1
Exponential and Logarithmic Expressions
1.1
Properties of Logarithms
The logarithm is an exponent. The properties learned regarding exponents
apply to logarithms. For example, 2x · 2y = 2x+y . Since the bases match, the
exponents can be added. This property can be extended to all logarithms in
general.
The three properties, or identities, of logarithms are
1. logb (AB) = logb (A) + logb (B)
A
2. logb ( B
) = logb (A) − logb (B)
3. logb (AC ) = C logb A
Following is a proof as to why the first property is true. The remaining two are
left as exercises. The key is to notice that the numbers A and B can each be
written as an exponential. For example 8 can be written in base 2 as 23 , or 10
can be written in base 2 as 23.32192 . Since each identity is in base b, the problem
will begin by writing both A and B in base b.
Proof of 1, for those that want to know why the properties are true. Let
A = bx and B = by where x and y are any real numbers. Then,
logb (AB) = logb (bx by )
= logb b(x+y)
Consider the problem of simplify logb bc . To solve this think of the problem as
logb bc = t, where t is some other variable. Then, logb bc = t ⇐⇒ bt = bc . Thus
t = c. This indicates that when the base of the logarithm equals the argument
of the logarithm, the result is the exponent. With that in mind,
logb b(x+y) = (x + y)
Finally, we know that an exponential equation has a logarithmic equivalent, so
A = bx ⇐⇒ logb A = x and B = by ⇐⇒ logb B = y respectively, so
x + y = logb A + logb B, as desired.
Proving property 2 and 3 are left to the reader, but the following examples
illustrate using the properties.
1
Example. Use the properties of logarithms to write the logarithm
log3 (3x2 )
as individual variables or numbers.
Solution.
log3 (3x2 ) = log3 (3) + log3 (x2 ) by property 1.
= 1 + log3 (x2 ), since log3 3 = 1
= 1 + 2 log3 x, by property 3.
Example. Use the properties of logarithms to write the logarithm log
ab2
100c3
as individual variables or numbers.
Solution. When the base of the logarithm is not written, it is implied to be
10.
ab2
log
= log(ab2 ) − log(100c3 ), by property 2.
100c3
= log(a) + log(b2 ) − log(100) + log(c3 ) , by property 1.
= log(a) + 2 log(b) − log(100) − log(c3 ),
by property 3 and distributing the −1.
= log(a) + 2 log(b) − log(102 ) − log(c3 ), write 100 in base 10.
= log(a) + 2 log(b) − 2 log(10) − 3 log(c),
by property 3 applied to the last two terms.
= log(a) + 2 log(b) − 2(1) − 3 log(c),
because log 10 = 1.
= log(a) + 2 log(b) − 2 − 3 log(c)
In the above problem, all three properties were used and one logarithm was
simplified in detail. With practice and understand, one may find that fewer
steps are necessary; however, do not rush the process.
Practice Problems 1. Use the properties of logarithms to write each expression in terms of the logarithms of individual variables or numbers.
2
logb (y 4 )
1. logb (5x)
2. logb (2xy)
3.
4. logb (32 x5 )
5. logb (x2 y 3 z)
6. logb ( 5x
3 )
x
7. logb ( yz
)
3
10. logb ( x6y )
√
13. logb ( x)
√ 16. logb 4yx
q 19. logb 3 yx2
1.2
8. logb ( yx3 )
11. logb xy1 3
√
14. logb (x2 y)
√ xy
17. logb z2
9. logb ( 21 )
2
12. logb xy
z3
2y
15. logb √
x
p
18. logb ( x5 )
p 20. logb 2 yz
Properties of Logarithms to Evaluate an Expression
Example. Assume that logb (2) = 0.12, logb (3) = 0.19, and logb (5) = 0.28.
Simplify logb (30).
Solution. At first glance there is no information provided regarding 30; however, notice that 30 = 2 · 3 · 5. Use property 1 above to write the logb 30 as the
logarithm of its factors. Then,
logb 30 = logb (6 · 5),
= logb 6 + logb 5, by property 1.
