Mastering Operations with Logarithms: Key Concepts Explained

Property	Reason
$lo g_{b} b = 1$	A number raised to the power of $1$ is equal to itself.
$lo g_{b} 1 = 0$	A number raised to the power of $0$ is equal to $1 .$

Property

Reason

lo g_{b} b = 1

A number raised to the power of

1

is equal to itself.

lo g_{b} 1 = 0

A number raised to the power of

0

is equal to

1 .

$lo g_{b} a = c \Leftrightarrow a = b^{c}$
$a = b^{c} Substitute a = b^{l o g_{b} a}$
$lo g_{b} a = c Substitute lo g_{b} b^{c} = c$

lo g_{b} a = c \Leftrightarrow a = b^{c}

a = b^{c} Substitute a = b^{l o g_{b} a}

lo g_{b} a = c Substitute lo g_{b} b^{c} = c

a Consider the Product Property of Logarithms.

lo g_{b} m n = lo g_{b} m + lo g_{b} n

Knowing that

lo g_{2} 3 \approx 1.585

and that

lo g_{2} 5 \approx 2.322,

this property can be used to evaluate the given expression. Start by rewriting

15

as the product of

3

and

5 .

lo g_{2} 6

SplitIntoFactors

Split into factors

lo g_{2} (3 \cdot 5)

$lo g_{2} (m n) = lo g_{2} (m) + lo g_{2} (n)$

lo g_{2} 3 + lo g_{2} 5

$lo g_{2} 3 \approx 1.585$ , $lo g_{2} 5 \approx 2.322$

1.585 + 2.322

AddTerms

Add terms

3.907

b Because of the quotient inside the logarithm, the Quotient Property of Logarithms will be used this time.

lo g_{b} \frac{m}{n} = lo g_{b} m - lo g_{b} n

Also, as a consequence of the definition of a logarithm, it is known that

lo g_{b} 1 = 0

for any positive number

b

different than

1 .

Using this information, the above property, and the fact that

lo g_{2} 5 \approx 2.322,

the given expression can be evaluated.

lo g_{2} \frac{1}{5}

$lo g_{2} (\frac{a}{b}) = lo g_{2} (a) - lo g_{2} (b)$

lo g_{2} 1 - lo g_{2} 5

$lo g_{2} (1) = 0$

0 - lo g_{2} 5

$lo g_{2} 5 \approx 2.322$

0 - 2.322

SubTerm

Subtract term

- 2.322

c Recall the Power Property of Logarithms and the Product Property of Logarithms.

Power : Product : lo g_{b} m^{n} = n lo g_{b} m lo g_{b} m n = lo g_{b} m + lo g_{b} n

Also, as a consequence of the definition of a logarithm, it is known that

lo g_{b} b = 1

for any positive number

b

different than

1 .

Using this information, the above properties, and the fact that

lo g_{2} 3 \approx 1.585,

the given expression can be evaluated.

lo g_{2} 24

SplitIntoFactors

Split into factors

lo g_{2} (8 \cdot 3)

$lo g_{2} (m n) = lo g_{2} (m) + lo g_{2} (n)$

lo g_{2} 8 + lo g_{2} 3

WritePow

Write as a power

lo g_{2} 2^{3} + lo g_{2} 3

$lo g_{2} (a^{m}) = m \cdot lo g_{2} (a)$

3 lo g_{2} 2 + lo g_{2} 3

$lo g_{2} (2) = 1$

3 (1) + lo g_{2} 3

IdPropMult

\IdPropMult

3 + lo g_{2} 3

$lo g_{2} 3 \approx 1.585$

3 + 1.585

AddTerms

Add terms

4.585

Definition	$lo g_{b} a = c \Leftrightarrow a = b^{c}$
Identity Derived From the Definition	$lo g_{b} b = 1$
Identity Derived From the Definition	$lo g_{b} 1 = 0$
Product Property of Logarithms	$lo g_{b} m n = lo g_{b} m + lo g_{b} n$
Quotient Property of Logarithms	$lo g_{b} \frac{m}{n} = lo g_{b} m - lo g_{b} n$
Power Property of Logarithms	$lo g_{b} m^{n} = n lo g_{b} m$

Definition

lo g_{b} a = c \Leftrightarrow a = b^{c}

Identity Derived From the Definition

lo g_{b} b = 1

Identity Derived From the Definition

lo g_{b} 1 = 0

Product Property of Logarithms

lo g_{b} m n = lo g_{b} m + lo g_{b} n

Quotient Property of Logarithms

lo g_{b} \frac{m}{n} = lo g_{b} m - lo g_{b} n

Power Property of Logarithms

lo g_{b} m^{n} = n lo g_{b} m

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Discussion

Logarithm

Why

Example

Evaluating and Rewriting Logarithmic Expressions

Hint

Solution

Pop Quiz

Evaluating Logarithms

Pop Quiz

Rewriting Logarithmic and Exponential Equations

Discussion

Properties of Logarithms

Rule

Product Property of Logarithms

Proof

Rule

Quotient Property of Logarithms

Proof

Rule

Power Property of Logarithms

Proof

Example

Using Properties to Evaluate Expressions

Hint

Solution

Example

Using Properties to Expand Expressions

Hint

Solution

Example

Using Properties to Condense Expressions

Hint

Solution

Closure

Inverse Operations

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