MATH 136
The Natural Logarithm Function
Exercises
1. Solve for x and simplify completely:
1
(a) 8 e −3x =
8
€
(b) 3 e −x / 2 = 15
(d) 4 ln(x + 6) = 2
(e) ln(4 − 3x) = −5
€
€
2. Evaluate f ′(x) :
 x5 
(b) f (x) = ln 

 sec x 
3
(a) f (x) = 4x ln(6x)

(d) f (x) = ln 

(c) 50e x / 4 = 100
 cos 2 x 

(c) f (x) = ln  8
 x tan 4 x 
 x3 

(e) f (x) = ln
2
 (4 − x) 
x 3 sin x 

cos x 
€
(f) f (x) =
€
ln(4 x)
x5
Solutions
1. (a) e
−3x
=
1
 1
gives −3 x = ln   . Thus,
64
64
1/ 3
1  1
 1  −1
 1
 1
  = ln 4 .
x = − ln
− ln
=
=
−
ln
=
ln
 64 
4
4
3  64 
---------------------------------------------------------------------------------------------------------------------
x
 1
−2
= ln5 . Thus, x = −2 ln 5 = ln 5 = ln   .
25
2
--------------------------------------------------------------------------------------------------------------------(b) e
− x/ 2
= 5 gives −
x
= ln2 and x = 4 ln2 = ln2 4 = ln16 .
4
--------------------------------------------------------------------------------------------------------------------(c) 50e x / 4 €
= 100 gives e x / 4 = 2. So
€
(d) 4 ln(x + 6) €
So x + 6 = e 1/2 and x = e1/ 2 − 6 .
= 2 gives ln(x + 6) = 1 / 2 . €
€
--------------------------------------------------------------------------------------------------------------------−5
€ and x = 4 − e .
(e) ln(4 − 3x) = −5 gives 4 − 3x = e−5 . So 4 − e−5 = 3x
3
---------------------------------------------------------------------------------------------------------------------
€
 6
€
€
2. (a) Apply product rule: f ′(x) = 12x 2 ln(6x) + €
4x 3   = 12x 2 ln(6x) + 4x 2 .
6x
---------------------------------------------------------------------------------------------------------------------
5 sec x tan x
5
= − tan x .
−
x
sec x
x
---------------------------------------------------------------------------------------------------------------------
(b) f (x) = 5 ln x − ln sec x . Then f ′(x) =
(c) f (x) = 2 ln( cos x ) − 8 ln x − 4 ln( tan x ) .
(− sin x) 8
sec2 x
8 4sec 2 x
= −2tan x − −
.
− −4
cos x
x
tan x
x
tan x
--------------------------------------------------------------------------------------------------------------------1
1
(d) f (x) = ln x + ln sin x − ln cos x . Then
2
3
1 1 1 cos x −sin x 1 1
−
=
+ cot x + tan x .
f ′(x) = × + ×
2 x 3 sin x cos x 2x 3
-------------------------------------------------------------------------------------------------------------------- x3 
3
2
3
2
 = 3ln x − 2ln 4 − x . Then f ′(x) = −
(e) f (x) = ln
.
(−1) = +
2
x 4−x
x 4−x
 (4 −€x) 
--------------------------------------------------------------------------------------------------------------------4
× x 5 − ln(4 x) × 5x 4 1− 5ln(4 x)
= €6
(f) f ′(x) = 4 x
10
x
x
Then f ′(x) = 2
€
€