= logb (2 · 3) + logb 5
= logb 2 + logb 3 + logb 5, by property 1.
=0.12 + 0.19 + 0.28, using the given values.
=0.31 + 0.28
=0.59
Example. Assume
that logb (2) = 0.12, logb (3) = 0.19, and logb (5) = 0.28.
q
5
Simplify logb ( 2 ).
Solution. The strategy is to use the properties of logarithms to separate the
3
logarithm into any of the given values.Then,
r !
21
5
5
logb
= logb
2
2
1
5
= logb
, by property 3.
2
2
1
= (logb 5 − logb 2) , by property 2.
2
1
= (0.28 − 0.12)
2
1
= (0.16)
2
=0.08
Example. Assume that logb (2) = 0.12, logb (3) = 0.19, and logb (5) = 0.28.
Simplify logb (0.1).
Solution. Sometimes a fraction is easier to work with than a decimal. The
value of 0.1 can be written as the fraction of one-tenth, and the 10 in the
denominator is a product of 2 and 5. Then,
1
logb (0.1) = logb
10
−1
= logb (10)
= − 1 logb (10)
= − 1 logb (2 · 5)
= − 1 (logb 2 + logb 5)
= − 1(0.12 + 0.28)
= − 0.4
Practice Problems 2.Given that logb (2) = 0.12, logb (3) = 0.19, and logb (5) =
0.28, and using the properties of logarithms evaluate the following logarithms.
21. logb (6)
23. logb 35
24. logb
25. logb 9
26. logb 20
27. logb
√
22. logb 10
15
2
√
5
29. logb 0.01
28. logb
3
2
30. logb 0.003
4
1.3
Logarithms as a Single Logarithm
Recall the properties of logarithms.
1. logb (AB) = logb (A) + logb (B)
A
2. logb ( B
) = logb (A) − logb (B)
3. logb (AC ) = C logb A
In these exercises the goal is to utilize the left side of the equation to write the
logarithms as a single logarithm.
Example. Use the properties of logarithms to write logb x + 3 logb y − logb z as
a single logarithmic expression.
Solution.
logb x + 3 logb y − logb z = logb x + logb y 3 − logb z,
by property 3 applied to the second term.
= logb (xy 3 ) − logb z,
by property 1 applied to the first two terms.
3
xy
= logb
,
z
by property 2.
Example. Use the properties of logarithms to write
single logarithmic expression.
1
3
logb x − 3 logb z as a
Solution.
1
1
logb x − 3 logb z = logb x 3 − logb z 3
3
by property 3 applied to both terms.
1
x3
z3
by property 2.
√
3
x
= logb
z3
= logb
5
Practice Problems 3. Use the properties of logarithms to write each combination of individual logarithms as a single logarithmic expression.
31. logb 2 + logb x
32. logb (2x − 1) − logb (x − 5)
33. logb a + logb y − logb x
34. logb x − logb y − logb z
35. 4 logb y + logb z
36. logb 6 + 2 logb y
37. 3 logb x − 2 logb y
38.
1
2
logb x − logb 3
+ logb (y))
40.
1
2
logb a − 3 logb z
1
2
42.
2
3
logb (2) − 2 logb x − logb (y)
39.
1
2 (logb (a)
41. 2 logb x +
43.
1
3
logb y − 3 logb z
logb (a) + 3 logb (3) − 2 logb (x)
44. logb (x − 2) + logb (x + 3)
45. logb (x + 1) + logb (x − 1)
46. logb (3x + 2) + logb (2x − 1) − logb x
6
Solutions
Practice Problems 1.
