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CIE A-Level Maths Study Notes

2.2.3 Solving Equations with Logarithms

In logarithmic equations, where the unknown is an exponent, are fundamental. These equations require a deep understanding of logarithms and exponents, and their interplay is crucial for solving complex mathematical problems.

Understanding Logarithmic Equations

Logarithmic equations, where the unknown is in the exponent, are solved by exploiting the relationship between logarithms and exponents. This involves using the inverse nature of logarithms to 'undo' exponential functions and applying logarithmic laws to simplify and solve the equations.

Key Components:

1. Exponential Functions:

Represented as axa^x, where aa is a positive constant and xx is the exponent.

2. Logarithmic Functions:

Denoted as loga(x)\log_a(x), these are the inverse of exponential functions, answering "To what power must aa be raised to obtain xx?".

3. Logarithmic Laws:

Critical for simplifying and solving logarithmic equations.

  • Product Law: loga(xy)=loga(x)+loga(y)\log_a(xy) = \log_a(x) + \log_a(y)
  • Quotient Law: loga(xy)=loga(x)loga(y)\log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y)
  • Power Law: loga(xb)=bloga(x)\log_a(x^b) = b \log_a(x)
  • Change of Base Formula: logb(x)=loga(x)loga(b).\log_b(x) = \frac{\log_a(x)}{\log_a(b)}.

Solving Strategy:

1. Isolate the exponential part of the equation.

2. Apply Logarithms on both sides, using a common base like 10 or ee.

3. Utilize Logarithm Laws to simplify the equation into a more solvable form.

4. Solve for the Variable using algebraic techniques.

Examples:

Example 1: Solve 25^x < 1

  • Convert to Logarithmic Form: 2525 as 525^2, so 5^{2x} < 1.
  • Apply Logarithms: \log5^{2x} < \log(1).
  • Simplify Using Logarithm Laws: 2x \log5 < 0 since log1=0.\log1 = 0.
  • Solve for xx: As \log5 > 0, x < 0.

Answer: x < 0.

Example 2: Solve 3^{2x} < 5^{3x-1}

  • Apply Logarithms: \log3^{2x} < \log5^{3x-1}.
  • Use Logarithm Laws: Simplify to 2xlog32x \log3 < 3x1log5.3x-1 \log5.
  • Rearrange and Solve: 2x \log(3) - 3x \log(5) &lt; -\log(5), simplifying to x(2\log(3) - 3\log(5)) &lt; -\log(5)

Answer: Solve forx x: x &lt; \frac{-\log(5)}{2\log(3) - 3\log(5)} = 0.6117.

Example 3: Solve 23x=82^{3x} = 8

  • Recognize the Equation: 23x=8.2^{3x} = 8.
  • Simplify the Equation: 8 as 23,2^3, so 23x=23.2^{3x} = 2^3.
  • Equating Exponents: As bases are the same, 3x=3.3x = 3.
  • Solve for x:x: x=33=1.x = \frac{3}{3} = 1.

Answer: x=1.x = 1.

Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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