Logarithm Laws and Properties Revision Notes for HSC SSCE Mathematics Advanced

Logarithm Laws and Properties

Introduction

Logarithm laws and properties are essential tools for working with logarithmic expressions. These laws allow you to simplify complex logarithmic expressions, expand single logarithms into multiple terms, or combine multiple logarithmic terms into a single expression. Understanding these properties comes from recognising that logarithms are the inverse of exponential functions, and their laws are derived from the laws of indices.

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The power of logarithm laws lies in their ability to transform multiplication into addition, division into subtraction, and powers into multiplication - making complex calculations much more manageable!

Special logarithm properties

Several fundamental properties arise directly from the definition of logarithms. Recall that if $a^x = b$ , then $\log_a(b) = x$ , where $a > 0$ , $a \neq 1$ , and $b > 0$ .

Property 1: Logarithm of a power of the base

$\log_a(a^x) = x$

where $a$ is the base ( $a > 0$ , $a \neq 1$ ) and $x$ is any real number.

This property tells us that when we take the logarithm of the base raised to any power, the result is simply that power. This makes sense because we're asking "what power of $a$ gives us $a^x$ ?" and the answer is obviously $x$ .

Property 2: Base raised to a logarithm

$a^{\log_a(x)} = x$

where $x$ is the argument ( $x > 0$ ).

This property demonstrates that exponentiating the base by a logarithm returns the original argument. It's the inverse relationship: if $\log_a(x) = y$ , then by definition $a^y = x$ , so substituting gives $a^{\log_a(x)} = x$ .

Property 3: Logarithm of the base

$\log_a(a) = 1$

This follows because $a^1 = a$ . We're asking "what power of $a$ equals $a$ ?" and the answer is $1$ .

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This property is particularly useful for quick evaluations. Remember: the logarithm of the base always equals 1, regardless of which base you're working with!

Property 4: Logarithm of one

$\log_a(1) = 0$

This follows because $a^0 = 1$ for any base $a$ . We're asking "what power of $a$ equals $1$ ?" and the answer is always $0$ .

Property 5: Logarithm of a reciprocal

$\log_a\left(\frac{1}{x}\right) = -\log_a(x)$

To understand this property, recall that $\frac{1}{x} = x^{-1}$ . If $\log_a(x) = y$ , then $a^y = x$ . This means $a^{-y} = \frac{1}{x}$ , and therefore $\log_a\left(\frac{1}{x}\right) = -y = -\log_a(x)$ .

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Critical insight: These five properties all stem directly from the definition of logarithms and the relationship between logarithms and exponentials. Mastering these properties is essential before moving on to the logarithm laws!

Worked example: Using special properties

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Worked Example 1: Evaluating a Simple Logarithm

Evaluate $\log_5(5)$

Solution:

Using Property 3, we know that the logarithm of the base equals 1.

$\log_5(5) = 1$

This is because $5^1 = 5$ .

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Worked Example 2: Verifying the Reciprocal Property

Show that $\log_3\left(\frac{1}{9}\right) = -\log_3(9)$ without using the reciprocal property.

Solution:

We can verify this by evaluating each side separately using the definition of logarithms and index laws.

Left side:

$\log_3\left(\frac{1}{9}\right) = \log_3(3^{-2})$

Using index laws to rewrite $\frac{1}{9}$ as $3^{-2}$ :

$= -2$

Using Property 1 where $\log_a(a^x) = x$ .

Right side:

$-\log_3(9) = -\log_3(3^2)$

Expressing $9$ as $3^2$ :

$= -2$

Using Property 1 where $\log_a(a^x) = x$ .

Conclusion:

Therefore, $\log_3\left(\frac{1}{9}\right) = -\log_3(9)$ as both sides equal -2.

Logarithm laws

The three main logarithm laws are derived from the laws of indices. These laws work for any base $a > 0$ where $a \neq 1$ , and they apply to positive arguments.

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Each logarithm law has a direct connection to index laws:

Product law ↔ $a^m \times a^n = a^{m+n}$
Quotient law ↔ $a^m \div a^n = a^{m-n}$
Power law ↔ $(a^m)^n = a^{mn}$

Product law

$\log_a(xy) = \log_a(x) + \log_a(y)$

where $x$ and $y$ are positive numbers.

The product law tells us that the logarithm of a product equals the sum of the logarithms. This is incredibly useful for breaking down complex expressions or for simplifying products inside logarithms.

Derivation:

Let $m = \log_a(x)$ and $n = \log_a(y)$ . This means $x = a^m$ and $y = a^n$ .

$xy = a^m \times a^n = a^{m+n}$

Applying the index law for multiplication.

Taking the logarithm base $a$ of both sides:

$\log_a(xy) = \log_a(a^{m+n})$

Using Property 1 where $\log_a(a^k) = k$ :

$\log_a(xy) = m + n$

Substituting back $m = \log_a(x)$ and $n = \log_a(y)$ :

$\log_a(xy) = \log_a(x) + \log_a(y)$

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Memory aid: "Logs make multiplication into addition" - When you see multiplication inside a logarithm, you can split it into separate logarithms that you add together.

