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FAMOUS LOG PROPERTIES
THE IDEA
In the last section, we introduced some of the motivation behind the log functions, roughly
speaking, these were invented to ease computation. At the heart of such utility is, for example,
the fact that to multiply numbers with the same base it is enough [and it is easier] to just add
the exponents.
Here we formally state this and other famous log properties. We prove at lease some cases of these
properties and we consider many examples which make use of such properties. We will assume
the domain for all our log functions1 is R>0 , the co-domain is R, and all bases are non-negative
real numbers, and we will assume all variables come from the corresponding sets.
Famous Log Properties
Some properties about logs, assume a, b, c, k ∈ R>0 and n ∈ R:
• Log of a Product [LP=SL] :
• Change of Base Property [CBP] :
logc (ab) = logc a + logc b
logb k =
• Bring It Down [BID] :
logc an = n logc a
logc k
logc b
• Composition Inverses [CInv.] :
• Log of a Quotient [LQ=DL] :
a
= logc a − logc b
logc
b
logb (b)k = k
and
blogb k = k
EXAMPLEs Using [LP=SL] to EXPAND
Often it is useful to have full command of all log properties to expand expression into simpler
terms. Assume all input values on all log functions represent positive real numbers.
1. Expand using [LP=SL]:
log2 (4 · x)
Solution:
log2 (4 · x) = log2 (4) + log2 (x)
(LP=SL)
2. Expand using [LP=SL]:
log2 (4 · 32)
Solution:
1
1
It is possible to extend the domain to complex numbers, but such extension is beyond the scope of our class
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log2 (4 · 32) = log2 (4) + log2 (32)
(LP=SL)
EXAMPLEs Using [LP=SL] to CONDENSE
1. Condense using [LP=SL]:
log2 (4) + log2 (x)
Solution:
log2 (4) + log2 (x) = log2 ((4) · (x))
= log2 4x
(LP=SL)
(BI)
2. Condense using [LP=SL]:
log2 (4) + log2 (32)
Solution:
log2 (4) + log2 (32) = log2 ((4) · (32))
= log2 128
(LP=SL)
(BI)
3. Condense
log(x2 − 1) − log(x − 1)
solution:
x2 −1
x−1
log (x−1)(x+1)
x−1
log(x2 − 1) − log(x − 1) = log
=
= log (x + 1)
LQ=DL
alg
alg, so long as x 6= 1
EXAMPLEs Using [BID] to Simplify
1. Expand using [LP=SL]:
log2 (4 · x)
Solution:
log2 (4 · x) = log2 (4) + log2 (x)
2
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2. Expand using [LP=SL]:
log2 (4 · 32)
Solution:
log2 (4 · 32) = log2 (4) + log2 (32)
(LP=SL)
EXAMPLE calculate NON-FAMOUS logs
Suppose we wanted to calculate ln 23, the natural way to do this is by using calculator with a
built in ln function. Similarly for any common-base log such as log 57. However, it get more
interesting when we need to calculate a log to some other non-famous base such as base 3. The
key to calculate any log base 3 is to convert it to base 10 or base e using the change of base
formula.
log3 100 =
ln 100
ln 3
CBP
≈
4.605
1.099
calc.
≈ 4.191
3
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Assume a, b, c, k ∈ R>0 and n ∈ R:
1. Expand using [LP=SL]:
log3 (6 · x)
Solution:
log3 (6 · x) = log3 (6) + log3 (x)
(LP=SL)
2. Expand using [LP=SL]:
log12 (4 · 32)
Solution:
log12 (4 · 32) = log12 (4) + log12 (32)
(LP=SL)
3. Expand using [LP=SL]:
log5 x3 · y 7
Solution:
log5 x3 · y 7 = log5 x3 + log5 y 7
(LP=SL)
4. Expand using [LP=SL]:
loge 5 · y 2
Solution:
loge 5 · y 2 = loge (5) + loge y 2
(LP=SL)
(recall notation: loge x = ln x)
4
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5. Expand using [LP=SL]:
log3 23 · 57
Solution:
log3 23 · 57 = log3 23 + log3 57
(LP=SL)
6. Expand using [LP=SL]:
log5 4 · x2
Solution:
log5 4 · x2 = log5 (4) + log5 x2
(LP=SL)
7. Use CBF and a scientific calculator to compute
log3 (10)
Solution: We could use the natural base 10 since the log function is on most scientific calculators...
