Logarithms and indices for CDS

Indices and logarithms are two ways of writing the same fact. 2⁵ = 32 says "2 multiplied by itself 5 times is 32". log₂ 32 = 5 says "the power of 2 that gives 32 is 5". Every rule for logarithms is a rule for indices, turned around. If you remember that, you never need to memorise the log laws separately.

CDS questions in this area are usually short: simplify a fractional power, solve an exponential equation, combine logs into one, or use given values of log 2 and log 3. They are quick marks for a candidate who knows the laws and a steady source of errors for one who half-remembers them.

Laws of indices

For a, b > 0 and any powers m and n:

Fractional powers

a to the power 1/n is the nth root of a. a to the power m/n is (ⁿ√a)ᵐ, the nth root raised to the power m.

• 27 to the power 2/3 = (∛27)² = 3² = 9
• 81 to the power 3/4 = (⁴√81)³ = 3³ = 27
• 32 to the power −2/5 = 1 ÷ (⁵√32)² = 1/4

Take the root first. It keeps the numbers small.

Exponential equations

If aˣ = aʸ with a > 0 and a ≠ 1, then x = y. So the method is to write both sides as powers of the same base. When the equation has both aˣ and a²ˣ, substitute y = aˣ to turn it into a quadratic.

Logarithms

logₐ N = x means aˣ = N, where a > 0, a ≠ 1 and N > 0. In CDS, "log" with no base usually means base 10. Check the question.

Domain check

The number inside a log must be positive. After solving a log equation, reject any root that makes an argument zero or negative. This is where many candidates pick a wrong option that "satisfies" their algebra.

Counting digits

For a number N ≥ 1, the number of digits = (whole-number part of log N) + 1. Why: a number with d digits lies between 10ᵈ⁻¹ and 10ᵈ, so its log lies between d − 1 and d. With log 2 = 0.3010, log (2⁵⁰) = 50 × 0.3010 = 15.05, so 2⁵⁰ has 16 digits.

Values worth knowing (base 10): log 2 ≈ 0.3010, log 3 ≈ 0.4771, log 5 = 1 − log 2 ≈ 0.6990, log 7 ≈ 0.8451.

Worked questions

Question 1: Simplify 81 to the power 3/4 × 32 to the power −2/5.

• 81 to the power 3/4 = 3³ = 27, and 32 to the power −2/5 = 1/4.
• The product = 27/4.

Question 2: Solve 4ˣ − 3 × 2ˣ⁺¹ + 8 = 0.

• Write 4ˣ = (2ˣ)² and 2ˣ⁺¹ = 2 × 2ˣ. Put y = 2ˣ: y² − 6y + 8 = 0.
• (y − 2)(y − 4) = 0, so 2ˣ = 2 or 2ˣ = 4.
• x = 1 or x = 2.

Question 3: Show that log (75/16) − 2 log (5/9) + log (32/243) = log 2.

• 2 log (5/9) = log (25/81).
• The expression = log [(75/16) × (81/25) × (32/243)].
• (75 × 81 × 32) ÷ (16 × 25 × 243) = 194,400 ÷ 97,200 = 2. So the value is log 2.

Question 4: Solve log₂ x + log₂ (x − 2) = 3.

• log₂ [x(x − 2)] = 3, so x(x − 2) = 8 and x² − 2x − 8 = 0.
• (x − 4)(x + 2) = 0, so x = 4 or x = −2.
• x = −2 makes log₂ x undefined, so x = 4 only.

Question 5: Find log₂ 3 × log₃ 4 × log₄ 5 × … × log₁₅ 16.

• By change of base, each term is log (next number) ÷ log (current number).
• Everything cancels except log 16 ÷ log 2 = log₂ 16 = 4.

Question 6: Given log 2 = 0.3010 and log 3 = 0.4771, find log 72.

• 72 = 2³ × 3², so log 72 = 3 log 2 + 2 log 3.
• = 0.9030 + 0.9542 = 1.8572.

Practice set

1. Find 16 to the power 3/4.
2. Find log₅ 125.
3. Find log 4 + log 25 (base 10).
4. Solve 3²ˣ = 81.
5. Find log₃ (1/81).
6. If logₓ 64 = 3, find x.
7. Given log 3 = 0.4771, how many digits does 3²⁰ have?
8. If 2ˣ = 3ʸ = 6⁻ᶻ, find 1/x + 1/y + 1/z.

Answers

1. 8. (⁴√16)³ = 2³.
2. 3. 5³ = 125.
3. 2. log (4 × 25) = log 100.
4. x = 2. 3²ˣ = 3⁴, so 2x = 4.
5. −4. 1/81 = 3⁻⁴.
6. 4. x³ = 64.
7. 10 digits. 20 × 0.4771 = 9.542, and the whole-number part plus 1 is 10.
8. 0. Let each equal k. Then 2 = k to the power 1/x, 3 = k to the power 1/y and 6 = k to the power −1/z. Since 6 = 2 × 3, 1/x + 1/y = −1/z.

What to do next

• Write each log law next to the index law it comes from
• Practise ten fractional powers, always taking the root first
• After every log equation, check that each argument is positive
• Revise quadratic equations, which exponential equations often turn into, and the number system