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MATH 115
QUIZ6-SAMPLE
October 31, 2016
Laws of exponentials and logarithms. ( true/false and multiple choice.) Derivatives
of log and exponential functions. (Single answer. ) Review: Section 5.1: 1-8, Section
5.1: 9-16, Section 5.1: 17-26, Section 5.2: 17-20, Section 5.2:21-28, Section 5.2: 35-44,
Section 5.2: 57-64, Section 4.4: 1-28 and Section 4.5:1-40.
5
1. (
xn y m
) =
z k x2
(A)
x5n−2 y 5m
z 5k
(B)
x5n−10 y 5m
z 5k
(C)
x2n+2 y 5k
zk
Solution: B
−2
2. ((x2 y 5 )3 ))
(A) x6 y 15
(B)
x10
y 25
(C)
1
x12 y 30
Solution: C
3. Solve
1
= 27x then x =
36
(A) − 2
(B) 2
(C) − 3
(D) 3
Solution: That is, 3−6 = (33 )x so 3−6 = 33x .
3x = −6 gives x = −2
A
4. Solve 3x−1 = 35x+9 , then x =
(A) − 2
(B) − 2.5
(C) 2
(D) 2.5
Solution: B
5. Solutions to e2x − 3ex + 2 = 0 are
(A)x = 1, 2
(B) x = 0, 1
(C) x = 2, 3
(D) x = 0, ln 3
(E) x = 0, ln 2
Solution: Let y = ex , then y 2 − 3y + 2 = 0.
y = 1, 2
Replace y by ex :
ex = 1 gives x = 0
ex = 2 gives x = ln(2)
E
6. ln (
ex
√
)=
3
x5 x2 + 1
(A) ln(ex )−5 ln(x)+(1/3) ln(x2 +1)
(B) x−5 ln(x)−(1/3)(2x)−(1/3)
(C) x−5 ln(x)−(1/3) ln(x2 +1)
Solution: C
7. ln (
ex
)=
ex + 1
(A)1
(B) 2ex + 1
(C) x − ln(ex + 1)
Solution: C
8. ln (xe) =
(A) ln(x) + e (B) ln(x) + 1
(C) ex
Solution: B
9. Solution to
2
= 1 is
1 + 3e−5t
(A) −
Solution:
ln(3)
5
(B)
ln(3)
5
1 + 3e−5t
=1
2
so (1 + 3e−5t ) = 2 so e−5t =
1
3
−5t = ln( 13 )
−5t = − ln(3)
A
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(C) ln(3)
(D) 0
1. Find the derivative of the following functions.
(a) f (x) = x2 e5x .
2
(b) g(x) = ex
e2x − e−2x
(c) h(x) =
2
ex + e−x
(d) j(x) =
ex
(e) n(x) = e3x+2
(f) k(x) = xe−x
2
Solution:
(a) One product rule, then a chain rule.
f ′ (x) = 2xe5x + 5x2 e5x
(b) Chain rule: g ′ (x) = 2xex .
2
(c) h′ (x) = (e2x + e2x )
(ex − e−x )(ex ) − (ex )(ex + e−x ) e2x − 1 − 1 − e2x −2
=
= 2x = −2e−2x
e2x
e2x
e
or Rewrite the original: j(x) = (ex +e−x )(e−x ) = 1+e−2x and take the derivative j ′ (x) = −2e−2x
(d) Quotient rule: j ′ (x) =
(e) A chain rule.n(x) = 3e3x+2
(f) A product rule with a chain rule for the second part. k ′ (t) = e−x − 2x2 e−x
2
2. Find the derivative of the following functions:
(a) f (x) = ln(x3 + 4x2 + 5)
(b) g(x) = ln( xx+2
2 +1 )
x
(c) h(x) =
ln(x)
(d) n(x) = (ln(x))2
Solution:
(a) Chain rule. f (x) =
3x2 + 8x
(x3 + 4x2 + 5)
(b) Use your logarithmic laws to rewrite: g(x) = ln(x + 2) − ln(x2 + 1)
1
2x
Then two chain rules: g ′ (x) =
−
x + 2 x2 + 1
(c) Quotient rule. h′ (x) =
ln(x) − x(1/x)
x−1
=
(ln(x))2
(ln(x))2
(d) Chain rule with log inside. n′ (x) =
2 ln(x)
x
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2
Short Notes
• Laws of Exponents
Let a and b be positive numbers and let x and y be real numbers. Then,
1. bx .by = bx+y
bx
2. y = bx−y
b
3. (bx )y = bxy
4. (ab)x = ax bx
a x ax
5. ( ) = x
b
b
• Laws of Logarithms
If m and n are positive numbers and b > 0, b ≠ 1, then
1. logb mn = logb m + logb n
2. logb
m
n
= logb m − logb n
n
3. logb m = n logb m
4. logb 1 = 0
5. logb b = 1
• Laws of exponents in English. (These only serve to help you to memorize the laws.)
1. Product of two exponentials with like base is the base to sum of the exponents.
2. Quotient of two exponentials with like base is the base to difference of the exponents.
3. To power to power is to product of the powers.
4. Product to a power is first to the power times the second to the power.
5. Quotient to a power is top to the power divided by the bottom to the power.
• Laws of logarithm in English (These only serve to help you to memorize the laws.)
1. log of product is equal to sum of logs.
2. log of quotients is equal to difference of logs.
3. log of a number to a power is the power times the log of the number.
4. log of 1 is zero.
5. log of base is 1.
• The Chain Rule for Exponential and logarithmic Functions
If f (x) is a differentiable function, then
●
d
(ef (x) )
dx
= ef (x) f ′ (x)
●
d
[ln f (x)]
dx
=
f ′ (x)
f (x)
(f (x) > 0)
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