Logarithm Properties

Grade 11 · algebra · 85 practice problems · read aloud

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📊 Logarithm Properties

Logarithms are the inverse operations of exponents. They help us solve equations where the unknown appears in the exponent, and they're essential in many real-world applications like measuring earthquake intensity (Richter scale), sound (decibels), and pH in chemistry.

Key Properties & Formulas

  • Product Rule: logₐ(mn) = logₐm + logₐn
  • Quotient Rule: logₐ(m/n) = logₐm - logₐn
  • Power Rule: logₐ(mⁿ) = n·logₐm
  • Change of Base: logₐb = logₓb/logₓa

Step-by-Step Problem Solving

  1. Identify which properties apply to the expression
  2. Apply one property at a time, working from the inside out
  3. Simplify using basic arithmetic where possible
  4. Check your work by verifying domain restrictions

Example 1: Expand log₂(8x³/y)

Step 1: Apply quotient rule: log₂(8x³) - log₂y

Step 2: Apply product rule: log₂8 + log₂(x³) - log₂y

Step 3: Apply power rule & simplify: 3 + 3log₂x - log₂y

Example 2: Condense 2lnx - ½lny

Step 1: Apply power rule: ln(x²) - ln(√y)

Step 2: Apply quotient rule: ln(x²/√y)

⚠️ Common Mistakes

  • Distributing logs: log(m+n) ≠ logm + logn
  • Forgetting domain: logₐx only exists when x > 0
  • Misapplying power rule: logₐmⁿ ≠ (logₐm)ⁿ
  • Base confusion: log without base usually means log₁₀ (common log)

💡 Tips & Tricks

  • Memory aid: "Multiplication adds, division subtracts, exponents come out front"
  • Domain check: Always verify arguments are positive before starting
  • Change of base: Use when you need to calculate logs with different bases
  • Natural log shortcut: ln(eˣ) = x and e^(lnx) = x

Practice Suggestions

Start with expanding and condensing expressions, then move to solving logarithmic equations. Try these:

  • Expand: log₃(27x²y⁵)
  • Condense: 2logx + 3log(x+1) - ½logy
  • Solve: log₂(x+3) + log₂(x-1) = 3

Practice with both common logs (base 10) and natural logs (base e) to build fluency.

Practice problems

6 of the 85, worked through step by step — try them before opening the answer.

1 log₂(32x) + log₂(x) = 7

Hint: Use the product property of logarithms to combine the two logs into one, then rewrite the equation in exponential form.

Show the answer

Answer: x = 2

  1. Apply the product property: log₂(32x) + log₂(x) = log₂(32x * x) = log₂(32x²).
  2. The equation becomes log₂(32x²) = 7.
  3. Rewrite in exponential form: 2⁷ = 32x².
  4. Compute 2⁷ = 128, so 128 = 32x².
  5. Divide both sides by 32: x² = 128/32 = 4.
  6. Take the square root: x = 2 (since x > 0 for logarithms to be defined).

The answer is x = 2.

2 log₂(8x) + log₂(x/2) = 6

Hint: Combine the two logarithms using the product rule, then rewrite the equation in exponential form to solve for x.

Show the answer

Answer: x = 4

  1. Use the product rule: log₂(8x) + log₂(x/2) = log₂((8x)(x/2)) = log₂(4x²).
  2. The equation becomes log₂(4x²) = 6.
  3. Rewrite in exponential form: 2⁶ = 4x².
  4. Compute 2⁶ = 64, so 64 = 4x².
  5. Divide both sides by 4: 16 = x².
  6. Take the square root: x = 4 (since x > 0 for the logarithm to be defined).

The answer is x = 4.

3 log₂(8x) + log₂(x/2) = 7

Hint: Use the product rule to combine the two logarithms into a single logarithm, then rewrite the equation in exponential form to solve for x.

Show the answer

Answer: x = 4

  1. Apply the product rule: log₂(8x) + log₂(x/2) = log₂((8x)(x/2)) = log₂(4x²).
  2. The equation becomes log₂(4x²) = 7.
  3. Rewrite in exponential form: 2⁷ = 4x².
  4. Compute 2⁷ = 128, so 128 = 4x².
  5. Divide both sides by 4: x² = 32.
  6. Take the square root: x = √32 = 4√2.
  7. Since the domain requires x > 0, the solution is x = 4√2.

The answer is x = 4√2.

4 log₃(27x) + log₃(x/9) = 5

Hint: Use the product and quotient properties of logarithms to combine the two log terms into a single logarithm, then rewrite the equation in exponential form to solve for x.

Show the answer

Answer: x = 9

  1. Apply the product property: log₃(27x) = log₃(27) + log₃(x).
  2. Apply the quotient property: log₃(x/9) = log₃(x) - log₃(9).
  3. The equation becomes: log₃(27) + log₃(x) + log₃(x) - log₃(9) = 5.
  4. Evaluate log₃(27) = 3 because 3³ = 27. Evaluate log₃(9) = 2 because 3² = 9.
  5. Substitute: 3 + 2log₃(x) - 2 = 5, which simplifies to 1 + 2log₃(x) = 5.
  6. Subtract 1 from both sides: 2log₃(x) = 4.
  7. Divide by 2: log₃(x) = 2.
  8. Rewrite in exponential form: x = 3² = 9.

The answer is x = 9.

5 log₂(32x) + log₂(x/8) = 6

Hint: Use the product and quotient rules to combine the logarithms into a single logarithm, then rewrite in exponential form to solve for x.

Show the answer

Answer: x = 4

  1. Apply the product rule: log₂(32x) + log₂(x/8) = log₂(32x * x/8) = log₂(4x²)
  2. Set equal to 6: log₂(4x²) = 6
  3. Rewrite in exponential form: 2⁶ = 4x² → 64 = 4x²
  4. Divide both sides by 4: 16 = x²
  5. Take the square root: x = 4 (since x > 0 for the logarithm to be defined)

The answer is x = 4.

6 log₃(27x) + log₃(x/9) = 7

Hint: Combine the two logarithms using the product rule: log₃(A) + log₃(B) = log₃(AB). Then rewrite the logarithmic equation in exponential form to solve for x.

Show the answer

Answer: x = 81

  1. Apply the product rule: log₃(27x) + log₃(x/9) = log₃((27x)(x/9)).
  2. Simplify inside the logarithm: (27x)(x/9) = 27x²/9 = 3x². So the equation becomes log₃(3x²) = 7.
  3. Rewrite in exponential form: 3⁷ = 3x².
  4. Compute 3⁷ = 2187, so 2187 = 3x².
  5. Divide both sides by 3: x² = 729.
  6. Take the square root: x = √729 = 27.
  7. Check for extraneous solutions: The original logarithms require 27x > 0 and x/9 > 0, so x > 0. x = 27 is valid.

The answer is x = 27.

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