Enrichment: More on Logarithms Revision Notes for Grade 12 NSC Matric Mathematics

Enrichment: More on Logarithms

Laws of logarithms

Understanding logarithmic laws is essential for working with complex logarithmic expressions. These laws allow us to break down complicated logarithms into simpler parts and solve logarithmic equations more efficiently.

Product law

The product law states that the logarithm of a product equals the sum of the logarithms of the individual factors.

Formula: $\log_a(xy) = \log_a(x) + \log_a(y)$ where $x > 0$ and $y > 0$

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The product law is one of the most fundamental logarithmic laws. It transforms multiplication inside a logarithm into addition outside the logarithm, making complex calculations much simpler.

Proof explanation: If we let $\log_a(x) = m$ and $\log_a(y) = n$ , then by definition:

$x = a^m$
$y = a^n$

When we multiply x and y: $xy = a^m \times a^n = a^{(m+n)}$

Converting back to logarithmic form: $\log_a(xy) = m + n = \log_a(x) + \log_a(y)$

This means the logarithm of a product equals the sum of the logarithms of its factors.

Quotient law

The quotient law states that the logarithm of a quotient equals the difference of the logarithms of the numerator and denominator.

Formula: $\log_a(\frac{x}{y}) = \log_a(x) - \log_a(y)$ where $x > 0$ and $y > 0$

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Remember the domain restrictions: both x and y must be positive for logarithms to be defined. This applies to all logarithmic operations.

Proof explanation: Using similar reasoning as the product law, if $\log_a(x) = m$ and $\log_a(y) = n$ :

$x = a^m$
$y = a^n$

When we divide x by y: $\frac{x}{y} = \frac{a^m}{a^n} = a^{(m-n)}$

Converting back to logarithmic form: $\log_a(\frac{x}{y}) = m - n = \log_a(x) - \log_a(y)$

This shows that the logarithm of a quotient equals the difference of the logarithms of the numerator and denominator.

Worked examples

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Worked Example 1: Applying the product law

Question: Simplify $\log 5 + \log 2 - \log 30$

Solution:

Step 1: Apply the logarithmic law to combine the first two terms

Since the product of 5 and 2 equals 10, we can write:

$\log 5 + \log 2 - \log 30 = (\log 5 + \log 2) - \log 30 = \log(5 \times 2) - \log 30 = \log 10 - \log 30$

Step 2: Simplify using known values

Since $\log 10 = 1$ :

$\log 10 - \log 30 = 1 - \log 30$

Step 3: Expand the last term to simplify further

$1 - \log 30 = 1 - \log(3 \times 10) = 1 - (\log 3 + \log 10) = 1 - (\log 3 + 1) = 1 - \log 3 - 1 = -\log 3$

Final answer: $\log 5 + \log 2 - \log 30 = -\log 3$

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Worked Example 2: Applying the quotient law

Question: Simplify $\log 40 - \log 4 + \log_5(\frac{8}{5})$

Solution:

Step 1: Apply the logarithmic law to the first two terms

Since both terms have the same base, we can use the quotient law:

$\log 40 - \log 4 + \log_5(\frac{8}{5}) = \log(\frac{40}{4}) + \log_5(\frac{8}{5}) = \log 10 + \log_5(\frac{8}{5})$

Step 2: Simplify the common logarithm

Since $\log 10 = 1$ :

$\log 10 + \log_5(\frac{8}{5}) = 1 + \log_5(\frac{8}{5})$

Step 3: Expand the last term using the quotient law

$1 + \log_5(\frac{8}{5}) = 1 + (\log_5 8 - \log_5 5) = 1 + \log_5 8 - 1 = \log_5 8$

Final answer: $\log 40 - \log 4 + \log_5(\frac{8}{5}) = \log_5 8$

Simplification of logarithms

When simplifying logarithmic expressions, we can use various algebraic techniques combined with logarithmic laws. The key is recognising patterns and applying the appropriate laws systematically.

