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Laws of Logarithms

CSEC Additional Mathematics Essential Knowledge: Logarithms are the inverse operations of exponents. They transform multiplicative relationships into additive ones, making complex calculations simpler. Understanding the laws of logarithms is crucial for solving exponential equations, simplifying expressions, and tackling real-world applications like compound interest, population growth, and sound intensity.

Key Concept: If \(a^x = b\) (where \(a > 0\), \(a \neq 1\), and \(b > 0\)), then the logarithm of \(b\) to base \(a\) is \(x\), written as \(\log_a b = x\). Logarithms “undo” exponents: \(\log_a(a^x) = x\) and \(a^{\log_a x} = x\).

Part 1: Understanding the Logarithm-Exponent Relationship

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The Fundamental Connection

\(a^x = b\)
Exponential Form
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\(\log_a b = x\)
Logarithmic Form
\(2^3 = 8\)
Exponential Statement
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\(\log_2 8 = 3\)
Logarithmic Equivalent
“Log base 2 of 8 equals 3”
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Critical Restrictions

For \(\log_a b\) to be defined:

  1. Base \(a > 0\) and \(a \neq 1\) (Positive, not equal to 1)
  2. Argument \(b > 0\) (Positive only)

Why? Logarithms are the inverse of exponents. Since \(a^x\) is always positive for \(a > 0\), we can only take logs of positive numbers. Base 1 gives a horizontal line, which isn’t invertible.

Two Fundamental Identities:

\(\log_a a = 1\) (since \(a^1 = a\))
\(\log_a 1 = 0\) (since \(a^0 = 1\))
๐Ÿ“ Example 1: Converting Between Forms

Convert between exponential and logarithmic forms:

(a)
\(5^2 = 25\) โ†’ \(\log_5 25 = 2\)
(b)
\(\log_3 81 = 4\) โ†’ \(3^4 = 81\)
(c)
\(10^{-2} = 0.01\) โ†’ \(\log_{10} 0.01 = -2\)
(d)
\(\log_2 \frac{1}{8} = -3\) โ†’ \(2^{-3} = \frac{1}{8}\)

Part 2: The Three Fundamental Laws of Logarithms

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The Core Laws for Manipulation

Law 1: Product Law

\(\log_a (mn) = \log_a m + \log_a n\)

Rule: The logarithm of a product equals the sum of the logarithms.

Why it works: If \(\log_a m = x\) and \(\log_a n = y\), then \(m = a^x\) and \(n = a^y\). So \(mn = a^x \times a^y = a^{x+y}\), thus \(\log_a (mn) = x+y = \log_a m + \log_a n\).

Example:

\(\log_2 (8 \times 4) = \log_2 8 + \log_2 4 = 3 + 2 = 5\)

Check: \(8 \times 4 = 32\), and \(\log_2 32 = 5\) since \(2^5 = 32\) โœ“

Law 2: Quotient Law

\(\log_a \left(\frac{m}{n}\right) = \log_a m – \log_a n\)

Rule: The logarithm of a quotient equals the difference of the logarithms.

Why it works: If \(\log_a m = x\) and \(\log_a n = y\), then \(m = a^x\) and \(n = a^y\). So \(\frac{m}{n} = \frac{a^x}{a^y} = a^{x-y}\), thus \(\log_a \left(\frac{m}{n}\right) = x-y = \log_a m – \log_a n\).

Example:

\(\log_3 \left(\frac{81}{9}\right) = \log_3 81 – \log_3 9 = 4 – 2 = 2\)

Check: \(\frac{81}{9} = 9\), and \(\log_3 9 = 2\) since \(3^2 = 9\) โœ“

Law 3: Power Law

\(\log_a (m^n) = n \log_a m\)

Rule: The logarithm of a power equals the exponent times the logarithm of the base.

Why it works: If \(\log_a m = x\), then \(m = a^x\). So \(m^n = (a^x)^n = a^{nx}\), thus \(\log_a (m^n) = nx = n \log_a m\).