log10 (10)
log10 (3)
log(10)
=
log(3)
1
≈
0.4771
≈ 2.0959
log3 (10) =
(CBF)
(def of natural log ’log’)
(Calc)
(Calc)
ALTERNATIVELY,
5
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We could also use the natural base e since the ln function is also on most scientific calculators...
loge (10)
loge (3)
ln(10)
=
ln(3)
2.3026
≈
1.0986
≈ 2.0959
log3 (10) =
(CBF)
(def of ln)
(Calc)
(Calc)
8. Use CBF and a scientific calculator to compute
log5 (10)
Solution: We could use the natural base 10 since the log function is on most scientific calculators...
log10 (10)
log10 (5)
log(10)
=
log(5)
1
≈
0.699
≈ 1.4307
log5 (10) =
(CBF)
(def of natural log ’log’)
(Calc)
(Calc)
ALTERNATIVELY,
We could also use the natural base e since the ln function is also on most scientific calculators...
loge (10)
loge (5)
ln(10)
=
ln(5)
2.3026
≈
1.6094
≈ 1.4307
log5 (10) =
(CBF)
(def of ln)
(Calc)
(Calc)
9. Use CBF and a scientific calculator to compute
log3 (1000)
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Solution: We could use the natural base 10 since the log function is on most scientific calculators...
log10 (1000)
log10 (3)
log(1000)
=
log(3)
3
≈
0.4771
≈ 6.2878
log3 (1000) =
(CBF)
(def of natural log ’log’)
(Calc)
(Calc)
ALTERNATIVELY,
We could also use the natural base e since the ln function is also on most scientific calculators...
loge (1000)
loge (3)
ln(1000)
=
ln(3)
6.9078
≈
1.0986
≈ 6.2878
log3 (1000) =
(CBF)
(def of ln)
(Calc)
(Calc)
10. Use CBF and a scientific calculator to compute
log2 (500)
Solution: We could use the natural base 10 since the log function is on most scientific calculators...
log10 (500)
log10 (2)
log(500)
=
log(2)
2.699
≈
0.301
≈ 8.9664
log2 (500) =
(CBF)
(def of natural log ’log’)
(Calc)
(Calc)
ALTERNATIVELY,
7
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We could also use the natural base e since the ln function is also on most scientific calculators...
loge (500)
loge (2)
ln(500)
=
ln(2)
6.2146
≈
0.6931
≈ 8.9664
log2 (500) =
(CBF)
(def of ln)
(Calc)
(Calc)
11. Use CBF and a scientific calculator to compute
log7 (10)
Solution: We could use the natural base 10 since the log function is on most scientific calculators...
log10 (10)
log10 (7)
log(10)
=
log(7)
1
≈
0.8451
≈ 1.1833
log7 (10) =
(CBF)
(def of natural log ’log’)
(Calc)
(Calc)
ALTERNATIVELY,
We could also use the natural base e since the ln function is also on most scientific calculators...
loge (10)
loge (7)
ln(10)
=
ln(7)
2.3026
≈
1.9459
≈ 1.1833
log7 (10) =
(CBF)
(def of ln)
(Calc)
(Calc)
12. Use CBF and a scientific calculator to compute
log30 (10)
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Solution: We could use the natural base 10 since the log function is on most scientific calculators...
log10 (10)
log10 (30)
log(10)
=
log(30)
1
≈
1.4771
≈ 0.677
(CBF)
log30 (10) =
(def of natural log ’log’)
(Calc)
(Calc)
ALTERNATIVELY,
We could also use the natural base e since the ln function is also on most scientific calculators...
loge (10)
loge (30)
ln(10)
=
ln(30)
2.3026
≈
3.4012
≈ 0.677
log30 (10) =
(CBF)
(def of ln)
(Calc)
(Calc)
13. Use CBF and a scientific calculator to compute
log5 (1000)
Solution: We could use the natural base 10 since the log function is on most scientific calculators...
log10 (1000)
log10 (5)
log(1000)
=
log(5)
3
≈
0.699
≈ 4.2922
log5 (1000) =
(CBF)
(def of natural log ’log’)
(Calc)
(Calc)
ALTERNATIVELY,
9
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We could also use the natural base e since the ln function is also on most scientific calculators...
