 2
9x3– x
C)
9x – 2
9x2– x
D)
27x – 2
9x2– x
Solve the problem.
139)
The demand function for a certain product is given by
D(p) =600e–0.1p,
where p is price per unit. Recall that total revenue is given by R(p) = pD(p). At what price per unit
p will the revenue be maximum?
139)
A)
$20
B)
$9
C)
$5
D)
$10
Differentiate.
140)
x3 ln x
140)
A)
(3x2+ 1) ln x
B)
3x2 ln x +x2
C)
x2 ln x +x2
D)
3x2 ln x
E)
none of these
Solve the equation for x.
141)
ln x =6
141)
A)
x =e6
B)
x = ln 6
C)
x =1,000,000
D)
x =6e
Simplify.
142)
(t2)x·(t4)x·(t1/3)x
142)
A)
(t19/3x)3x
B)
t(8/3)x
C)
(t8)x/3
D)
(t19/3)x
Use logarithmic differentiation to find dy/dx.
143)
y =24–x
143)
A)
24–x
B)
–24–x
C)
– ln 24 (24–x)
D)
ln 24 (24–x)
Solve the equation.
144)
2(7 + 3x) =1
4
144)
A)
–3
B)
1
C)
1
2
D)
3
Solve for x.
145)
ln x + ln x8=9
145)
A)
e8
9
B)
e
C)
e9
D)
e9/8
146)
Find the values of x at which the function f(x) =e–2x + 2x has a possible relative maximum or
minimum point.
146)
A)
minimum at x = 0
B)
There are no relative maximum/minimum points.
C)
maximum at x =e
2
D)
maximum at x =0.69
2
E)
none of these
29
Differentiate.
147)
y = ln 6+x2
147)
A)
1
6+x2
B)
x
x2+6
C)
ln x
x2+6
D)
1
2(x2+6)
Simplify.
148)
(e2x)35
e1/2x
148)
A)
5e11/2x
B)
5e3x
C)
5e2x + 3
e1/2x
D)
5e3/2x + 3
Differentiate.
149)
y =5ex
2ex+ 1
149)
A)
5ex
(2ex+ 1)3
B)
5ex
(2ex+ 1)
C)
5ex
(2ex+ 1)2
D)
ex
(2ex+ 1)2
Solve the equation for x.
150)
ex=0.64
150)
A)
x = ln 0.64
B)
x = – ln 0.64
C)
x =log 30.64
D)
x = log 0.64
151)
If (ex)2·e2x · e =1
e2, find x.
151)
A)
–1 or –2
B)
–3
4
C)
–2
D)
–2
5
152)
At what value of x could the function f(x) =ln x + x
x have a possible relative maximum or
minimum?
152)
A)
x =e2
B)
x = e
C)
x = 1
D)
x =1
e
E)
none of these
Compute the given derivative.
153)
d
dx (ex)x = – 6
153)
A)
6e6
B)
e1/6
C)
1
e6
D)
–6
e6
Differentiate.
154)
f(t) = ln [(t6–5)(t5+3)]
154)
A)
1
(t6–5)(t5+3)
B)
6t5(t5+3) +5t4(t6–5)
(t6–5)(t5+3)
C)
ln[6t5(t5+3) +5t4(t6–5)]
D)
30t9
(t6–5)(t5+3)
31
155)
If 1
4
3x + 1 =26 – 2x, find x.
155)
A)
x =1
3
B)
x = 3
C)
x =1
2
D)
x = – 2
Differentiate.
156)
y =e7x/2
156)
A)
7
2xe7x/2
B)
e7x/2
C)
7
2e7x/2– 1
D)
7
2e7x/2
157)
y =e10 –5x
157)
A)
–5 ln (10 –5x)
B)
e–5
C)
10e10 –5x
D)
–5e10 –5x
Given ln 2 = 0.6931 and ln 5 = 1.6094, find the following.
158)
ln 5
8
158)
A)
–0.4699
B)
0.4699
C)
0.77401048
D)
3.6887
159)
Determine a function y = f(x) such that y’ =1
10 y and f(0) = – 3?
159)
A)
y = – 3e(1/10) x
B)
y =e(1/10) y – 3
C)
y =1
10 e–3x
D)
y =1
20 y2– 3
E)
none of these
Solve for x.
160)
3 + ln x = 0
160)
A)
x = ln 1
3
B)
x = ln(–3)
C)
x = – e3
D)
x =1
e3
E)
none of these
Differentiate.
161)
(ln(x2+ 2))3
161)
A)
e1
x2+ 2
2· 2x
B)
6x
x2+ 2 (ln(x2+ 2))2
C)
3(ln(2x))2
D)
1
(ln(x2+ 2))3· 2x
33
Solve the problem.
162)
Suppose that the demand function for x units of a certain item is p =100 +180 ln(x + 5)
x, where p is
the price per unit, in dollars. Find the marginal revenue.
162)
A)
dR
dx =180 [x – (x + 5) ln(x + 5)]
x2(x + 5)
B)
dR
dx =180[x –[ln(x + 5) ]2]
x2 ln(x + 5)
C)
dR
dx =100 +180
x + 5
D)
dR
dx =100 +180
ln(x + 5)
163)
The sales in thousands of a new type of product are given by S(t) =170 – 80e–0.9t, where t
represents time in years. Find the rate of change of sales at the time when t =7.
163)
A)
–38,462.6 thousand per year
B)
38,462.6 thousand per year
C)
–0.1 thousand per year
D)
0.1 thousand per year
164)
Estimate the slope of the curve y =ex at x = 0.
164)
A)
0
B)
e
C)
1
D)
ex
E)
none of these
Simplify.
165)
ln(x2– 2) + e · ln(x2– 2)
165)
A)
±3
B)
(ln + 1)(x2– 2)
C)
(1 + e)ln (x2– 2)
D)
ln(x2– 2) +(x2– 2)
34
C
Solve for x.
166)
ex2– 4 = 3
166)
A)
± (2 +ln 3)
B)
±4 + ln 3
C)
2 ±ln 3
D)
none of these
Find the derivative of the function.
167)
y = ln 4x2
167)
A)
2
x
B)
2x
x2+ 4
C)
8
x
D)
1
2x + 4
Solve for x.
168)
ln(x + 1) = 2 + ln x
168)
A)
e2– 1
B)
1
e2– 1
C)
e + 1
D)
e – 1
E)
none of these
Find the derivative of the function.
169)
y = ln x
8
169)
A)
1
x
B)
1
8x
C)
8
x
D)
1
x– ln 8
Answer Key
Testname: C4
Answer Key
Testname: C4
37
Answer Key
Testname: C4
Answer Key
Testname: C4
Answer Key
Testname: C4