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