loge (1000)
loge (5)
ln(1000)
=
ln(5)
6.9078
≈
1.6094
≈ 4.2922
log5 (1000) =
(CBF)
(def of ln)
(Calc)
(Calc)
14. Use CBF and a scientific calculator to compute
log7 (500)
Solution: We could use the natural base 10 since the log function is on most scientific calculators...
log10 (500)
log10 (7)
log(500)
=
log(7)
2.699
≈
0.8451
≈ 3.1937
log7 (500) =
(CBF)
(def of natural log ’log’)
(Calc)
(Calc)
ALTERNATIVELY,
We could also use the natural base e since the ln function is also on most scientific calculators...
loge (500)
loge (7)
ln(500)
=
ln(7)
6.2146
≈
1.9459
≈ 3.1937
log7 (500) =
(CBF)
(def of ln)
(Calc)
(Calc)
15. Expand using [LP=SL]:
log3
10
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Solution:
log3
2
x2
= log3 (2) − log3 x2
(LQ=dL)
16. Expand using [LP=SL]:
5
log3
7
Solution:
5
= log3 (5) − log3 (7)
log3
7
(LQ=dL)
17. Expand using [LP=SL]:
log3
x3
y2
Solution:
log3
x3
y2
= log3 x3 − log3 y 2
18. Expand using [LP=SL]:
log3
(LQ=dL)
xr z5
Solution:
log3
xr z5
= log3 (xr) − log3 z 5
(LQ=dL)
19. Expand using [LP=SL]:
log3
11
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Solution:
log3
5
11
= log3 (5) − log3 (11)
(LQ=dL)
20. Expand using [LP=SL]:
log3
x
y6
Solution:
log3
x
y6
= log3 (x) − log3 y 6
(LQ=dL)
21. Expand using [LP=SL]:
R
log3
z
Solution:
R
log3
= log3 (R) − log3 (z)
z
(LQ=dL)
22. Expand using [LP=SL]:
log5 (4 · 1)
Solution:
log5 (4 · 1) = log5 (4) + log5 (1)
(LP=SL)
23. Expand using [LP=SL]:
log5 (1 · 4)
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Solution:
log5 (1 · 4) = log5 (1) + log5 (4)
(LP=SL)
24. Expand using [LP=SL]:
log2 x5 · y 2
Solution:
log2 x5 · y 2 = log2 x5 + log2 y 2
(LP=SL)
25. Expand using [LP=SL]:
log6 5 · x2
Solution:
log6 5 · x2 = log6 (5) + log6 x2
(LP=SL)
26. Expand using [LP=SL]:
log3 (6 · xy)
Solution:
log3 (6 · xy) = log3 (6) + log3 (xy)
(LP=SL)
27. Expand using [LP=SL]:
log5 4 · x2 y 3
Solution:
log5 4 · x2 y 3 = log5 (4) + log5 x2 y 3
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28. Expand using [LP=SL]:
log5 4x2 · y 3
Solution:
log5 4x2 · y 3 = log5 4x2 + log5 y 3
(LP=SL)
29. Expand using [LP=SL]:
log3 23 · 57
Solution:
log3 23 · 57 = log3 23 + log3 57
(LP=SL)
30. Expand using [LP=SL]:
log2 x5 · y 2
Solution:
log2 x5 · y 2 = log2 x5 + log2 y 2
(LP=SL)
31. Expand using [LP=SL]:
log6 5 · x2
Solution:
log6 5 · x2 = log6 (5) + log6 x2
(LP=SL)
32. Expand using [LP=SL]:
log6 ((x − 2) · (x + 2))
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Solution:
log6 ((x − 2) · (x + 2)) = log6 ((x − 2)) + log6 ((x + 2))
(LP=SL)
33. Expand using [LP=SL]:
log6 x2 · (x + 2)
Solution:
log6 x2 · (x + 2) = log6 x2 + log6 ((x + 2))
(LP=SL)
34. Expand using [LP=SL]:
log2 x5 · (3x + 1)
Solution:
log2 x5 · (3x + 1) = log2 x5 + log2 ((3x + 1))
