In Class Problem Set #20
CSE 1400 and MTH 2051
Fall 2012
Logarithms and Exponentials
1. Answer True or False. Explain your answer.
(a) 2a 2b = 2a+b .
Answer: True. Multiplying a 2’s by the product of b 2’s is the
same as multiplying a + b 2’s together.
(b) 2a 2b = 2ab .
Answer: False. As a counterexample, 23 24 = (2 · 2 · 2)(2 · 2 · 2 ·
2) = 27 6= 212 . The correct rule is 2a 2b = 2a+b
(c) 2 · 2n = 4n .
Answer: False. As a counterexample, 2 · 23 = 2 · (2 · 2 · 2) =
24 6= 43 = (4 · 4 · 4). The correct rule is 2a 2b = 2a+b
(d) 2n 2n = 22n .
Answer: True
2
(e) 2n 2n = 2n .
Answer: False. As a counterexample, 23 23 = (2 · 2 · 2)(2 · 2 · 2) =
2
26 6= 23 = 29 . The correct rule is 2n 2n = 22n .
(f) (2a )b = 2ab .
Answer: True
b
(g) (2a )b = 2a .
3
Answer: False. As a counterexample, 23 = (2 · 2 · 2)(2 · 2 ·
3
2)(2 · 2 · 2) = 29 6= 23 = 227 . The correct rule is (2a )b = 2ab .
2
(h) (2n )n = 2n .
Answer: True
n
(i) (2n )n = 2n .
Answer: False
2. Evaluate the logarithm base 2 function lg x at the given value for
x.
(a) lg(256)
Answer: lg(256) = 8 because 28 = 256.
(b) lg(0.25)
Answer: lg(0.25) = −2 because 2−2 = 0.25.
(c) lg(0.125)
Answer: lg(0.125) = −3 because 2−3 = 0.125.
in class problem set #20
(d) lg(1/1024)
Answer: lg(1/1024) = −10 because 2−10 = 1/1024.
√
(e) lg( 3 16)
√
√
Answer: lg( 3 16) = 4/3 because 24/3 = 3 16.
√
(f) lg( 5 8)
√
√
Answer: lg( 5 8) = 3/5 because 23/5 = 5 8.
√
(g) lg 3 32
Answer: The logarithm base 2 of the cube root of 32 is 5/3
√
because 25/3 = 3 32.
√
(h) log16 3 32
Answer: The logarithm base 16 of the cube root of 32 is 5/12.
Note that
√
1
1 5
5
3
log16 32 = log16 32 = · =
3
3 4
12
(i) lg(2n! )
Answer: lg(2n! ) = n!
3. Compute the following values.
(a) 10lg x for x = 1, 2, 4, 8
Answer:
10lg 1 = 1,
10lg 2 = 10,
10lg 4 = 100,
10lg 8 = 1000
4lg 10 = 100,
8lg 10 = 1000
(b) xlg 10 for x = 1, 2, 4, 8
Answer:
1lg 10 = 1,
2lg 10 = 10,
(c) Is it True or False that 10lg x = xlg 10 ? Explain your answer.
Answer: It is True. Notice that If y = 10lg x , then taking the
logarithm base 2 of both sides of the equation yields
lg y = lg 10lg x = lg x lg 10
Therefore, raising the the sides as powers of 2
y = 2lg y = 2lg x lg 10 = xlg 10
2
in class problem set #20
(d) Show that alogb x = xlogb a .
Answer: The log base b of alogb x is logb a logb x. The log base b
of xlogb a is logb x logb a. Clearly, the two logarithms are equal
and since the logarithm function is one-to-one, the original
values, alogb x and xlogb a are equal.
4. True or False: Since 210 = 1024 is approximately equal to 103 =
1000, the log base 2 of 10 is approximately equal to 3 and 1/3.
Explain your answer.
Answer: This is True. Compute the approximations
103 ≈ 210
lg(103 ) ≈ lg(210 )
3 lg(10) ≈ 10
lg(10) ≈
10
3
5. Use the fact that 27 = 128 is approximately equal to 53 = 125 to
approximate the value of log5 (2).
Answer: Since 27 ≈ 53 we have log5 (27 ) = 7 log5 (2) ≈ log5 (53 ) = 3
or log5 (2) ≈ 3/7.
6. Show the following are True.
(a) The log base b of x is 1 over the log base x of b.
logb ( x ) =
1
logx (b)
Answer: If y = logb ( x ), then by = x. Take the logarithm base x
of this last equation to get
logx (by ) = y logx (b) = 1
so that
y=
1
= logb ( x )
logx (b)
(b) The log base b to the n of x is 1 over n times the log base b of
x.
1
logbn ( x ) = logb ( x )
n
Answer:
1
1
1
logb ( x ) =
n
n logx (b)
1
=
logx (bn )
= logbn ( x )
3
in class problem set #20
(c) The log base b of x is the log base c of x divided by the log
base c of b.
logc ( x )
logb ( x ) =
logc (b)
Answer: If y = logb ( x ), then by = x. Take the logarithm base c
of this last equation to get
logc ( x ) = logc (by ) = y logc (b)
so that
y=
logc ( x )
logc (b)
4