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Chapter 5: Exponential and Logarithmic Functions 5.5: Properties and Laws of Logarithms. Essential Question: What are the three properties that simplify logarithmic expressions? Describe how to use them. 5.5: Properties and Laws of Logarithms. Basic Properties of Logarithms - PowerPoint PPT Presentation
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Chapter 5: Exponential and Logarithmic Chapter 5: Exponential and Logarithmic FunctionsFunctions5.5: Properties and Laws of Logarithms5.5: Properties and Laws of LogarithmsEssential Question: What are the three properties that simplify logarithmic expressions? Describe how to use them.
5.5: Properties and Laws of Logarithms5.5: Properties and Laws of Logarithms
Basic Properties of Logarithms◦ Logarithms are only defined for positive real
numbers Not possible for 10 or e to be taken to an exponent
and result in a negative number
◦ Log 1 = 0 and ln 1 = 0 100 = 1 & e0 = 1
◦ Log 10k = k and ln ek = k log10104 = k 10k = 104
k = 4
◦ 10log v = v and eln v = v 10log 22 = v log10v = log 22 v = 22
5.5: Properties and Laws of Logarithms5.5: Properties and Laws of Logarithms
Solving Equations by Using Properties of Logarithms◦ln(x + 1) = 2
Method #1 e2 = x + 1 e2 – 1 = x x ≈ 6.3891
Method #2 eln(x + 1) = e2
x + 1 = e2
See method #1 above
5.5: Properties and Laws of Logarithms5.5: Properties and Laws of Logarithms
Product Law of Logarithms◦Law of exponents states bmbn = bm+n
◦Because logarithms are exponents: log (vw) = log v + log w ln (vw) = ln v + ln w Proof:
vw = 10log v • 10log w = 10log v + log w
vw = 10log vw
Taking from above: 10log v + log w = 10log vw
log v + log w = log vw Proof of ln/e works the same way
5.5: Properties and Laws of Logarithms5.5: Properties and Laws of Logarithms
Product Law of Logarithms (Application)◦Given that log 3 = 0.4771 and log 11 =
1.0414find log 33 log 33 = log (3 • 11)
= log 3 + log 11= 0.4771 + 1.0414= 1.5185
◦Given that ln 7 = 1.9459 and ln 9 = 2.1972find ln 63 ln 63 = ln (7 • 9)
= ln 7 + ln 9= 1.9459 + 2.1972= 4.1431
5.5: Properties and Laws of Logarithms5.5: Properties and Laws of Logarithms
Quotient Law of Logarithms◦Law of exponents states ◦Because logarithms are exponents:
log ( ) = log v – log w
ln ( ) = ln v – ln w
◦Proof is the same as the Product Law
v
w
v
w
mm n
n
bb
b
5.5: Properties and Laws of Logarithms5.5: Properties and Laws of Logarithms
Quotient Law of Logarithms (Application)◦Given that log 28 = 1.4472 and log 7 =
0.8451find log 4 log 4 = log (28 / 7)
= log 28 – log 7= 1.4472 – 0.8451= 0.6021
◦Given that ln 18 = 2.8904 and ln 6 = 1.7918find ln 3 ln 3 = ln (18 / 6)
= ln 18 – ln 6= 2.8904 – 1.7918= 1.0986
5.5: Properties and Laws of Logarithms5.5: Properties and Laws of Logarithms
Power Law of Logarithms◦Law of exponents states (bm)k = bmk
◦Because logarithms are exponents: log (vk) = k log v ln (vk) = k ln v Proof:
v = 10log v → vk = (10log v)k = 10k log v
vk = 10log vk
Taking from above: 10k log v = 10log vk
k log v = log vk
Proof of ln/e works the same way
5.5: Properties and Laws of Logarithms5.5: Properties and Laws of Logarithms
Power Law of Logarithms (Application)◦Given that log 6 = 0.7782 find log
log = log 6½
= ½ log 6= ½ (0.7782)= 0.3891
◦Given that ln 50 = 3.9120 find ln
66
3 50133
13
13
ln 50 ln50
ln50
(3.9120)
1.3040
5.5: Properties and Laws of Logarithms5.5: Properties and Laws of Logarithms
Simplifying Expressions◦Write as a single logarithm:
ln 3x + 4 ln x – ln 3xy4
4
5
5
4
ln 3 4ln ln 3 ln3 ln ln 3
ln(3 ) ln 3
ln3 ln3
3ln3
ln
x x xy x x xy
x x xy
x xy
x
xy
x
y
5.5: Properties and Laws of Logarithms5.5: Properties and Laws of Logarithms
Simplifying Expressions◦Write as a single logarithm:
12 1
4
11 42
2 24
2
21 12 4
21 12 4
1 12 4
1 1 12 4 2
14
14
ln ln ln ln
ln ln
ln ln
ln ln ln
ln ln 2ln
ln ln ln
ln
x xex ex
x x
x ex
x ex
x e x
x e x
x e x
e
5.5: Properties and Laws of Logarithms5.5: Properties and Laws of Logarithms
Assignment◦Page 369◦Problems 1-25, odd problems◦Show work