Example:

\(\log_2 (8^3) = 3 \log_2 8 = 3 \times 3 = 9\)

Check: \(8^3 = 512\), and \(\log_2 512 = 9\) since \(2^9 = 512\) โœ“

Memory Aid: “Log of product = SUM of logs”, “Log of quotient = DIFFERENCE of logs”, “Log of power = POWER times log”. Remember the order: Multiplication โ†’ Addition, Division โ†’ Subtraction, Exponentiation โ†’ Multiplication.

Part 3: Special Cases and Change of Base

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Important Special Results

Reciprocal Law

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

Special case of Quotient Law with \(m = 1\)

Root Law

\(\log_a (\sqrt[n]{m}) = \frac{1}{n} \log_a m\)

Special case of Power Law with exponent \(\frac{1}{n}\)

Base Swap Identity

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

Logarithm and exponent cancel each other

Exponent Identity

\(\log_a (a^x) = x\)

Logarithm “undoes” the exponential

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Change of Base Formula

Sometimes we need to evaluate logarithms with bases not on our calculator (which typically only has \(\log_{10}\) or \(\ln\)). The change of base formula allows us to convert between bases:

\(\log_a b = \frac{\log_c b}{\log_c a}\)

Where \(c\) is any valid base (usually 10 or \(e\) for calculators).

๐Ÿ“ Example 2: Using Change of Base

Evaluate \(\log_5 125\) using common logarithms (\(\log_{10}\)).

1
Apply change of base: \(\log_5 125 = \frac{\log_{10} 125}{\log_{10} 5}\)
2
Calculate: \(\log_{10} 125 โ‰ˆ 2.09691\)
3
Calculate: \(\log_{10} 5 โ‰ˆ 0.69897\)
4
Divide: \(2.09691 รท 0.69897 โ‰ˆ 3\)
5
Check: \(5^3 = 125\), so \(\log_5 125 = 3\) โœ“

Alternative Forms: The change of base formula can also be written as \(\log_a b = \frac{1}{\log_b a}\) or \(\log_a b = \frac{\ln b}{\ln a}\). All are equivalent and useful in different situations.

Part 4: Applying the Laws – Simplification & Equations

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Practical Applications

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Strategy for Simplifying Logarithmic Expressions
1
Express all numbers as powers of the base if possible
2
Apply the power law to bring exponents to the front
3
Use product/quotient laws to combine/split logs
4
Simplify using \(\log_a a = 1\) and \(\log_a 1 = 0\)
5
Combine like terms if possible
๐Ÿ“ Example 3: Simplifying Complex Expression

Simplify \(\log_2 8 + \log_2 4 – \log_2 16\)

1
Apply product law to first two terms: \(\log_2 (8 \times 4) – \log_2 16\)
2
Calculate product: \(\log_2 32 – \log_2 16\)
3
Apply quotient law: \(\log_2 \left(\frac{32}{16}\right)\)
4
Simplify: \(\log_2 2\)
5
Final answer: 1 (since \(\log_2 2 = 1\))
๐Ÿ“ Example 4: Solving Logarithmic Equations

Solve for \(x\): \(\log_3 (x + 1) + \log_3 (x – 1) = 1\)

1
Combine using product law: \(\log_3 [(x+1)(x-1)] = 1\)
2
Expand: \(\log_3 (x^2 – 1) = 1\)
3
Convert to exponential form: \(x^2 – 1 = 3^1\)
4
Simplify: \(x^2 – 1 = 3\)
5
Rearrange: \(x^2 = 4\)
6
Solve: \(x = \pm 2\)
7
Check domain: \(x+1 > 0\) and \(x-1 > 0\) โ‡’ \(x > 1\)
8
Thus \(x = -2\) is invalid. Final answer: \(x = 2\)

Critical Check: Always verify solutions satisfy the domain restrictions (arguments > 0). Many students lose marks by forgetting this step!

