11.4 Properties of Logarithms
The next thing we want to do is talk about some of the properties that are inherent to logarithms.
Properties of Logarithms
1. loga(uv) = loga u + loga v
2. loga(u/v) = loga u – loga v
n
3. logau = n logau
1. ln(uv) = ln u + ln v
2. ln(u/v) = ln u – ln v
n
3. ln u = n ln u
There are several applications to these properties that we will need later. Some of which will
appear in the next section.
However, currently we just want a familiarity with the properties and we want to perform two basic
types problems: expanding and condensing logarithms.
Before we get to the expanding and condensing problems, we want to first notice that the
properties take the operations of multiplying, dividing or exponents from the inside of a logarithms
and turn them into adding, subtracting or coefficients on the outside of the logarithm, or vice
versa.
Example 1:
Condense the following logarithms.
a. 4 ln 2 + 2 ln x – ln y
b. 1/5 [ln x – 2ln (x + 4)]
c. ¼ logx 3 – ½ logx y – logx z
Solution:
By condense the log, we really mean write it as a single logarithm with coefficient of 1. So we will
need to use the properties above to condense these logarithms.
a. The first thing we must do is move the coefficients from the front into the exponents by using
property 3. This gives us
4
2
4 ln 2 + 2 ln x – ln y = ln 2 + ln x – ln y
Now by properties 1 and 2 we can condense the addition and subtractions into
multiplication and division. We get
4
2
4 2
ln 2 + ln x – ln y = ln 2 x – ln y
So, 4 ln 2 + 2 ln x – ln y = ln
= ln
24 x 2
y
= ln
16 x 2
y
16 x 2
y
b. This time we need to follow the order of operations a little closer. So we must start inside the
brackets and work our way out. This means that the 1/5 will be the last thing we take care of.
So proceeding like part a. we condense the coefficient then the subtraction as follows
2
1/5 [ln x – 2ln (x + 4)] = 1/5 [ln x – ln (x + 4) ]
x
= 1/5 ln
 x  4 2
Now we can deal with the 1/5. Pulling it up as an exponent we get
1/5 ln
x
 x  4 2
= ln

= ln
5

1
x
 x  4 2
x
 x  4 2
5
th
since 1/5 power is the same as a 5 root.
So 1/5 [ln x – 2ln (x + 4)] = ln
c.
5
x
 x  4 2
Finally, we proceed as we did in the two previous examples. We will turn the coefficients into
exponents and then condense the subtraction. We get
1/4
1/2
¼ logx 3 – ½ logx y – logx z = logx 3 – logx y – logx z
1
34
- logx z
1
y 2
= logx
1
34
1
y 2
= logx
which we need to simplify.
z
So we recall to simplify a complex fraction we multiply numerator
1/2
and denominator by the lcd. That is y here. This gives us
1
34
1
1
34
y 2
logx
= logx 1
z
y 2z
4
= logx
4
So ¼ logx 3 – ½ logx y – logx z = logx
3
z y
3
z y
Example 2:
Expand the following logarithms.
a.
ln 3
x2
x 1
b. log4[4x(x - 5)]
2
c. log3
9
x2 y
Solution:
By expand the logarithm, we basically mean we want to write it out as it was given to us in
example 1. That is, if there is any property that can be done two write the logarithm, then we
must do it. We basically perform expanding by doing condensing backwards.
 x2 
x2

a. First, we need to change the radical into a rational exponent. ln 3
 ln 
x 1
 x  1
1
3
Now we pull the exponent out to the front as a coefficient and use property 2 to turn the
division inside the ln into a subtraction outside. We must be careful to leave the coefficient
on the entire expression. This gives us
 x2 

