MATH 31B BRIDGE PROGRAM: HW 3 SOLUTIONS
JOE HUGHES
5
6. log2 (8 3 ) =
5
3
log2 (8) =
5
3
· 3 = 5.
1
7. 64 = 43 , so 4 = 64 3 . Therefore log64 4 = 31 .
15. 7log7 (29) = 29 because 7x and log7 x are inverses.
2
21. If 2x −2x = 8, then after taking the logarithm to base 2 of each side, we get x2 −2x = 3.
Therefore x = 3 or x = −1.
24. The equation log3 y + 3 log3 (y 2 ) = 14 can be written as 7 log3 (y) = 14 by the rules of
logarithms. Therefore log3 (y) = 2, or y = 32 = 9.
25. The simplest example is probably a = 1, b = e. Then ln(ab) = ln(e) = 1, but
ln(a) ln(b) = 0 · 1 = 0.
29.
d
dx x ln(x)
39.
d
dx
= ln(x) +
ln(ln(x)) =
68. If y =
x(x+1)3
,
(3x−1)2
x
x
1 d
ln(x) dx
= ln(x) + 1.
ln(x) =
1
x ln(x) .
then
ln(y) = ln(x) + 3 ln(x + 1) − 2 ln(3x − 1)
Therefore
d
x(x + 1)3 h 1
3
6 i
dy
=y
ln(y) =
+
−
dx
dx
(3x − 1)2 x x + 1 3x − 1
87. Take u = 2x + 4, so du = 2 dx and
Z
Z
dx
1
du
1
1
=
= ln(u) + C = ln(2x + 4) + C
2x + 4
2
u
2
2
89. Set u = t2 + 4, then du = 2t dt and
Z
Z
t
1
du
1
1
dt =
= ln(u) + C = ln(t2 + 4) + C
2
t +4
2
u
2
2
90. Set u = x3 + 2, then du = 3x2 dx and
Z
Z
x2
1
du
1
1
dx =
= ln(u) + C = ln(x3 + 2) + C
x3 + 2
3
u
3
3
1
2
102.
JOE HUGHES
R
3x dx =
3x = eln(3)x .
3x
ln(3)
+ C, either by remembering the rule for
R
bx dx or by writing