๐Ÿ“ Example 5: Solving Exponential Equations Using Logs

Solve for \(x\): \(3^{2x-1} = 5\)

1
Take log of both sides: \(\log_{10} (3^{2x-1}) = \log_{10} 5\)
2
Apply power law: \((2x-1)\log_{10} 3 = \log_{10} 5\)
3
Isolate \(x\): \(2x-1 = \frac{\log_{10} 5}{\log_{10} 3}\)
4
Calculate: \(\frac{\log_{10} 5}{\log_{10} 3} โ‰ˆ \frac{0.69897}{0.47712} โ‰ˆ 1.46497\)
5
Solve: \(2x = 1.46497 + 1 = 2.46497\)
6
Final answer: \(x โ‰ˆ 1.2325\)

Part 5: Natural Logarithms and Special Bases

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Common Logarithms (log) vs Natural Logarithms (ln)

Common Logarithm

Notation: \(\log x\) or \(\log_{10} x\)

Base: 10

Meaning: \(\log x = y\) means \(10^y = x\)

Usage: pH scale, Richter scale, decibels

\(\log 100 = 2\)

Natural Logarithm

Notation: \(\ln x\)

Base: \(e โ‰ˆ 2.71828\)

Meaning: \(\ln x = y\) means \(e^y = x\)

Usage: Calculus, growth/decay, compound interest

\(\ln e^3 = 3\)

Relationship

Conversion: \(\ln x = \frac{\log x}{\log e}\)

Approximation: \(\ln x โ‰ˆ 2.3026 \log x\)

Inverse: \(e^{\ln x} = x\) and \(\ln(e^x) = x\)

Special value: \(\ln 1 = 0\), \(\ln e = 1\)

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Important Properties of Natural Logarithms

All logarithm laws apply to natural logs:

\(\ln(ab) = \ln a + \ln b\)
\(\ln\left(\frac{a}{b}\right) = \ln a – \ln b\)
\(\ln(a^n) = n \ln a\)
\(\ln 1 = 0, \quad \ln e = 1, \quad e^{\ln x} = x\)

Part 6: CSEC Past Paper Questions

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Exam-Style Questions

CSEC May/June 2019 Paper 2

Question: Given that \(\log_a 2 = p\) and \(\log_a 3 = q\), express in terms of \(p\) and \(q\):

(a) \(\log_a 6\)

(b) \(\log_a \left(\frac{8}{9}\right)\)

(c) \(\log_a \sqrt{12}\)

(a)

\(\log_a 6 = \log_a (2 \times 3) = \log_a 2 + \log_a 3 = p + q\)

(b)

\(\log_a \left(\frac{8}{9}\right) = \log_a 8 – \log_a 9\)

\(= \log_a (2^3) – \log_a (3^2) = 3\log_a 2 – 2\log_a 3\)

\(= 3p – 2q\)

(c)

\(\log_a \sqrt{12} = \log_a (12^{\frac{1}{2}}) = \frac{1}{2} \log_a 12\)

\(= \frac{1}{2} \log_a (4 \times 3) = \frac{1}{2} [\log_a 4 + \log_a 3]\)

\(= \frac{1}{2} [\log_a (2^2) + q] = \frac{1}{2} [2\log_a 2 + q]\)

\(= \frac{1}{2} [2p + q] = p + \frac{q}{2}\)

CSEC January 2018 Paper 2

Question: Solve the equation \(\log_2 (x + 3) + \log_2 (x – 1) = 3\)

1
Combine logs: \(\log_2 [(x+3)(x-1)] = 3\)
2
Convert to exponential: \((x+3)(x-1) = 2^3 = 8\)
3
Expand: \(x^2 + 2x – 3 = 8\)
4
Rearrange: \(x^2 + 2x – 11 = 0\)
5
Solve quadratic: \(x = \frac{-2 \pm \sqrt{4 + 44}}{2} = \frac{-2 \pm \sqrt{48}}{2}\)
6
Simplify: \(x = \frac{-2 \pm 4\sqrt{3}}{2} = -1 \pm 2\sqrt{3}\)
7
Check domain: \(x+3 > 0\) and \(x-1 > 0\) โ‡’ \(x > 1\)
8
\(-1 – 2\sqrt{3} โ‰ˆ -4.46\) (reject), \(-1 + 2\sqrt{3} โ‰ˆ 2.46\) (accept)
9
Final answer: \(x = -1 + 2\sqrt{3}\)

