Math 105 Spring 2012
Answers for Review Problems for Chapter 5
1. The graph is
2. (a) The graph is
(b) The graph is
(c) The graph is
(d) We need a such that 2x + 2 = −x + a when x = 1. Plugging x = 1 into
2x + 2 = −x + a gives 4 = −1 + a, from which we can get a = 5.
3. (a) 3 − x = e1
(b) 18 = 3x
4. (a) x − 5 = log3 7
(b) x2 = log2 5
5. (a) 3
(b)
1
2
(c) 0
6. (a) 2 ln x + ln y + 2 ln z
(b) log a + log b + log c − log d
√
3
(c) log a2 b = 31 log(a2 b) = 31 (log(a2 ) + log b) = 13 (2 log a + log b)
7. (a) ln(a3 b4 )
3x (b) log
or equivalently log 23 x
2
√x √x √
1
(c) 2 ln x − ln y − 3 ln z = ln x − ln y − ln z 3 = ln
− ln z 3 = ln
y
yz 3
8. (a) x = − 21
(b) x = −2
(c) We use logarithms:
6x = 21
ln(6x ) = ln 21
x ln 6 = ln 21
ln 21
x=
ln 6
(d) First, divide both sides by 4, and then use logarithms as in (c):
4e0.1x = 20
e0.1x = 5
ln(e0.1x ) = ln 5
(0.1x) ln(e) = ln 5
0.1x = ln 5 (because ln(e) = 1)
x = 10 ln 5
(e) Use logarithms to get the variables out of the exponents, then solve the resulting
linear equation:
3x−1 = 2x
ln(3x−1 ) = ln(2x )
(x − 1) ln 3 = x ln 2
x ln 3 − ln 3 = x ln 2
x ln 3 + x ln 2 = ln 3
ln 3
≈ 0.613
x=
ln 3 + ln 2
√
9. (a) x = 81/3 = 3 8
(b) Since ln(3x − 5) = ln x, we have 3x − 5 = x, from which we can get x =
(c) Using the properties of logarithms, we have
5
2
ln(x − 1) + ln(x + 2) = ln 4
ln((x − 1)(x + 2)) = ln 4
Equating the expressions inside the logarithms gives
(x − 1)(x + 2) = 4
x2 + x − 2 = 4
x2 + x − 6 = 0
(x + 3)(x − 2) = 0
x = −3, x = 2
We need to check these solutions in the original equation.
For x = −3, we have
LHS = ln(−3 − 1) + ln(−3 + 2)
= ln(−4) + ln(−1)
We cannot take the logarithm of a negative number, so x = −3 does not work.
For x = 2, we have
LHS = ln(2 − 1) + ln(2 + 2)
= ln(1) + ln(4)
= ln(4)
So, x = 2 works, and is the only solution.
(d) Equating the expressions inside the logarithms gives
x2 − x − 7 = x + 1
x2 − 2x − 8 = 0
(x − 4)(x + 2) = 0
x = 4, x = −2
As in (c), we need to check these solutions.
For x = 4, we have
LHS = log(42 − 4 − 7)
= log(5)
RHS = log(4 + 1)
= log(5)
So, x = 4 works.
For x = −2, we have
LHS = log((−2)2 − (−2) − 7)
= log(−1)
RHS = log(−2 + 1)
= log(−1)
We cannot take the logarithm of a negative number, so x = −2 does not work.
Thus, x = 4 is the only solution.