(LP=SL)
35. Condense using [LP=SL]:
log3 23 + log3 22
Solution:
log3 23 + log3 22 = log3 (23 ) · (22 )
= log3 32
(LP=SL)
(BI)
36. Condense using [LP=SL]:
log3 25 + log3 25
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Solution:
log3 25 + log3 25 = log3 (25 ) · (25 )
= log3 1024
(LP=SL)
(BI)
37. Condense using [LP=SL]:
log3 x3 + log3 x2
Solution:
log3 x3 + log3 x2 = log3 (x3 ) · (x2 )
= log3 x5
(LP=SL)
(BI)
38. Condense using [LP=SL]:
log3 x5 + log3 x5
Solution:
log3 x5 + log3 x5 = log3 (x5 ) · (x5 )
= log3 x10
(LP=SL)
(BI)
39. Condense using [LP=SL]:
log5 x2 − 1 + log5 x2 + 1
Solution:
log5 x2 − 1 + log5 x2 + 1 = log5 (x2 − 1) · (x2 + 1)
= log5 x4 − 1
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40. Condense using [LP=SL]:
log3 (2x + 3) + log3 (2x − 3)
Solution:
log3 (2x + 3) + log3 (2x − 3) = log3 ((2x + 3) · (2x − 3))
= log3 4x2 − 9
(LP=SL)
(BI)
41. Condense using [LP=SL]:
log5 (x − 1) + log5 (x + 3)
Solution:
log5 (x − 1) + log5 (x + 3) = log5 ((x − 1) · (x + 3))
= log5 x2 + 2x − 3
(LP=SL)
(BI)
42. Condense using [LP=SL]:
log5 (2x − 1) + log5 (5x + 3)
Solution:
log5 (2x − 1) + log5 (5x + 3) = log5 ((2x − 1) · (5x + 3))
= log5 10x2 + x − 3
(LP=SL)
(BI)
43. Condense using [LP=SL]:
log5 (x − 1) + log5 x3 + x2 + x + 1
Solution:
log5 (x − 1) + log5 x3 + x2 + x + 1 = log5 (x − 1) · (x3 + x2 + x + 1)
= log5 x4 − 1
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44. Condense using [LP=SL]:
log5 (x − 1) + log5 x4 + x3 + x2 + x + 1
Solution:
log5 (x − 1) + log5 x4 + x3 + x2 + x + 1 = log5 (x − 1) · (x4 + x3 + x2 + x + 1) (LP=SL)
(BI)
= log5 x5 − 1
45. Condense using [LP=SL]:
loge (x − 1) + loge x2 + x + 1
Solution:
loge (x − 1) + loge x2 + x + 1 = loge (x − 1) · (x2 + x + 1)
= loge x3 − 1
(LP=SL)
(BI)
(recall notation: loge x = ln x)
46. Condense using the log properties where appropriate:
log2 x2 y 3 + log2 x2 y 3
Solution:
log2 x2 y 3 + log2 x2 y 3 = log2 x2 y 3 · x2 y 3
= log2 x2 x2 y 3 y 3
= log2 x2+2 y 3+3
4 6
= log2 x y
(LP=SL)
(CoLM, ALM)
(JAE)
(BI)
OR an alternative idea that helps practice BID...
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(LP=SL)
log2 x2 y 3 + log2 x2 y 3 = log2 x2 + log2 y 3 + log2 x2 + log2 y 3
= 2 log2 (x) + 3 log2 (y) + 2 log2 (x) + 3 log2 (y)
(BID)
= 2 log2 (x) + 2 log2 (x) + 3 log2 (y) + 3 log2 (y)
(ALA, CoLA)
= (2 + 2) log2 (x) + (3 + 3) log2 (y)
(DL)
= 4 log2 (x) + 6 log2 (y)
(BI)
(BID, ’backwards’)
= log2 x4 + log2 y 6
4 6
(LP=SL)
= log2 x y
47. Condense using the log properties where appropriate:
log2 x2 y 2 + log2 x5 y −3
Solution:
log2 x2 y 2 + log2 x5 y −3 = log2 x2 y 2 · x5 y −3
= log2 x2 x5 y 2 y −3
= log2 x2+5 y 2+−3
7 −1
= log2 x y
(LP=SL)
(CoLM, ALM)
(JAE)
(BI)
OR an alternative idea that helps practice BID...
