Skip to main content

Logarithms

Subject: Additional Mathematics
Topic: 5
Cambridge Code: 4037 / 0606


Definition of Logarithm​

Logarithm - The inverse function of an exponential function

Relationship to Exponentials​

If ax=ba^x = b, then log⁡ab=x\log_a b = x

In other words: log⁡ab=x  ⟺  ax=b\log_a b = x \iff a^x = b

Examples​

  • log⁡28=3\log_2 8 = 3 because 23=82^3 = 8 ✓
  • log⁡10100=2\log_{10} 100 = 2 because 102=10010^2 = 100 ✓
  • log⁡525=2\log_5 25 = 2 because 52=255^2 = 25 ✓

Key Points​

  • aa is the base (must be positive, a≠1a \neq 1)
  • bb is the argument (must be positive)
  • xx is the logarithm (can be any real number)

Common Logarithms​

Base 10 Logarithm​

log⁡b=log⁡10b\log b = \log_{10} b

Usually written without the base (implied base 10)

Natural Logarithm​

ln⁡b=log⁡eb\ln b = \log_e b

where e≈2.718e \approx 2.718 (base of natural logarithms)

Examples​

  • log⁡100=log⁡10100=2\log 100 = \log_{10} 100 = 2
  • log⁡1=0\log 1 = 0 (any base)
  • ln⁡e=1\ln e = 1

Laws of Logarithms​

Law 1: Product Rule​

log⁡a(xy)=log⁡ax+log⁡ay\log_a(xy) = \log_a x + \log_a y

Example: log⁡2(8×4)=log⁡28+log⁡24=3+2=5\log_2(8 \times 4) = \log_2 8 + \log_2 4 = 3 + 2 = 5 ✓ (Check: 8×4=32=258 \times 4 = 32 = 2^5)

Law 2: Quotient Rule​

log⁡a(xy)=log⁡ax−log⁡ay\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y

Example: log⁡10(10010)=log⁡10100−log⁡1010=2−1=1\log_{10}\left(\frac{100}{10}\right) = \log_{10} 100 - \log_{10} 10 = 2 - 1 = 1 ✓ (Check: 10010=10\frac{100}{10} = 10)

Law 3: Power Rule​

log⁡a(xn)=nlog⁡ax\log_a(x^n) = n \log_a x

Example: log⁡2(82)=2log⁡28=2×3=6\log_2(8^2) = 2\log_2 8 = 2 \times 3 = 6 ✓ (Check: 82=64=268^2 = 64 = 2^6)

Law 4: Change of Base​

log⁡ax=log⁡bxlog⁡ba\log_a x = \frac{\log_b x}{\log_b a}

Or more commonly: log⁡ax=ln⁡xln⁡a or log⁡ax=log⁡xlog⁡a\log_a x = \frac{\ln x}{\ln a} \text{ or } \log_a x = \frac{\log x}{\log a}

Example: Find log⁡58\log_5 8 log⁡58=ln⁡8ln⁡5=2.0791.609≈1.292\log_5 8 = \frac{\ln 8}{\ln 5} = \frac{2.079}{1.609} \approx 1.292

Special Cases​

log⁡a1=0(any base)\log_a 1 = 0 \quad \text{(any base)} log⁡aa=1(any base)\log_a a = 1 \quad \text{(any base)} alog⁡ax=xa^{\log_a x} = x log⁡a(ax)=x\log_a(a^x) = x


Solving Logarithmic Equations​

Type 1: Direct Application​

Equations where you apply the definition

Example: Solve log⁡2x=5\log_2 x = 5

x=25=32x = 2^5 = 32

Type 2: Using Laws of Logarithms​

Example: Solve log⁡x+log⁡4=2\log x + \log 4 = 2

log⁡(4x)=2\log(4x) = 2 4x=102=1004x = 10^2 = 100 x=25x = 25

Type 3: Complex Equations​

Example: Solve 2ln⁡x−ln⁡4=ln⁡82\ln x - \ln 4 = \ln 8

ln⁡(x2)−ln⁡4=ln⁡8\ln(x^2) - \ln 4 = \ln 8 ln⁡(x24)=ln⁡8\ln\left(\frac{x^2}{4}\right) = \ln 8 x24=8\frac{x^2}{4} = 8 x2=32x^2 = 32 x=32=42x = \sqrt{32} = 4\sqrt{2} (taking positive root)


