Elementary Algebra

Section 4.3

Properties of Logarithms

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Properties of Logarithms

Consider logb x = m and logb y = n By definition

bm = x and bn = y

xy = (bm)(bn) = bm + n

So logb (xy) = logb (bm + n)

Product Rule for Logarithms

logb xy = logb x + logb y

for any positive real numbers b, x, y with b ≠ 1

= m + n

= logb x + logb y

WHY?

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Properties of Logarithms

Examples

1. log4 (3 7) = log4 3 + log4 7

2. log8 10 + log8 3 = log8 (10 3) = log8 30

3. loga x2 = loga xx

= loga x + loga x

= 2 loga x

4. 1 + log 2 + log x + log 2x2

5. ln (x – 1) + ln (x + 1)

= 1 + log 4x3

= ln (x2 – 1)

= ln (x – 1)(x + 1)( )

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Again consider

logb x = m and logb y = n , for x, y, b positive, b ≠ 1

Thus bm = x and bn = y

Quotient Rule for Logarithms

for any positive real numbers b, x, y with b ≠ 1

Properties of Logarithms

logb = logb x – logb yxy

xylogb =

bm

bnlogb = logb bm – n

= m – n

= logb x – logb y

WHY?

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Examples

1.

2.

3. Solve for x :

Properties of Logarithms

94( )log7 = log7 9 – log7 4

316( )log4 = log4 3 – log4 16 = log4 3 – 2

log (x + 3)

log (x + 1)= 2

log (x + 3) log (x + 1)= 2

log (x + 1)= 2

x + 3 = (x + 1)2

0 = x2 + x – 2

= x2 + 2x+ 1

= (x + 2)(x – 1)

Solution set: { –2 , 1 }

WHY?

WHY?

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Properties of Logarithms

Consider

logb x = m for x, b positive, b ≠ 1

Thus bm = x

and xr = bmr

So logb xr = logb (brm)

Power Rule for Logarithms for any positive real numbers b, x with b ≠ 1

logb xr = r logb x

and (bm)r = xr

= brm

= rm

= r logb x

, for any real r

WHY?

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= log5(x + 1)3

Examples

1.

2.

3. 3 log5(x + 1)

4. x log 2

5. ln 1

6. ln 0

Properties of Logarithms

log352 = 2 log35

loga x4 = 4 loga x

= log 2x

= loge1 = 0

= ?

Question:Is 0 in the domain of any logarithm function ?What does this tell you about ln 0 ?

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Properties of Logarithms

Property Recognition

Rewrite as a logarithm of a single expression :

1. log 4 + log 7

2. log 35 – log 7

3. ln 5e – ln

4. log 5e – log

5. logb x5 – logb x3 + logb x2

6.

7. logb (x y) in terms of logb x and logb y

1

20e( )e

20( )

logb x

logb y

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Properties of Logarithms

More Examples

1. Rewrite in expanded form: log4 (3x + 7)

Cannot be written in expanded form !

2. TRUE or FALSE : log6 36 – log6 6 = log6 30

Rewriting: log6 (36/6) = log6 (6 5)

log6 6 = log6 6 + log6 5

0 = log6 5

Since log6 1 = 0 then log6 1 = log6 5

This implies that 1 = 5 ... a CONTRADICTION !!

Hence the given statement is FALSE !

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Properties of Logarithms

More Examples

3. TRUE or FALSE : log3 (log2 8)log7 49

log8 64=

log3 (3)log7 72

log8 82=

12 log7 7

2 log8 8=

12 (1)

2 (1)=

1 = 1

So, the given statement is TRUE !!

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Bases for Logarithms

ConversionsCan we use logb x to find loga x ?

Let loga x = y

By definition ay = x

logb (ay) = logb x

y logb a = logb x

(loga x)(logb a) = logb x

Thus

loga xlogb x

logb a= logb x

loga x

loga b=OR

NOTE:logb x

logb alogb x – logb a≠

applying logb ...

applying power rule

replacing y

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Conversion Examples Find log3 17 on your calculator

Having trouble ?

Bases for Logarithms

log3 17log 17

log 3=1.23044

0.477121= ≈ 2.5789

OR

log3 17ln 17

ln 3=2.8332

1.0986= ≈ 2.5789

... if you can

Let’s try using a little math first ...

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Solve1. Find x to the nearest whole number

e.02x = 192

ln(e.02x) = ln(192)

(.02x)ln e = ln(192)

.02x = 5.2575

x ≈ 262.9 ≈ 263

2. Find x exactly

log3 (x + 1)5 = 3

(x + 1)5 = 27

x + 1 = 271/5

x = –1 + 271/5

More Equations

3log (x+1)35 = 33

Solution set: { 263 }

Solution set: { –1 + 271/5 }

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Solve3. Find x exactly

log8 ( 2x + 5) + log8 3 = log8 33

log8 ( 2x + 5) + log8 3 = log8 3 + log8 11

log8 ( 2x + 5) = log8 11

2x + 5 = 11

x = 3

4. Find x exactly

log3 2x – log3 (3x + 15) = –2

More Equations

Solution set: { 3 }

Solution set: { 1 }

3x + 15

2x( )log3 = –2

3x + 15

2x= 3–2

18x = 3x + 15x = 1

= log8 (3 ∙ 11)

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Think about it !

Module 4

Section 4.5

Properties of Logarithms