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Number and Algebra - Logarithms and their properties

Grade 10IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Relationship between Exponents and Logarithms: A logarithm is the inverse operation of exponentiation. If ax=ba^x = b, then x=log⁡abx = \log_a b. Visually, imagine the base aa 'moving across' to the other side to support the logarithm, while the exponent xx drops down to become the subject of the equation.

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Properties of Logarithmic Graphs: The graph of y=log⁡axy = \log_a x (where a>1a > 1) always passes through the point (1,0)(1, 0) and has a vertical asymptote at the y-axis (x=0x=0). Visually, the curve rises steeply from the bottom near the y-axis and then flattens out as it moves to the right, staying entirely in the first and fourth quadrants because the argument xx must be positive.

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The Product and Quotient Laws: These laws state that log⁡a(xy)=log⁡ax+log⁡ay\log_a(xy) = \log_a x + \log_a y and log⁡a(xy)=log⁡ax−log⁡ay\log_a(\frac{x}{y}) = \log_a x - \log_a y. Visually, these properties allow us to 'stretch' a single condensed log expression into a horizontal sum or difference of simpler parts, or 'compress' them back into one.

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The Power Law: The exponent of the argument in a logarithm can be moved to the front as a coefficient: log⁡a(xk)=klog⁡ax\log_a(x^k) = k \log_a x. In terms of manipulation, this transforms an exponential relationship into a linear multiplication, which is essential for solving equations where the unknown is in the exponent.

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Common and Natural Logarithms: Logarithms with base 1010 are known as common logarithms and are typically written simply as log⁡x\log x. Logarithms with the irrational base e≈2.718e \approx 2.718 are natural logarithms, written as ln⁡x\ln x. These are the standard functions found on scientific calculators.

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Change of Base Formula: To calculate a logarithm with a base other than 1010 or ee, use the formula log⁡ab=log⁡cblog⁡ca\log_a b = \frac{\log_c b}{\log_c a}. Visually, this represents the ratio between the logarithms of the argument and the old base calculated in a new, more convenient base.

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Identity and Zero Rules: For any valid base aa, log⁡aa=1\log_a a = 1 (because a1=aa^1 = a) and log⁡a1=0\log_a 1 = 0 (because a0=1a^0 = 1). On a graph, this explains why every basic log function crosses the x-axis at x=1x=1 and has a height of 11 when xx equals the base.

📐Formulae

ax=b  ⟺  x=log⁡aba^x = b \iff x = \log_a b

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

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

log⁡a(xk)=klog⁡ax\log_a (x^k) = k \log_a x

log⁡aa=1\log_a a = 1

log⁡a1=0\log_a 1 = 0

log⁡ab=log⁡cblog⁡ca\log_a b = \frac{\log_c b}{\log_c a}

log⁡a(ax)=x\log_a (a^x) = x

💡Examples

Problem 1:

Solve for xx: 2log⁡3x−log⁡3(x−2)=22\log_3 x - \log_3 (x-2) = 2

Solution:

  1. Use the power law to move the coefficient: log⁡3x2−log⁡3(x−2)=2\log_3 x^2 - \log_3 (x-2) = 2
  2. Use the quotient law to combine the logs: log⁡3(x2x−2)=2\log_3 (\frac{x^2}{x-2}) = 2
  3. Convert to exponential form: x2x−2=32\frac{x^2}{x-2} = 3^2
  4. Simplify: x2x−2=9\frac{x^2}{x-2} = 9
  5. Multiply by the denominator: x2=9(x−2)  ⟹  x2=9x−18x^2 = 9(x-2) \implies x^2 = 9x - 18
  6. Rearrange into a quadratic: x2−9x+18=0x^2 - 9x + 18 = 0
  7. Factor: (x−6)(x−3)=0(x-6)(x-3) = 0
  8. Potential solutions: x=6x=6 or x=3x=3. Both are valid as they result in positive arguments for the original logarithms.

Explanation:

This problem uses log laws to condense the left side into a single logarithm, allowing us to 'undo' the log by converting it into index form and solving the resulting quadratic equation.

Problem 2:

Solve for xx: 52x−1=125^{2x-1} = 12, giving your answer to 3 significant figures.

Solution:

  1. Take the natural log (ln⁡\ln) of both sides: ln⁡(52x−1)=ln⁡(12)\ln(5^{2x-1}) = \ln(12)
  2. Use the power law to bring the exponent down: (2x−1)ln⁡5=ln⁡12(2x-1) \ln 5 = \ln 12
  3. Divide by ln⁡5\ln 5: 2x−1=ln⁡12ln⁡52x - 1 = \frac{\ln 12}{\ln 5}
  4. Add 1 to both sides: 2x=ln⁡12ln⁡5+12x = \frac{\ln 12}{\ln 5} + 1
  5. Divide by 2: x=12(ln⁡12ln⁡5+1)x = \frac{1}{2}(\frac{\ln 12}{\ln 5} + 1)
  6. Calculate the value: x≈12(1.54395+1)≈1.27197x \approx \frac{1}{2}(1.54395 + 1) \approx 1.27197
  7. Final answer: x≈1.27x \approx 1.27

Explanation:

When the variable is in the exponent and the bases cannot be made equal, we take the logarithm of both sides to extract the exponent using the power law.

Logarithms and their properties - Revision Notes & Key Formulas | IB Grade 10 Maths