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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.