ICSE Class 9 Maths: Laws of Indices & Logarithms Explained

Ever wondered how a huge number like 2^10 becomes just 1024, or how you can turn a messy multiplication into a neat addition? That magic comes from the laws of indices and logarithms – the shortcuts every ICSE 9th‑grader needs.

💡 In Simple Words: Indices (or exponents) tell you how many times to multiply a number by itself. Logarithms are the reverse – they ask, “To what power must we raise a base to get this number?” Both follow easy rules that let you shrink big calculations into bite‑size steps.

What are indices? (aka exponents)

An index is just a tiny number written up‑right that says “multiply the base that many times”. For example, 3³ means 3 × 3 × 3 = 27. Think of it like stacking identical Lego bricks: the index tells you how many bricks you stack.

Basic index laws you must know

• Multiplication rule: a^m × a^n = a^(m+n). Like adding the heights of two stacks of the same kind of bricks.
• Division rule: a^m ÷ a^n = a^(m‑n). Removing bricks from a taller stack.
• Power‑of‑a‑power rule: (a^m)^n = a^(m·n). Stacking a stack on top of itself n times.
• Power‑of‑a‑product rule: (ab)^n = a^n b^n. Distribute the exponent to each factor, just like spreading paint equally on two canvases.
• Zero exponent rule: a^0 = 1 (provided a ≠ 0). Anything to the power zero is like an empty stack – you still have one whole piece of “nothing”.
• Negative exponent rule: a^(‑n) = 1 / a^n. Turning a tall stack upside‑down.

Worked example: simplifying with index laws

Simplify [(2³ × 2⁴) ÷ 2²]⁵.

1. First use the multiplication rule inside the brackets: 2³ × 2⁴ = 2^(3+4) = 2⁷.
2. Now divide by 2²: 2⁷ ÷ 2² = 2^(7‑2) = 2⁵.
3. Apply the power‑of‑a‑power rule: (2⁵)⁵ = 2^(5·5) = 2²⁵.

So the whole expression collapses to 2²⁵ – a huge number, but we got there with just three quick steps.

What are logarithms?

A logarithm asks the opposite question of an exponent: log_b a = c means “b raised to what power gives a?” If you’ve ever used a map’s scale to figure out how far you’ve walked, you’re doing a log‑like conversion.

Logarithm properties (laws)

• Product rule: log_b (xy) = log_b x + log_b y. Multiplying inside becomes adding outside.
• Quotient rule: log_b (x⁄y) = log_b x – log_b y. Dividing inside becomes subtracting outside.
• Power rule: log_b (x^n) = n·log_b x. Bring the exponent down as a multiplier.
• Change‑of‑base rule: log_b a = log_k a ÷ log_k b. Switch the base to any convenient number k (often 10 or e).
• Log of 1: log_b 1 = 0 because any non‑zero base to the power 0 equals 1.
• Log of the base: log_b b = 1 because b¹ = b.

Worked example: solving with log rules

Solve for x: 5^{2x‑3} = 125.

1. Write 125 as a power of 5: 125 = 5³.
2. Now the equation is 5^{2x‑3} = 5³. Since the bases match, equate exponents: 2x‑3 = 3.
3. Solve: 2x = 6 ⇒ x = 3.

Alternatively, using logs: take log₅ on both sides → (2x‑3)·log₅5 = log₅125 → (2x‑3)·1 = 3 → x = 3.

Quick comparison: Indices vs Logarithms

📝 Likely Exam Questions

1. Question: Simplify (3⁴ × 3⁻²) ÷ 3³.
   Answer: Use multiplication rule: 3⁴ × 3⁻² = 3^(4‑2) = 3². Then divide: 3² ÷ 3³ = 3^(2‑3) = 3⁻¹ = 1/3.
2. Question: If log₂ x = 5, find the value of x.
   Answer: Convert to exponent form: 2⁵ = x ⇒ x = 32.
3. Question: Express log₁₀ 50 as a sum of log₁₀ 5 and log₁₀ 10 using log laws.
   Answer: 50 = 5 × 10, so log₁₀ 50 = log₁₀ 5 + log₁₀ 10 = log₁₀ 5 + 1.
4. Question: Solve 4^{x+1} = 64.
   Answer: Write 64 as 4³ (since 4³ = 64). Hence x+1 = 3 ⇒ x = 2.
5. Question: Using the change‑of‑base rule, evaluate log₅ 20 to two decimal places (use log₁₀ values log₁₀ 20 ≈ 1.3010, log₁₀ 5 ≈ 0.6990).
   Answer: log₅ 20 = log₁₀20 ÷ log₁₀5 ≈ 1.3010 / 0.6990 ≈ 1.86.