MAT 106 – 050 TRIGONOMETRY
TEST 2
PRACTICE
WING HONG TONY WONG
Problem 1.
(a) log3 27 =?
(b) log2 64 =?
(c) log4 8 =?
(d) log36 6 =?
1
=?
(e) log5
125
4
(f) log 2 =?
3 9
9
=?
(g) log 4
3 16
3
(h) log 9 =?
25 5
27
(i) log 81
=?
16 8
16
=?
(j) log 125
64 25
Solution.
(d) Let log36 6 = x. Then
36x = 6
62x = 6
2x = 1
1
x=
2
Date: Friday, October 23, 2015.
1
(j) Let log 125
64
16
= x. Then
25
x
125
16
=
64
25
3x 2
5
4
=
4
5
3x = −2
2
x=−
3
Problem 2.
(a) Solve for x if logx 64 = 3.
(b) Solve for x if log3 x = 4.
9
(c) Solve for x if logx
= −2.
16
5
1
(d) Solve for x if logx = .
4
2
4
2
(e) Solve for x if logx
=− .
25
3
Solution.
(b)
34 = x
x = 81
(c)
9
16
16
x2 =
9
4
or
x=
3
x−2 =
−
4
(rejected since x > 0)
3
(e)
4
25
2
25
x3 =
4
1
5
x3 =
2
125
x=
8
2
x− 3 =
2
Problem 3.
(a) Solve for x if log4 x ≥ log4 16.
(b) Solve for x if log 1 (x2 + 15) > log 1 (8x).
3
3
Solution.
(a) Since the base 4 is great than 1, by removing log4 on both sides, we have x ≥ 16.
1
(b) Since the base is less than 1, by removing log 1 on both sides, we have
3
3
x2 + 15 < 8x
x2 − 8x + 15 < 0
(x − 3)(x − 5) < 0
Hence, 3 < x < 5, or equivalently, x is in the interval (3, 5).
Problem 4.
(a) Solve for x if 2 log3 x + log3 5 = log3 80.
3 log5 x
= 2.
(b) Solve for x if
log5 8
Solution.
(a)
2 log3 x + log3 5 = log3 80
log3 x2 + log3 5 = log3 80
log3 5x2 = log3 80
5x2 = 80
x2 = 16
x = 4 or
− 4 (rejected since x > 0)
(b)
3 log5 x
=2
log5 8
log5 x3
=2
log5 8
log8 x3 = 2
82 = x3
x3 = 64
x=4
Problem 5. (10 points)
(a) Express ln 65 + 2 ln 6 − ln 39 as a single logarithm with coefficient 1.
2 log3 5 + 3 log3 2
(b) Express
with coefficient 1.
2 log3 4
3
Solution.
(a)
ln 65 + 2 ln 6 − ln 39
= ln
65 × 62
= ln(5 × 12) = ln 60
39
(b)
2 log3 5 + 3 log3 2
2 log3 4
log 52 + log3 23
= 3
log3 42
log (25 × 8)
= 3
log3 16
= log16 200
Problem 6. (10 point)
(a) Solve for x if 27x+3 = 81x+1 .
(b) Solve for x if 27x+3 > 81x+1 .
Solution.
(a)
27x+3 = 81x+1
33x+9 = 34x+4
3x + 9 = 4x + 4
x=5
(b)
27x+3 > 81x+1
33x+9 > 34x+4
3x + 9 > 4x + 4
x<5
Problem 7. (10 point)
2x−1 x+1
1
1
(a) Solve for x if
=
.
8
16
2x−1 x+1
1
1
(b) Solve for x if
≥
.
8
16
4
Solution.
(a)
2x−1 x+1
1
1
=
8
16
6x−3 4x+4
1
1
=
2
2
6x − 3 = 4x + 4
7
x=
2
(b)
2x−1 x+1
1
1
≥
8
16
6x−3 4x+4
1
1
≥
2
2
6x − 3 ≤ 4x + 4
7
x≤
2
Problem 8. (10 points)
(a) Sketch the graphs of y = log3 x, y = log3 (x + 2), y = log3 (−x + 2), and y = − log3 (x + 2).
(b) Sketch the graphs of y = log 2 x, y = 3 + log 2 x, y = 3 + log 2 (−x), and y = 3 − log 2 x.
3
3
3
3
Problem 9. (10 point)
(a) Solve for x if log3 (x + 1) = 4.
(b) Solve for x if log3 (x + 1) ≤ 4.
Solution.
(a)
log3 (x + 1) = 4
34 = x + 1
x = 80
(b)
log3 (x + 1) ≤ 4
3log3 (x+1) ≤ 34
x + 1 ≤ 81
x ≤ 80
5
Problem 10. (5 point)
(a) Solve for x if log 1 (x − 4) = 2.