Useful logarithmic values

These standard values are essential to memorise:

$\log 1 = 0$
$\log 10 = 1$
$\log 100 = 2$
$\log 1000 = 3$
$\log(\frac{1}{10}) = -1$
$\log 0.1 = -1$
$\log 0.01 = -2$
$\log 0.001 = -3$

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Worked Example 3: Complex simplification

Question: Simplify (without a calculator): $3 \log 3 + \log 125$

Solution:

Step 1: Apply logarithmic laws to simplify

$3 \log 3 + \log 125 = 3 \log 3 + \log 5^3 = 3 \log 3 + 3 \log 5 = 3(\log 3 + \log 5) = 3 \log(3 \times 5) = 3 \log 15$

Step 2: State the final answer

Since we cannot simplify $\log 15$ further using standard values, the expression $3 \log 3 + \log 125 = 3 \log 15$ .

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All algebraic manipulation techniques (factorising, expanding, etc.) also apply to logarithmic expressions. Always consider the number of terms to determine the best approach for simplification.

Solving logarithmic equations

When solving logarithmic equations, the key strategy involves converting between logarithmic and exponential forms. This allows us to isolate the variable and find the solution.

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Worked Example 4: Basic logarithmic equation

Question: Solve for p: $18 \log p - 36 = 0$

Solution:

Step 1: Make $\log p$ the subject of the equation

$18 \log p - 36 = 0$

$18 \log p = 36$

$\log p = \frac{36}{18} = 2$

Step 2: Convert from logarithmic to exponential form

If $\log p = 2$ , then $p = 10^2 = 100$

Step 3: State the final answer

$p = 100$

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Worked Example 5: Exponential equation requiring logarithms

Question: Solve for n (correct to the nearest integer): $(1.02)^n = 2$

Solution:

Step 1: Convert from exponential to logarithmic form

$(1.02)^n = 2$ means $n = \log_{1.02} 2$

Step 2: Use change of base formula to solve

$n = \frac{\log 2}{\log 1.02} = 35.00...$

Step 3: Round to the nearest integer

$n = 35$

Summary of key concepts

Understanding logarithms involves recognising their relationship with exponentials and applying the fundamental laws correctly. Here are the essential points:

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Key Points to Remember:

Basic relationship:

The logarithm of a number (x) with base (a) equals the exponent (y) needed to raise the base to equal that number
If $x = a^y$ , then $y = \log_a(x)$ , where $a > 0$ , $a \neq 1$ and $x > 0$
Logarithms and exponentials are inverse functions

Key logarithmic laws:

Product law: $\log_a(xy) = \log_a(x) + \log_a(y)$ ( $x > 0$ and $y > 0$ )
Quotient law: $\log_a(\frac{x}{y}) = \log_a(x) - \log_a(y)$ ( $x > 0$ and $y > 0$ )
Power law: $\log_a(x^b) = b \log_a(x)$ ( $x > 0$ )
Change of base: $\log_a(x) = \frac{\log_b(x)}{\log_b(a)}$ ( $b > 0$ and $b \neq 1$ )

Special forms:

Common logarithm: $\log a$ means $\log_{10} a$
Natural logarithm: $\ln$ uses base $e$
Special values: $a^0 = 1$ means $\log_a 1 = 0$ ; $a^1 = a$ means $\log_a a = 1$

Problem-solving approach:

Identify which logarithmic law applies
Apply the law systematically
Use known logarithmic values where possible
Convert between logarithmic and exponential forms when solving equations
Check answers for reasonableness

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Remember These Essential Points:

The product law turns multiplication inside logarithms into addition outside: $\log_a(xy) = \log_a(x) + \log_a(y)$
The quotient law turns division inside logarithms into subtraction outside: $\log_a(\frac{x}{y}) = \log_a(x) - \log_a(y)$
When solving logarithmic equations, convert to exponential form to isolate the variable
Memorise key values like $\log 1 = 0$ , $\log 10 = 1$ , and $\log 100 = 2$ for quick simplification
Always check that your solutions make sense within the domain restrictions (arguments must be positive)

Enrichment: More on Logarithms (Grade 12 NSC Matric Mathematics): Revision Notes