ln 
 x 1
1
3
1
x2
 ln
3 x 1

Finally, we can pull the 2
nd


1
ln x 2  ln x  1
3
power to the front of the lnx. So we have
ln 3
x2
1
 2 ln x  ln x  1
x 1 3
Notice that we cannot do any expanding to ln (x + 1) since the properties clearly state
that it is addition outside not addition inside that can be changed.
b. Again we will start with pulling the exponent to the front as a coefficient, then use property 2
to expand the multiplications out.
2
log4[4x(x - 5)] = 2 log4[4x(x - 5)]
= 2 (log4 4 + log4 x + log4 (x - 5))
Now we notice that we have the expression log4 4. If we evaluate this we get 1.
Therefore what we really have is
2
log4[4x(x - 5)] = 2 (1 + log4 x + log4 (x - 5))
= 2 + 2log4 x + 2 log4 (x - 5)
c.
Lastly, we will change the radical to a rational exponent and go about the problem in the
same fashion as the previous examples.
9
log3
x
2
= log3
y
9
1
x y2
2
2 1/2
= log3 9 – log3 x y
2
1/2
= log3 9 – [log3 x + log3 y ]
= log3 9 – [2 log3 x + ½ log3 y]
= 2 – 2 log3 x - ½ log3 y
11.4 Exercises
Condense.
1. log2 x + log2 y
4. ln u + ln x
7. 2 log4 p + log4(q + 1)
10. 5 log3(a – 1) – log3 b
13. log2 3 + log2(x + 4) – log2 y
16. 3 log u + 3 log v + 2 log w
19. ½ log2 x – 2 log2(y + 1)
2. log3 y – log3 z
5. ln x – ln y
8. log5 x – 4 log5 m
11. 4 log5 b + 2 log5 c
14. log5 4 – log5 a + log5 b
17. log7 n – 4 log7 3 + 2 log7 t
20. ln(p + r) + 1 3 ln(q – r)
3. 3 log5 v
6. 7 log4 x
9. ln x – 4 ln y
12. 3 ln(a + c) + 4 ln(c – d)
15. 2 ln x + 3 ln y + ln z
18. 3 log3 y – 2 log3 z + log3 x
21. ¼ ln(x – y) – 3 ln y
22. 2(log4 x + log4 y) – ½ log4 z
23. 1 3 log2 n + 2(log2 m – 3 log2 l)
24. 4(ln x + ln y + ln z)
25. ½(ln x + ln y – ln z)
26. 4 log6 u – ½ log6 v + 2 log6 r
27. 3 log3 l + 1 3 log3 w - 2 log3 h
28. log2 x – log2(y – 1) – log2(z + 3)
29. 4 log5(p + q) – log5(q – r) – log5 r
31.
1
33.
1
35.
1
30. ½ (ln(u+1) – ln(v-1)) +
1
3
3
(log4 5 + log4(x + y)) – 2 log4 7
32. 2 log2 x + 3 log2 y – ½ (log2 a + log2 b)
3
(log4 x + log4 y) – 2 (log4 a + log4 b)
34. ½ log3 u +
log2 x – 2 (log2 n – 4 log2 m)
36. 2 log3 x – ½ (log3 y + 5 log3 z)
3
37. ½ (ln x + 3 ln y) –
1
3
ln z
38.
1
3
1
3
(log3 v – log3 w) + log3 x
[ln u – 2 ln v – ln x]
39. ln(a + 1) – 3 ln(b – 1) –3 (ln(c + 2) – ln(d – 2))
40. 2 ln(w – 5) – ½ (ln(x + y) – ln(x – y)) – 3 ln z
1
Expand.
41.
ln w
log 2 xy
42.
log 3
x
y
43.
log 4 x 3
x
y
44.
log 5 nm
45.
ln
47.
log 5 x 2 y
48.
log 2 yz 2
50.
u2
ln 3
v
51.
53.
log 7
56.
log 3 xy
57.
ln x y
59.
log 3
u v
w2
60.
log 4
62.
log 3
65.
log 3
68.
log 3
71.
uv 2
log 3
w
x2
54. log 6
pq 2
x
yz
n
3
m r
3
x 1
27 x 2
ln 3
66.
log 5
69.
log 2
y
ln ex  1
72.

77.

ln ex  4e
80.
ln
log 4 x 2  1
2
49.
ln
52.
x3 y
log 4 2
z
55.
log 2 x  y
58.
ln
61.
log 4
64.
log 2 2 xx  1
67.
log 4 16uu  v 
70.
ln e 2 x
73.
log 6 216x  h
76.
log 2 4 x 2  12 x  8
79.
log 2
pq
5
xy
x 1
3
2
49
ab
log 7
Factor the inside of the logarithm. Then expand.
74.
ln x 7

x
y2
y
u
2
3
n
xy
ab
63.
46.

75.
ln x 2  2 x  1
78.
log 2
32 x
x2 1
xy
z
2


x 2  2x  1
16
x 2  2x  1
x 2  2x  1
log A  a , log B  b , and logC  c to evaluate the
logarithmic expression in terms of a , b , and c .
2
3
81. log AB
82. log B
83. log A C
For Question 81 – 85 use the fact that
84.
log
 
AB
C2
85.
log
 
A
B 3C
For questions 86 – 90 use the fact that logb 5  161
. ,
evaluate the following logarithmic expressions.
86. logb 15
87. logb 49
89.
log

3
b 5
90.
log
 
49
b 125
to
logb 3  0.792 , and logb 7  1404
.
88.
logb 105