(LP=SL)
log2 x2 y 2 + log2 x5 y −3 = log2 x2 + log2 y 2 + log2 x5 + log2 y −3
= 2 log2 (x) + 2 log2 (y) + 5 log2 (x) + −3 log2 (y)
(BID)
= 2 log2 (x) + 5 log2 (x) + 2 log2 (y) + −3 log2 (y) (ALA, CoLA)
= (2 + 5) log2 (x) + (2 + −3) log2 (y)
(DL)
= 7 log2 (x) + − 1 log2 (y)
(BI)
(BID, ’backwards’)
= log2 x7 + log2 y −1
7 −1
(LP=SL)
= log2 x y
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48. Condense using the log properties where appropriate:
log2 x3 y −5 + log2 x−4 y −4
Solution:
log2 x3 y −5 + log2 x−4 y −4 = log2 x3 y −5 · x−4 y −4
= log2 x3 x−4 y −5 y −4
= log2 x3+−4 y −5+−4
−1 −9
= log2 x y
(LP=SL)
(CoLM, ALM)
(JAE)
(BI)
OR an alternative idea that helps practice BID...
(LP=SL)
log2 x3 y −5 + log2 x−4 y −4 = log2 x3 + log2 y −5 + log2 x−4 + log2 y −4
= 3 log2 (x) + −5 log2 (y) + −4 log2 (x) + −4 log2 (y)
(BID)
= 3 log2 (x) + −4 log2 (x) + −5 log2 (y) + −4 log2 (y)
(ALA, CoLA)
= (3 + −4) log2 (x) + (−5 + −4) log2 (y)
= − 1 log2 (x) + − 9 log2 (y)
−1
+ log2 y −9
= log2 x
−1 −9
= log2 x y
(DL)
(BI)
(BID, ’backwards’)
(LP=SL)
49. Condense using the log properties where appropriate:
log2 x2 y −2 + log2 x2 y 4
Solution:
log2 x2 y −2 + log2 x2 y 4 = log2 x2 y −2 · x2 y 4
= log2 x2 x2 y −2 y 4
= log2 x2+2 y −2+4
4 2
= log2 x y
20
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OR an alternative idea that helps practice BID...
(LP=SL)
log2 x2 y −2 + log2 x2 y 4 = log2 x2 + log2 y −2 + log2 x2 + log2 y 4
= 2 log2 (x) + −2 log2 (y) + 2 log2 (x) + 4 log2 (y)
(BID)
= 2 log2 (x) + 2 log2 (x) + −2 log2 (y) + 4 log2 (y) (ALA, CoLA)
= (2 + 2) log2 (x) + (−2 + 4) log2 (y)
(DL)
= 4 log2 (x) + 2 log2 (y)
(BI)
(BID, ’backwards’)
= log2 x4 + log2 y 2
4 2
(LP=SL)
= log2 x y
50. Condense using the log properties where appropriate:
log2 x4 y 3 + log2 x2 y −5
Solution:
log2 x4 y 3 + log2 x2 y −5 = log2 x4 y 3 · x2 y −5
= log2 x4 x2 y 3 y −5
= log2 x4+2 y 3+−5
6 −2
= log2 x y
(LP=SL)
(CoLM, ALM)
(JAE)
(BI)
OR an alternative idea that helps practice BID...
(LP=SL)
log2 x4 y 3 + log2 x2 y −5 = log2 x4 + log2 y 3 + log2 x2 + log2 y −5
= 4 log2 (x) + 3 log2 (y) + 2 log2 (x) + −5 log2 (y)
(BID)
= 4 log2 (x) + 2 log2 (x) + 3 log2 (y) + −5 log2 (y) (ALA, CoLA)
= (4 + 2) log2 (x) + (3 + −5) log2 (y)
(DL)
= 6 log2 (x) + − 2 log2 (y)
(BI)
(BID, ’backwards’)
= log2 x6 + log2 y −2
6 −2
(LP=SL)
= log2 x y
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ALGEBRA
Sec. 05
MathHands.com
Márquez
51. Condense using the log properties where appropriate:
log2 x8 y −3 + log2 x−7 y 4
Solution:
log2 x8 y −3 + log2 x−7 y 4 = log2 x8 y −3 · x−7 y 4
= log2 x8 x−7 y −3 y 4
= log2 x8+−7 y −3+4
1 1
= log2 x y
(LP=SL)
(CoLM, ALM)
(JAE)
(BI)
OR an alternative idea that helps practice BID...