Solving Exponential Equations​

Use logarithms to solve equations with variables in exponent

Example 1: Solve 3x=203^x = 20

Taking logarithm of both sides: log⁡(3x)=log⁡20\log(3^x) = \log 20 xlog⁡3=log⁡20x \log 3 = \log 20 x=log⁡20log⁡3=1.3010.477≈2.727x = \frac{\log 20}{\log 3} = \frac{1.301}{0.477} \approx 2.727

Example 2: Solve 2x+1=72^{x+1} = 7

log⁡(2x+1)=log⁡7\log(2^{x+1}) = \log 7 (x+1)log⁡2=log⁡7(x+1)\log 2 = \log 7 x+1=log⁡7log⁡2=0.8450.301≈2.807x+1 = \frac{\log 7}{\log 2} = \frac{0.845}{0.301} \approx 2.807 x≈1.807x \approx 1.807


Logarithmic Functions and Graphs​

Properties of f(x)=log⁡axf(x) = \log_a x (where a>1a > 1)​

  • Domain: x>0x > 0
  • Range: All real numbers
  • Asymptote: Vertical line at x=0x = 0
  • x-intercept: (1,0)(1, 0)
  • Increasing: As xx increases, yy increases
  • Inverse: f−1(x)=axf^{-1}(x) = a^x (exponential)

Graph Features​

  • Passes through (1,0)(1, 0): log⁡a1=0\log_a 1 = 0
  • Passes through (a,1)(a, 1): log⁡aa=1\log_a a = 1
  • Curve above x-axis for x>1x > 1
  • Curve below x-axis for 0<x<10 < x < 1
  • Vertical asymptote at x=0x = 0

Key Points to Remember​

  1. Logarithm is inverse of exponential
  2. log⁡ab=x\log_a b = x means ax=ba^x = b
  3. Four main laws: product, quotient, power, change of base
  4. Logarithms only defined for positive arguments
  5. Change of base to evaluate logs with any base
  6. Use logs to solve exponential equations

Worked Examples​

Example 1: Simplify Using Laws​

Simplify log⁡327+log⁡39−log⁡33\log_3 27 + \log_3 9 - \log_3 3

=log⁡3(27×9÷3)=log⁡3(81)=log⁡3(34)=4= \log_3(27 \times 9 \div 3) = \log_3(81) = \log_3(3^4) = 4

Example 2: Solve Logarithmic Equation​

Solve log⁡(2x−1)=log⁡5+log⁡3\log(2x-1) = \log 5 + \log 3

log⁡(2x−1)=log⁡15\log(2x-1) = \log 15 2x−1=152x - 1 = 15 x=8x = 8

Example 3: Solve Exponential Equation​

Solve 5x=1005^x = 100 (give answer to 3 d.p.)

ln⁡(5x)=ln⁡100\ln(5^x) = \ln 100 xln⁡5=ln⁡100x \ln 5 = \ln 100 x=ln⁡100ln⁡5=4.6051.609=2.861x = \frac{\ln 100}{\ln 5} = \frac{4.605}{1.609} = 2.861


Practice Questions​

  1. Evaluate:

    • log⁡232\log_2 32
    • log⁡42\log_4 2
    • log⁡273\log_{27} 3
  2. Simplify:

    • log⁡50+log⁡2−log⁡5\log 50 + \log 2 - \log 5
    • 2ln⁡3+ln⁡2−ln⁡62\ln 3 + \ln 2 - \ln 6
  3. Solve:

    • log⁡x16=4\log_x 16 = 4
    • 2x=502^x = 50
    • ln⁡(x+1)+ln⁡2=ln⁡8\ln(x+1) + \ln 2 = \ln 8

Revision Tips​

  • Learn the four laws thoroughly
  • Remember domain restriction: argument must be positive
  • Use change of base for any base
  • Logarithms useful for solving exponential equations
  • Graphs are inverse of exponential functions