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Course: Algebra 2 > Unit 8

Lesson 2: Properties of logarithms

Justifying the logarithm properties

Study the proofs of the logarithm properties: the product rule, the quotient rule, and the power rule.

In this lesson, we will prove three logarithm properties: the product rule, the quotient rule, and the power rule. Before we begin, let's recall a useful fact that will help us along the way.

$\log_{b} (b^{c}) = c$ ‍

In other words, a logarithm in base

b

reverses the effect of a base

b

power!

Keep this in mind as you read through the proofs that follow.

Product Rule: $\log_{b} (M N) = \log_{b} (M) + \log_{b} (N)$ ‍

Let's start by proving a specific case of the rule — the case when

M = 4

N = 8

, and

b = 2

Substituting these values into

\log_{b} (M N)

, we see:

$\begin{aligned} \log_{2} (4 \cdot 8) & = \log_{2} (2^{2} \cdot 2^{3}) & 2^{2} = 4 and 2^{3} = 8 \\ = \log_{2} (2^{2 + 3}) & a^{m} \cdot a^{n} = a^{m + n} \\ = 2 + 3 & \log_{b} (b^{c}) = c \\ = \log_{2} (4) + \log_{2} (8) & Since 2 = \log_{2} (4) and 3 = \log_{2} (8) \end{aligned}$ ‍

And so we have that

\log_{2} (4 \cdot 8) = \log_{2} (4) + \log_{2} (8)

While this only verifies one case, we can follow this logic to prove the product rule in general.

Notice, that writing

4

and

8

as powers of

2

was key to the proof. So in general, we'd like

M

and

N

to be powers of the base

b

. To do this, we can let

M = b^{x}

and

N = b^{y}

for some real numbers

x

and

y

Then by definition, it is also true that

\log_{b} (M) = x

and

\log_{b} (N) = y

Now we have:

$\begin{aligned} \log_{b} (M N) & = \log_{b} (b^{x} \cdot b^{y}) & Substitution \\ = \log_{b} (b^{x + y}) & Properties of exponents \\ = x + y & \log_{b} (b^{c}) = c \\ = \log_{b} (M) + \log_{b} (N) & Substitution \end{aligned}$ ‍

Quotient Rule: $\log_{b} (\frac{M}{N}) = \log_{b} (M) - \log_{b} (N)$ ‍

The proof of this property follows a method similar to the one used above.

Again, if we let

M = b^{x}

and

N = b^{y}

, then it follows that

\log_{b} (M) = x

and

\log_{b} (N) = y

We can now prove the quotient rule as follows:

$\begin{aligned} \log_{b} (\frac{M}{N}) & = \log_{b} (\frac{b^{x}}{b^{y}}) & Substitution \\ = \log_{b} (b^{x - y}) & Properties of exponents \\ = x - y & \log_{b} (b^{c}) = c \\ = \log_{b} (M) - \log_{b} (N) & Substitution \end{aligned}$ ‍

Power Rule: $\log_{b} (M^{p}) = p \log_{b} (M)$ ‍

This time, only

M

is involved in the property and so it is sufficient to let

M = b^{x}

, which gives us that

\log_{b} (M) = x

The proof of the power rule is shown below.

$\begin{aligned} \log_{b} (M^{p}) & = \log_{b} ({(b^{x})}^{p}) & Substitution \\ = \log_{b} (b^{x p}) & Properties of exponents \\ = x p & \log_{b} (b^{c}) = c \\ = \log_{b} (M) \cdot p & Substitution \\ = p \cdot \log_{b} (M) & Multiplication is commutative \end{aligned}$ ‍

Alternatively, we can justify this property by using the product rule.

For example, we know that

\log_{b} (M^{p}) = \log_{b} (M \cdot M \cdot … \cdot M)

, where

M

is multiplied by itself

p

times.

We can now use the product rule along with the definition of multiplication as repeated addition to complete the proof. This is shown below.

\begin{aligned} \log_{b} (M^{p}) & = \log_{b} (M \cdot M \cdot … \cdot M) & Definition of exponents \\ = \log_{b} (M) + \log_{b} (M) + … + \log_{b} (M) & Product rule \\ = p \cdot \log_{b} (M) & Repeated addition is multiplication \end{aligned}

And so you have it! We have just proven the three logarithm properties!

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Algebra 2

Course: Algebra 2 > Unit 8

Product Rule: logb⁡(MN)=logb⁡(M)+logb⁡(N)‍

Quotient Rule: logb⁡(MN)=logb⁡(M)−logb⁡(N)‍

Power Rule: logb⁡(Mp)=plogb⁡(M)‍

Want to join the conversation?

Product Rule: $\log_{b} (M N) = \log_{b} (M) + \log_{b} (N)$ ‍

Quotient Rule: $\log_{b} (\frac{M}{N}) = \log_{b} (M) - \log_{b} (N)$ ‍

Power Rule: $\log_{b} (M^{p}) = p \log_{b} (M)$ ‍