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CS 173: Discrete Structures, Fall 2011
Quiz 1 Review Solutions
1. Logarithms, Exponents, etc
Simplify the following expressions as much as possible, without using a calculator
(either hardware or software). Do not approximate. Express all rational numbers
as fractions.
Solution
(a) log⌊8.7⌋ (0.125⌈−2.3⌉ ) = log8 ((8−1 )−2 ) = log8 (82 ) = 2
(b) log512 (10243 × 40965) = log512 (230 · 260 ) = log512 (51210 ) = 10
(c)
−1+(log3 (log5 (125k )))
log9 k
(log3 9)(log 3 k)
=2
log3 k
(d)
(23 ×25 )10
512
=
(23+5 )10
29
=
−1+(log3 (k·log5 (125)))
log9 k
=
280
29
=
−1+log3 (3k)
log9 k
=
−1+log3 (3)+log3 k
log9 k
=
log3 k
log9 k
=
= 271
(e) (log2 13)(log13 2048) = (log2 13)(log13 211 ) = 11(log2 13)(log13 2). Using the changeof-base formula, we have (logx a)(loga b) = logx b, so this is 11 log2 2 = 11
(f)
log3 (81k )
7k
=
k log3 (34 )
7k
=
4
7
2. Algebra
(a) Suppose that x2 + 3x − 18 < 0. Which interval(s) of the real line must x belong
to?
(b) Solve the inequality |9 − 2x| < 15.
(c) Solve the inequality (x + y)2 ≤ x2 + y 2 .
(d) Suppose we have the equation 2x + 3 = m
. Solve for x in terms of m and n.
n
Your answer should be a single fraction (not the sum of two terms).
Solution:
(a) First, factor the polynomial: x2 + 3x − 18 < 0 = (x + 6)(x − 3). You could use
the quadratic formula but, in fact, this example is simple enough that it can be
factored by trial and error.
Then we can rewrite the inequality as (x + 6)(x − 3) < 0. We seek a range of
x-values such that this product is negative. Clearly this can occur when x+6 < 0
and x − 3 > 0, or when x + 6 > 0 and x − 3 < 0, but not when both terms are
positive or negative. Furthermore, x cannot be both greater than 3 and less than
−6, so the only region for which the expression is negative is −6 < x < 3.
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(b) We can remove the absolute value to get two equations: 9 − 2x < 15 and −15 <
9 − 2x. Moving terms from one side to another gives us −6 < 2x and 2x < 24.
Dividing by 2, we get that −3 < x < 12.
(c) First, notice that (x + y)2 = x2 + 2xy + y 2. So our inequality is equivalent to
x2 + 2xy + y 2 ≤ x2 + y 2, which is equivalent to 2xy ≤ 0, which is equivalent to
xy ≤ 0. This is true when (a) x or y is zero or (b) one of x and y is negative and
the other is not.
(d) We start with 2x + 3 =
x = m−3n
.
2n
m
.
n
This implies that 2x =
m
n
− 3. So 2x =
m−3n
.
n
So
3. Functions
Let F (x) = x − 6 and G(x) = x2 + 8. Compute the values of the following, simplifying
as much as possible.
(a) F (G(y)) = y 2 + 8 − 6 = y 2 + 2
(b) G(F (x)) = (x − 6)2 + 8 = x2 − 12x + 36 + 8 = x2 − 12x + 44
(c)
F (F (G(2)))
F (π)
=
(((4+8)−6)−6)
π−6
=0
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