Laws of Logarithms (AQA A-Level Mathematics): Revision Notes

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6.2.1 Laws of Logarithms

Laws of Logarithms

aabblog⁡a\log_alog⁡b\log_blog⁡(ab)\log(ab)log⁡(a)+log⁡(b)\log(a) + \log(b)
2.160.32220.77821.10041.1004
330.477100.47710.4771
π\pi2.40.49710.38020.87740.8774
2\sqrt{2}3\sqrt{3}0.15050.42250.57300.5730

Key Result

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log⁡(ab)≡log⁡(a)+log⁡(b)\log(ab) \equiv \log(a) + \log(b)

This suggests that:

Conversely,

10a×10b=10a+b10^a \times 10^b = 10^{a+b}

This just says the power of two things with the same base multiplied together is simply those two powers added.

Logarithms: Simplifying and Log Laws

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Examples:

  1. Simplifying a Logarithm:
log⁡10(10×1000)=log⁡10(101×103)=log⁡10(104)=4\log_{10}(10 \times 1000) = \log_{10}(10^1 \times 10^3) = \log_{10}(10^4) = 4
  1. Simplifying into a Single Logarithm:
log⁡3(10)+log⁡3(5)=log⁡3(10×5)=log⁡3(50)\log_{3}(10) + \log_{3}(5) = \log_{3}(10 \times 5) = \log_{3}(50)log⁡5(20)+log⁡5(14)=log⁡5(20×14)=log⁡5(5)=1\log_{5}(20) + \log_{5}\left(\frac{1}{4}\right) = \log_{5}\left(20 \times \frac{1}{4}\right) = \log_{5}(5) = 1log⁡10(20)+log⁡10(20)=log⁡10(20×20)=log⁡10(400)\log_{10}(20) + \log_{10}(20) = \log_{10}(20 \times 20) = \log_{10}(400)

Summary of Log Laws

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  1. Addition Rule:
log⁡(a)+log⁡(b)=log⁡(ab)\log(a) + \log(b) = \log(ab)
  1. Subtraction Rule:
log⁡(a)−log⁡(b)=log⁡(ab)\log(a) - \log(b) = \log\left(\frac{a}{b}\right)
  1. Power Rule:
blog⁡(a)=log⁡(ab)b \log(a) = \log(a^b)

Proof of Power Rule

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Starting with:

blog⁡(a)=log⁡(a)+log⁡(a)+⋯+log⁡(a)(b times)b \log(a) = \log(a) + \log(a) + \dots + \log(a) \quad \text{(b times)}

Using the addition rule, this simplifies to:

=log⁡(a×a×a×⋯×a)(b times)= \log(a \times a \times a \times \dots \times a) \quad \text{(b times)}=log⁡(ab) □= \log(a^b)\ \square

Laws of Logarithms

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2x×23=2x+32^x \times 2^3 = 2^{x+3} (When we multiply two things together with the same base, we add powers) 2x÷2y=2x−y2^x \div 2^y = 2^{x-y} (When we divide, we subtract the powers)

Using Laws of Logarithms

Express in the form log⁡an\log_a n:

  1. a.
log⁡a(4)+log⁡a(7)=log⁡a(4×7)=log⁡a(28)(Addition Rule)\log_a(4) + \log_a(7) = \log_a(4 \times 7) = \log_a(28) \quad \text{(Addition Rule)}
  1. b.
log⁡a(10)−log⁡a(5)=log⁡a(105)=log⁡a(2)(Subtraction Rule)\log_a(10) - \log_a(5) = \log_a\left(\frac{10}{5}\right) = \log_a(2) \quad \text{(Subtraction Rule)}
  1. c.
2log⁡a(6)=log⁡a(62)=log⁡a(36)(Power Rule)2\log_a(6) = \log_a(6^2) = \log_a(36) \quad \text{(Power Rule)}
  1. d.
log⁡a(9)−log⁡a(13)=log⁡a(9×3)=log⁡a(27)(Subtraction Rule)\log_a(9) - \log_a\left(\frac{1}{3}\right) = \log_a(9 \times 3) = \log_a(27) \quad \text{(Subtraction Rule)}
  1. e.
12log⁡a(25)+2log⁡a(3)=log⁡a(2512)+log⁡a(32)=log⁡a(5)+log⁡a(9)=log⁡a(45)(Power Rule, Addition Rule)\frac{1}{2}\log_a(25) + 2\log_a(3) = \log_a(25^{\frac{1}{2}}) + \log_a(3^2) = \log_a(5) + \log_a(9) = \log_a(45) \quad \text{(Power Rule, Addition Rule)}
  1. f.
log⁡a(48)−32log⁡a(2)−12log⁡a(9)\log_a(48) - \frac{3}{2}\log_a(2) - \frac{1}{2}\log_a(9) =log⁡a(48)−log⁡a(23)−log⁡a(3)= \log_a(48) - \log_a(2^3) - \log_a(3) =log⁡a(488×3)=log⁡a(2)(Subtraction Rule)= \log_a\left(\frac{48}{8 \times 3}\right) = \log_a(2) \quad \text{(Subtraction Rule)}
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Example: Write as a sum of separate logarithms involving no indices:

log⁡(3a2b3)\log(3a^2b^3)