Quiz: Test Your Understanding

Laws of Logarithms Quiz
Question 1: Simplify \(\log_2 16 + \log_2 2 – \log_2 8\)
Answer:
Combine first two: \(\log_2 (16 \times 2) – \log_2 8 = \log_2 32 – \log_2 8\)
Apply quotient law: \(\log_2 \left(\frac{32}{8}\right) = \log_2 4 = 2\)
Final answer: 2
Question 2: Given that \(\log_{10} 2 โ‰ˆ 0.3010\) and \(\log_{10} 3 โ‰ˆ 0.4771\), find \(\log_{10} 18\)
Answer:
\(18 = 2 \times 3^2\)
\(\log_{10} 18 = \log_{10} (2 \times 3^2) = \log_{10} 2 + 2\log_{10} 3\)
\(= 0.3010 + 2(0.4771) = 0.3010 + 0.9542 = 1.2552\)
Final answer: 1.2552
Question 3: Solve for \(x\): \(\log_3 (2x – 1) = 2\)
Answer:
Convert to exponential: \(2x – 1 = 3^2 = 9\)
\(2x = 10\)
\(x = 5\)
Check domain: \(2(5) – 1 = 9 > 0\) โœ“
Final answer: \(x = 5\)
Question 4: Express as a single logarithm: \(2\log x – 3\log y + \frac{1}{2}\log z\)
Answer:
Apply power law: \(\log x^2 – \log y^3 + \log z^{\frac{1}{2}}\)
Combine using product/quotient laws: \(\log \left(\frac{x^2 \sqrt{z}}{y^3}\right)\)
Final answer: \(\log \left(\frac{x^2 \sqrt{z}}{y^3}\right)\)
Question 5: Solve for \(x\): \(5^{2x} = 7\), giving your answer to 3 decimal places
Answer:
Take logs: \(\log 5^{2x} = \log 7\)
\(2x \log 5 = \log 7\)
\(2x = \frac{\log 7}{\log 5}\)
\(x = \frac{\log 7}{2\log 5} โ‰ˆ \frac{0.8451}{2 \times 0.6990} โ‰ˆ \frac{0.8451}{1.3980} โ‰ˆ 0.6045\)
Final answer: \(x โ‰ˆ 0.605\) (to 3 decimal places)

๐ŸŽฏ Key Concepts Summary

  • Definition: \(\log_a b = x\) โ‡” \(a^x = b\)
  • Three Fundamental Laws:
    • Product Law: \(\log_a (mn) = \log_a m + \log_a n\)
    • Quotient Law: \(\log_a \left(\frac{m}{n}\right) = \log_a m – \log_a n\)
    • Power Law: \(\log_a (m^n) = n \log_a m\)
  • Special Values:
    • \(\log_a a = 1\)
    • \(\log_a 1 = 0\)
    • \(\log_a \left(\frac{1}{n}\right) = -\log_a n\)
  • Change of Base: \(\log_a b = \frac{\log_c b}{\log_c a}\)
  • Natural Logarithms: \(\ln x = \log_e x\), where \(e โ‰ˆ 2.71828\)
  • Domain Restrictions: For \(\log_a x\), must have \(a > 0, a โ‰  1\), and \(x > 0\)
  • Common CSEC Question Types:
    • Simplify logarithmic expressions using laws
    • Solve logarithmic equations
    • Solve exponential equations using logs
    • Express logs in terms of given variables
    • Change of base calculations
  • Exam Strategy:
    • Always check domain restrictions when solving equations
    • Remember \(\log_a x\) only exists for \(x > 0\)
    • When combining logs, watch signs carefully
    • For exponential equations, take logs of both sides
    • Show all steps for full method marks

CSEC Exam Strategy: When solving logarithmic equations: (1) Combine logs using the laws, (2) Convert to exponential form, (3) Solve the resulting equation, (4) ALWAYS check that solutions satisfy the domain (\(x > 0\) inside all logs). When simplifying: (1) Express numbers as powers when possible, (2) Apply power law first, then product/quotient laws, (3) Look for opportunities to use \(\log_a a = 1\) and \(\log_a 1 = 0\).

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