(LP=SL)
log2 x8 y −3 + log2 x−7 y 4 = log2 x8 + log2 y −3 + log2 x−7 + log2 y 4
= 8 log2 (x) + −3 log2 (y) + −7 log2 (x) + 4 log2 (y)
(BID)
= 8 log2 (x) + −7 log2 (x) + −3 log2 (y) + 4 log2 (y) (ALA, CoLA)
= (8 + −7) log2 (x) + (−3 + 4) log2 (y)
(DL)
= 1 log2 (x) + 1 log2 (y)
(BI)
(BID, ’backwards’)
= log2 x1 + log2 y 1
1 1
(LP=SL)
= log2 x y
52. PROVE LS=SL
If a, b > 0
log(ab) = log a + log b
22
math
hands
c
2007-2009
MathHands.com
ALGEBRA
Sec. 05
MathHands.com
Márquez
Solution:
let x = log a
then 10x = a
let y = log a
then 10y = b
then 10x · 10y = a · b
then 10x+y = ab
then log(ab) = x + y
then log(ab) = log a + log b
(can can call log a anything)
(def B of logs)
(can can call log b anything)
(def B of logs)
(sub)
(JAE)
(def B of logs)
(substitute x and y)
53. Although BID is true when n is any rational number, and a > 0, prove the special case of the BID
when n = 2.
log an = n log a
Solution:
log(a2 ) = log(a · a)
= log a + log a
= 2 log a
therefore,
log(a2 ) = 2 log a
(algebra)
(LP=SL)
(algebra)
(done!)
54. Although BID is true when n is any rational number, and a > 0, prove the special case of the BID
when n = 5.
log an = n log a
23
math
hands
c
2007-2009
MathHands.com
ALGEBRA
Sec. 05
MathHands.com
Márquez
Solution:
log(a5 ) = log(a · a · a · a · a)
= log a + log a + log a + log a + log a
= 5 log a
therefore,
log(a5 ) = 5 log a
(algebra)
(LP=SL)
(algebra)
(done!)
55. Although BID is true when n is any rational number, and a > 0, prove the special case of the BID
when n = 8.
log an = n log a
56. PROVE LQ=DL
Assume b 6= 0:
log(a/b) = log a − log b
Solution:
let x = log a
then 10x = a
let y = log a
then 10y = b
10x
a
then
=
10y
b
a
x−y
then 10
=
a b
=x−y
then log
ab then log
= log a − log b
b
(can can call log a anything)
(def B of logs)
(can can call log b anything)
(def B of logs)
(sub)
(exponent properties)
(def B of logs)
(substitute x and y)
57. PROVE CBP
If b, k, c > 0
logb k =
24
math
hands
logc k
logc b
c
2007-2009
MathHands.com
ALGEBRA
Sec. 05
MathHands.com
Márquez
Solution:
let x = logb k
then bx = k
then logc bx = logc k
then x logc b = logc k
logc k
then x =
logc b
logc k
finally logb k =
logc b
(can can call logb k anything)
(def B of logs)
1
(logc (t) is well defined , ie slap a logc on each side)
(BID)
(algebra)
(sub)
58. Prove If b, k > 0
logb bk = k
59. Prove If b, k > 0
blogb k = k
60. Use the log properties to condense the following expression as much as possible:
log x log 5
Solution: maybe
log 5log x
log xlog 5
could also be
61. Use the log properties to condense the following expression as much as possible:
log(x − 1) + log(x2 + x + 1)
62. Use the log properties to condense the following expression as much as possible:
log(x3 − 1) − log(x2 + x + 1)
63. Use the log properties to condense the following expression as much as possible:
3 log x − 5 log(x + 1) + 2 log x2
25
math
hands
c
2007-2009
MathHands.com
ALGEBRA
Sec. 05
MathHands.com
Márquez
64. Use the log properties to condense the following expression as much as possible:
3 log9 x + log3 x
65. Use the log properties to condense the following expression as much as possible:
5 log2 x + log16 x + 4 log4 x
26
math
hands
c
2007-2009
MathHands.com