3
(b) Solve for x if log 1 (x − 4) < 2.
3
Solution.
(a)
log 1 (x − 4) = 2
3
2
1
=x−4
3
1
x=4+
9
37
=
9
(b)
log 1 (x − 4) < 2
3
log 1 (x−4) 2
1
1
3
>
3
3
1
x−4>
9
37
x>
9
Problem 11. (5 points)
(a) Solve for the domain log5 (x + 5).
(b) Solve for the domain log2 (x2 + 5x + 4).
Solution.
(a)
x+5>0
x > −5
(b)
x2 + 5x + 4 > 0
(x + 4)(x + 1) > 0
Hence, x < −4 or x > −1. Therefore, the domain is (−∞, −4) ∪ (−1, ∞).
Problem 12. (5 points)
Find the inverse function of f (x) = 4x + 9.
6
Solution.
y = 4x + 9
x = 4y + 9
x − 9 = 4y
log4 (x − 9) = log4 4y
log4 (x − 9) = y
f −1 (x) = log4 (x − 9)
Problem 13. (10 points)
(a) Solve for x if 2(32x+1 ) + 9 = 13.
(b) Solve for x if 42x+3 = 7x−1 .
Solution.
(a)
2(32x+1 ) + 9 = 13
2(32x+1 ) = 4
32x+1 = 2
log3 32x+1 = log3 2
2x + 1 = log3 2
2x = log3 2 − 1
log3 2 − 1
x=
2
(b)
42x+3 = 7x−1
(2x + 3) log 4 = (x − 1) log 7
(2 log 4)x + 3 log 4 = (log 7)x − log 7
(2 log 4)x − (log 7)x = − log 7 − 3 log 4
(2 log 4 − log 7)x = − log 7 − 3 log 4
− log 7 − 3 log 4
x=
2 log 4 − log 7
Problem 14. (15 points)
(a) Solve for x if log3 (2x + 6) − log3 (x + 3) = log3 (x + 4).
(b) Solve for x if log(x − 2) − log(x + 4) = log x.
(c) Solve for x if ln(x + 6) − ln(x + 2) = ln x.
7
Solution.
(a)
log3 (2x + 6) − log3 (x + 3) = log3 (x + 4)
2x + 6
log3
= log3 (x + 4)
x+3
log3 2 = log3 (x + 4)
2=x+4
x = −2
Note that x = −2 makes every term in the original equation be well-defined. Hence, x = −2.
(b)
log(x − 2) − log(x + 4) = log x
x−2
log
= log x
x+4
x−2
=x
x+4
x − 2 = x2 + 4x
0 = x2 + 3x + 2
x = −1 or x = −2
Note that both x = −1 and x = −2 makes log x in the original equation be undefined.
Hence, there is no solution.
(c)
ln(x + 6) − ln(x + 2) = ln x
x+6
ln
= ln x
x+2
x+6
=x
x+2
x + 6 = x2 + 2x
0 = x2 + x − 6
0 = (x + 3)(x − 2)
x = −3 or x = 2
Note that x = −3 makes ln x in the original equation be undefined, while x = 2 makes every
term in the original equation be well-defined. Hence, x = 2.
Problem 15. (10 points)
(a) Solve for x if 3−2 log3 x = 25.
(b) Solve for x if e2x − 6ex − 40 = 0.
8
Solution.
(a)
3−2 log3 x = 25
−2
3log3 x
= 25
x−2 = 25
1
x2 =
25
1
1
x=
or x = − (rejected since x > 0)
5
5
(b)
e2x − 6ex − 40 = 0
(ex )2 − 6ex − 40 = 0
(ex − 10)(ex + 4) = 0
ex = 10 or ex = −4 (rejected since ex > 0)
ln ex = ln 10
x = ln 10
Answers
3
1
1
3
1d.
1e. −3
1f. 2
1g. −2
1h.
1i.
2
2
2
4
4
25
125
2
2a. 4
2b. 81
2c.
2d.
2e.
3a. x ≥ 16
1j. −
3
3
16
8
3b. 3 < x < 5 (or using interval notation (3, 5))
4a. 4
4b. 4
5a. ln 60
7
7
5b. log16 200
6a. 5
6b. x < 5
7a.
7b. x ≤
9a. 80
9b. x ≤ 80
2
2
37
37
10a.
10b. x >
11a. (−5, ∞)
11b. (−∞, −4) ∪ (−1, ∞)
9
9
log3 2 − 1
− log 7 − 3 log 4
12. f −1 (x) = log4 (x − 9)
13a.
13b.
14a. −2
2
2 log 4 − log 7
1
14b. No solution
14c. 2
15a.
15b. ln 10
5
1a. 3
1b. 6
1c.
9