Quotient law

$\log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y)$

where $x$ and $y$ are positive numbers.

The quotient law tells us that the logarithm of a quotient equals the difference of the logarithms. This helps us separate fractions inside logarithms into simpler terms.

Derivation:

Let $m = \log_a(x)$ and $n = \log_a(y)$ . This means $x = a^m$ and $y = a^n$ .

$\frac{x}{y} = \frac{a^m}{a^n} = a^{m-n}$

Applying the index law for division.

Taking the logarithm base $a$ of both sides:

$\log_a\left(\frac{x}{y}\right) = \log_a(a^{m-n})$

Using Property 1 where $\log_a(a^k) = k$ :

$\log_a\left(\frac{x}{y}\right) = m - n$

Substituting back $m = \log_a(x)$ and $n = \log_a(y)$ :

$\log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y)$

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Memory aid: "Logs make division into subtraction" - When you see division inside a logarithm, you can split it into separate logarithms that you subtract.

Power law

$\log_a(x^n) = n \log_a(x)$

where $x$ is a positive number and $n$ is any real number.

The power law tells us that the logarithm of a number raised to a power equals the power multiplied by the logarithm. This allows us to "bring down" exponents from inside logarithms, making expressions much easier to work with.

Derivation:

Let $m = \log_a(x)$ , so $x = a^m$ .

$x^n = (a^m)^n$

Substituting $x = a^m$ .

$x^n = a^{mn}$

Applying the index law for powers.

Taking the logarithm base $a$ of both sides:

$\log_a(x^n) = \log_a(a^{mn})$

Using Property 1 where $\log_a(a^k) = k$ :

$\log_a(x^n) = mn$

Substituting back $m = \log_a(x)$ :

$\log_a(x^n) = n \log_a(x)$

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Memory aid: "Logs bring powers to the front" - Any exponent inside a logarithm can be moved to the front as a coefficient.

Applying logarithm laws

Understanding when and how to apply these laws is crucial for simplifying logarithmic expressions. The key is recognizing patterns and knowing which law to apply in each situation.

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Strategy for simplifying logarithmic expressions:

Look for products inside logarithms → use the product law to split them
Look for quotients inside logarithms → use the quotient law to split them
Look for powers inside logarithms → use the power law to bring the exponent forward
Look for coefficients in front of logarithms → use the power law to move them inside
Try to express arguments as powers of the base to use Property 1

Worked example: Simplifying expressions

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Worked Example 3a: Combining Multiple Logarithms

Simplify $\log_2(16) + \log_2(4) - \log_2(8)$

Solution:

Step 1: Use the product law to combine the first two terms:

$\log_2(16) + \log_2(4) - \log_2(8) = \log_2(16 \times 4) - \log_2(8)$

Step 2: Calculate the product inside the brackets:

$= \log_2(64) - \log_2(8)$

Step 3: Now apply the quotient law:

$= \log_2\left(\frac{64}{8}\right)$

Step 4: Evaluate the division:

$= \log_2(8)$

Step 5: Express $8$ as a power of $2$ :

$= \log_2(2^3)$

Step 6: Apply Property 1:

$= 3$

Verification: We can verify this answer: $\log_2(16) = 4$ , $\log_2(4) = 2$ , $\log_2(8) = 3$ , so $4 + 2 - 3 = 3$ . ✓

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Worked Example 3b: Applying the Power Law

Evaluate $2\log_5(4)$

Solution:

Step 1: Apply the power law, writing the coefficient as an exponent:

$2\log_5(4) = \log_5(4^2)$

Step 2: Evaluate the power:

$= \log_5(16)$

Step 3: To find a numerical answer, we can use the change of base formula:

$= \frac{\log_{10}(16)}{\log_{10}(5)}$

Step 4: Evaluating and rounding:

$\approx 1.7227$

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Exam tip: When simplifying logarithmic expressions, look for opportunities to:

Combine terms using the product law (addition becomes multiplication)
Separate terms using the quotient law (subtraction becomes division)
Move coefficients inside using the power law
Express arguments as powers of the base to use Property 1

Working systematically through these steps will help you avoid common mistakes and arrive at the correct answer efficiently.

Remember!

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Key Points to Remember:

Special properties come directly from the logarithm definition: $\log_a(a^x) = x$ , $a^{\log_a(x)} = x$ , $\log_a(a) = 1$ , $\log_a(1) = 0$ , and $\log_a\left(\frac{1}{x}\right) = -\log_a(x)$ .
Product law: The logarithm of a product equals the sum of logarithms: $\log_a(xy) = \log_a(x) + \log_a(y)$ .
Quotient law: The logarithm of a quotient equals the difference of logarithms: $\log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y)$ .
Power law: The logarithm of a power brings the exponent to the front: $\log_a(x^n) = n\log_a(x)$ .
All logarithm laws require positive arguments and are derived from the laws of indices by applying the inverse relationship between logarithms and exponentials.

Logarithm Laws and Properties (HSC SSCE Mathematics Advanced): Revision Notes