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 logab=x\log_a b = x

In other words: logab=x    ax=b\log_a b = x \iff a^x = b

Examples

  • log28=3\log_2 8 = 3 because 23=82^3 = 8
  • log10100=2\log_{10} 100 = 2 because 102=10010^2 = 100
  • log525=2\log_5 25 = 2 because 52=255^2 = 25

Key Points

  • aa is the base (must be positive, a1a \neq 1)
  • bb is the argument (must be positive)
  • xx is the logarithm (can be any real number)

Common Logarithms

Base 10 Logarithm

logb=log10b\log b = \log_{10} b

Usually written without the base (implied base 10)

Natural Logarithm

lnb=logeb\ln b = \log_e b

where e2.718e \approx 2.718 (base of natural logarithms)

Examples

  • log100=log10100=2\log 100 = \log_{10} 100 = 2
  • log1=0\log 1 = 0 (any base)
  • lne=1\ln e = 1

Laws of Logarithms

Law 1: Product Rule

loga(xy)=logax+logay\log_a(xy) = \log_a x + \log_a y

Example: log2(8×4)=log28+log24=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

loga(xy)=logaxlogay\log_a\left(\frac{x}{y}\right) = \log_a x - \log_a y

Example: log10(10010)=log10100log1010=21=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

loga(xn)=nlogax\log_a(x^n) = n \log_a x

Example: log2(82)=2log28=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

logax=logbxlogba\log_a x = \frac{\log_b x}{\log_b a}

Or more commonly: logax=lnxlna or logax=logxloga\log_a x = \frac{\ln x}{\ln a} \text{ or } \log_a x = \frac{\log x}{\log a}

Example: Find log58\log_5 8 log58=ln8ln5=2.0791.6091.292\log_5 8 = \frac{\ln 8}{\ln 5} = \frac{2.079}{1.609} \approx 1.292

Special Cases

loga1=0(any base)\log_a 1 = 0 \quad \text{(any base)} logaa=1(any base)\log_a a = 1 \quad \text{(any base)} alogax=xa^{\log_a x} = x loga(ax)=x\log_a(a^x) = x


Solving Logarithmic Equations

Type 1: Direct Application

Equations where you apply the definition

Example: Solve log2x=5\log_2 x = 5

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

Type 2: Using Laws of Logarithms

Example: Solve logx+log4=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 2lnxln4=ln82\ln x - \ln 4 = \ln 8

ln(x2)ln4=ln8\ln(x^2) - \ln 4 = \ln 8 ln(x24)=ln8\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)=log20\log(3^x) = \log 20 xlog3=log20x \log 3 = \log 20 x=log20log3=1.3010.4772.727x = \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)=log7\log(2^{x+1}) = \log 7 (x+1)log2=log7(x+1)\log 2 = \log 7 x+1=log7log2=0.8450.3012.807x+1 = \frac{\log 7}{\log 2} = \frac{0.845}{0.301} \approx 2.807 x1.807x \approx 1.807


Logarithmic Functions and Graphs

Properties of f(x)=logaxf(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: f1(x)=axf^{-1}(x) = a^x (exponential)

Graph Features

  • Passes through (1,0)(1, 0): loga1=0\log_a 1 = 0
  • Passes through (a,1)(a, 1): logaa=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. logab=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 log327+log39log33\log_3 27 + \log_3 9 - \log_3 3

=log3(27×9÷3)=log3(81)=log3(34)=4= \log_3(27 \times 9 \div 3) = \log_3(81) = \log_3(3^4) = 4

Example 2: Solve Logarithmic Equation

Solve log(2x1)=log5+log3\log(2x-1) = \log 5 + \log 3

log(2x1)=log15\log(2x-1) = \log 15 2x1=152x - 1 = 15 x=8x = 8

Example 3: Solve Exponential Equation

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

ln(5x)=ln100\ln(5^x) = \ln 100 xln5=ln100x \ln 5 = \ln 100 x=ln100ln5=4.6051.609=2.861x = \frac{\ln 100}{\ln 5} = \frac{4.605}{1.609} = 2.861


Practice Questions

  1. Evaluate:

    • log232\log_2 32
    • log42\log_4 2
    • log273\log_{27} 3
  2. Simplify:

    • log50+log2log5\log 50 + \log 2 - \log 5
    • 2ln3+ln2ln62\ln 3 + \ln 2 - \ln 6
  3. Solve:

    • logx16=4\log_x 16 = 4
    • 2x=502^x = 50
    • ln(x+1)+ln2=ln8\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