A common mistake:

log⁡(3a2b3)=log⁡(3ab)+10log⁡(ab)(Incorrect)\log(3a^2b^3) = \log(3ab) + 10\log(ab) \quad \text{(Incorrect)}

Correct:

log⁡(3)+2log⁡(a)+3log⁡(b)\log(3) + 2\log(a) + 3\log(b)

Note: The power of 10 is only attached to b3b^3; it should not be applied to all terms.

Simplification of Logarithmic Expressions

The correct way to simplify:

log⁡(3ab9)=log⁡(3)+log⁡(a)+10log⁡(b)(Addition Rule, Power Rule)\log(3ab^9) = \log(3) + \log(a) + 10\log(b) \quad \text{(Addition Rule, Power Rule)}
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Example: Simplify 1+log⁡3(5)1 + \log_3(5) as a single logarithm: 11 is not in logarithmic form.

log⁡3(3)=1\log_3(3) = 1

Therefore,

1+log⁡3(5)=log⁡3(3×5)=log⁡3(15)1 + \log_3(5) = \log_3(3 \times 5) = \log_3(15)

Express the following in terms of pp and qq:

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Given that log⁡ax=p\log_a x = p and log⁡ay=q\log_a y = q: 6. (i) log⁡a(xy)\log_a (xy)

log⁡a(xy)=log⁡a(x)+log⁡a(y)=p+q\log_a (xy) = \log_a (x) + \log_a (y) = p + q
  1. (ii) log⁡a(x2⋅x3y)\log_a \left(\frac{x^2 \cdot x^3}{y}\right)
log⁡a(x2⋅x3y)=log⁡a(x2)+log⁡a(x3)−log⁡a(y)\log_a \left(\frac{x^2 \cdot x^3}{y}\right) = \log_a (x^2) + \log_a (x^3) - \log_a (y)=2log⁡a(x)+3log⁡a(x)−log⁡a(y)= 2\log_a (x) + 3\log_a (x) - \log_a (y)=2p+3p−q=5p−q= 2p + 3p - q = 5p - q
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Examples

Write as a single logarithms, simplifying where appropriate

  1. Simplification Using Addition Rule:
log⁡(2)+log⁡(5)=log⁡(10)=1\log(2) + \log(5) = \log(10) = 1
  1. Combining Logs:
ln⁡(x)+3ln⁡(y)=ln⁡(x)+ln⁡(y3)=ln⁡(x⋅y3)\ln(x) + 3 \ln(y) = \ln(x) + \ln(y^3) = \ln(x \cdot y^3)

(Here, Power Law and Addition Law are applied)

Simplifying Logarithmic Expressions

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Example Expression:

12log⁡(9)+3log⁡(2)\frac{1}{2} \log(9) + 3\log(2)

Step 1: Apply the Power Rule

12log⁡(9)=log⁡(912)and3log⁡(2)=log⁡(23)\frac{1}{2} \log(9) = \log(9^{\frac{1}{2}}) \quad \text{and} \quad 3\log(2) = \log(2^3)

Step 2: Substitute the values

=log⁡(3)+log⁡(8)= \log(3) + \log(8)

Step 3: Apply the Addition Rule

=log⁡(3×8)=log⁡(24)= \log(3 \times 8) = \log(24)

Logarithms: Correct Method for Solving Equations

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Example Problem:

Given the equation:

log⁡(x−5)−log⁡(x)=7\log(x - 5) - \log(x) = 7

Wrong Method:

Attempting to solve by "unlogging" each term separately:

10log⁡(x−5)−10log⁡(x)=10710^{\log(x-5)} - 10^{\log(x)} = 10^7

Reason:

1+2=3but101+102≠1031 + 2 = 3 \quad \text{but} \quad 10^1 + 10^2 \neq 10^3

Correct Method:

Gather all logarithms into one term before "unlogging":

log⁡(x−5x)=7\log\left(\frac{x-5}{x}\right) = 7

Applying the Subtraction Rule:

x−5x=107\frac{x-5}{x} = 10^7

Cross-multiply:

x−5=107xx - 5 = 10^7x

Isolating xx:

−5=107x−x-5 = 10^7x - x−5=9999999x⇒x=−59999999-5 = 9999999x \quad \Rightarrow \quad x = \frac{-5}{9999999}

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