3
Question: Expand ln( yx5 z7 ).
Solution:
3
ln( yx5 z7 )
= ln(x3 ) − ln(y 5 z 7 )
= ln(x3 ) − [ln(y 5 z 7 )]
= ln(x3 ) − [ln(y 5 ) + ln(z 7 )]
= ln(x3 ) − ln(y 5 ) − ln(z 7 )
= 3 ln(x) − 5 ln(y) − 7 ln(z)
Question: Express as a single logarithm: ln 3 +
1
3
ln 8.
Solution:
ln 3 + 31 ln 8
1
= ln 3 + ln(8 3 )
1
= ln(3(8 3 ))
√
= ln(3( 3 8))
= ln(3(2))
= ln(6)
Question: Find the limit: limx→∞ [ln(10 + x) − ln(4 + x)].
Solution:
limx→∞ [ln(10 + x) − ln(4 + x)]
Notice that just evaluating each limit and subtracting gives ∞ − ∞, which is undefined.
We’ll try and work around this as follows:
= limx→∞ [ln( 10+x
4+x )]
Since ln(x) is continuous for x > 0, the limit and the ln commute:
= [ln(limx→∞
10+x
4+x )]
Multiplying both numerator and denominator by
= [ln(limx→∞
lim
10
x +1
4
x +1
10
x +1
4
x→∞ x +1
= [ln( limx→∞
)]
)]
= [ln( 11 )]
= [ln(1)]
=0
1
1
x:
Question: Differentiate: f (x) = ln(sin2 x).
Solution:
f 0 (x) =
2
d
dx [ln(sin
x)]
f 0 (x) =
1
d
[sin2
sin2 x dx
f 0 (x) =
1
d
[sin x]
2 sin x dx
sin2 x
f 0 (x) =
1
2 sin x cos x
sin2 x
x]
Question: Use logarithmic differentiation to differentiate: y =
(x+1)4 (x−5)3
.
(x−3)8
Solution: 4
(x−5)3
y = (x+1)
(x−3)8
4
3
(x−5)
ln(y) = ln( (x+1)
)
(x−3)8
ln(y) = ln((x + 1)4 (x − 5)3 ) − ln((x − 3)8 )
ln(y) = ln((x + 1)4 ) + ln((x − 5)3 ) − ln((x − 3)8 )
ln(y) = 4 ln((x + 1)) + 3 ln((x − 5)) − 8 ln((x − 3))
d
dx [ln(y)]
=
d
dx [4 ln((x
d
dx [4 ln((x
+ 1)) + 3 ln((x − 5)) − 8 ln((x − 3))]
1 d
y dx [y]
=
1 d
y dx [y]
d
d
d
= 4 dx
[ln((x + 1))] + 3 dx
[ln((x − 5))] − 8 dx
[ln((x − 3))]
1 d
y dx [y]
1
d
1 d
1 d
= 4 (x+1)
dx [x + 1] + 3 x−5 dx [x − 5] − 8 x−3 dx [x − 3]
1 d
y dx [y]
1
1
1
= 4 (x+1)
+ 3 (x−5)
− 8 (x−3)
+ 1))] +
d
dx [3 ln((x
d
dx [y]
1
1
1
= y[4 (x+1)
+ 3 (x−5)
− 8 (x−3)
]
d
dx [y]
=
(x+1)4 (x−5)3
1
[4 (x+1)
(x−3)8
− 5))] −
d
dx [8 ln((x
− 3))]
1
1
+ 3 (x−5)
− 8 (x−3)
]
Question: Evaluate the integral:
R3
1
dx.
0 5x+1
Solution:
R3 1
dx
0 5x+1
Try a substitution u = 5x + 1; then
=
du
dx
= 5, and dx =
R x=3
1
du
x=0 5u
=
1
5
R x=3
=
1
5
[ln |u|]x=0
=
1
5
[ln |5x + 1|]x=0
1
du
x=0 u
x=3
x=3
= 15 {ln |16| − ln |1|}
=
1
5
ln |16|
2
du
5 :
=
1
5
ln 16
3