1. logb (5x) = logb 5 + logb x
2. logb (2xy) = logb 2 + logb x + logb y
3. logb (y 4 ) = 4 logb y
4. logb (32 x5 ) = 2 logb 3 + 5 logb x
5. logb (x2 y 3 z) = 2 logb x + 3 logb y + logb z
6. logb ( 5x
3 ) = logb 5 + logb x − logb 3
x
) = logb x − logb y − logb z
7. logb ( yz
8. logb ( yx3 ) = logb x − 3 logb y
9. logb ( 12 ) = − logb 2
11. logb xy1 3 = − logb x − 3 logb y
√
13. logb ( x) = 12 logb x
2y
= logb 2 + logb y − 12 logb x
15. logb √
x
√ xy
17. logb z2 = 12 logb x + 12 logb y − 2 logb z
q 19. logb 3 yx2 = 13 logb x − 32 logb y
3
10. logb ( x6y ) = 3 logb x + logb y − logb 6
2
12. logb xy
= logb x + 2 logb y − 3 logb z
z3
√
14. logb (x2 y) = 2 logb x + 21 logb y
√ 16. logb 4yx = 12 logb x − logb 4 − logb y
logb x −
1
2
logb 5
p 20. logb 2 yz = logb 2 +
1
2
logb y −
18. logb (
px
5)
=
1
2
Practice Problems 2.
21. logb (6) = logb (2 · 3) = logb 2 + logb 3 = 0.12 + 0.19 = 0.31
22. logb 10 = logb (2 · 5) = logb 2 + logb 5 = 0.12 + 0.28 = 0.40
23. logb 35 = logb 3 − logb 5 = 0.19 − 0.28 = −0.09
24. logb
15
2
= logb 15 − logb 2 = logb 3 + logb 5 − logb 2 = 0.19 + 0.28 − 0.12 = 0.35
25. logb 9 = logb 32 = 2 logb 3 = 2(0.19) = 0.38
26. logb 20 = logb (2 · 2 · 5) = logb 2 + logb 2 + logb 5 = 0.12 + 0.12 + 0.28 = 0.52
√
1
27. logb 5 = logb 5 2 = 12 logb 5 = 12 (0.28) = 0.14
√
28. logb
3
2
=
1
2
logb 3 − logb 2 = 12 (0.19) − 0.12 = −0.025
29. logb 0.01 = −2 logb (10) = −2(0.40) = −0.80
3
30. logb 0.003 = logb 1000
= logb 3 − (−3) logb 10 = 0.19 + 3(0.40) = 1.39
Practice Problems 3.
7
1
2
logb z
31. logb 2 + logb x = logb (2x)
32. logb (2x − 1) − logb (x − 5) = logb
2x−1
x−5
33. logb a + logb y − logb x = logb (ay) − logb x = logb
ay
x
34. logb x − logb y − logb z = logb x − (logb y + logb z) = logb x − logb (yz) = logb
x
yz
35. 4 logb y + logb z = logb y 4 + logb z = logb (y 4 z)
36. logb 6 + 2 logb y = logb 6 + logb y 2 = log(6y 2 )
3
37. 3 logb x − 2 logb y = logb x3 − logb y 2 = logb ( xy2 )
1
38.
1
2
39.
1
2 (logb (a)
40.
1
2
logb x − logb 3 = logb x 2 − logb 3 = logb
2
3
1
2
1
logb (ay) = logb (ay) 2 = logb
1
logb a − 3 logb z = logb a 2 − logb z 3 = logb
41. 2 logb x +
42.
+ logb (y)) =
1
2
1
a2
z3
= logb
ay
√
a
z3
1
√
x2 y
z3
2
logb (2) − 2 logb x − logb (y) = logb 2 3 − (logb x2 + logb y)
2
1
3
√
logb y − 3 logb z = logb x2 + logb y 2 − logb z 3 = logb
= logb 2 3 − (logb (x2 y)) = logb
43.
√
x
3
2
23
x2 y
= logb
√
3
4
x2 y
1
logb (a) + 3 logb (3) − 2 logb (x) = logb a 3 + logb 33 − logb x2
1
= logb (a 3 33 ) − logb x2 = logb
√
27 3 a
x2
44. logb (x − 2) + logb (x + 3) = logb [(x − 2)(x + 3)] = logb (x2 + x − 6)
45. logb (x + 1) + logb (x − 1) = logb (x2 − 1)
46. logb (3x + 2) + logb (2x − 1) − logb x = logb (6x2 + x − 2) − logb x = logb
8
6x2 +x−2
x