Grade 11 Indices and Logarithms I: Short Notes

11 ශ්‍රේණිය · 1 වාරය · 2 පරිච්ඡේදය · දර්ශක හා ලඝුගණක - I

20%

of Grade 11 term tests include it (12 of 61)

102

marks in past papers (15 questions)

12 min

full lesson with quick checks

🎯 What you'll learn

Use the laws of indices, including negative and fractional indices
Solve index equations by writing both sides as powers of the same base
Solve equations where the unknown is the base, like x^(2/3) = 9
Switch between index form and logarithm form
Solve logarithm equations using the laws of logarithms

The Vesak message

You send a Vesak dansala notice to 2 friends. Each of them forwards it to 2 more, and so on. After how many rounds does one round reach 1024 phones? You need 2ˣ = 1024. Solving equations where the unknown sits in the index is what this chapter is about.

The index laws you need

• aᵐ × aⁿ = aᵐ⁺ⁿ (multiply → add indices)• aᵐ ÷ aⁿ = aᵐ⁻ⁿ (divide → subtract)• (aᵐ)ⁿ = aᵐⁿ (power of a power → multiply)• a⁰ = 1• a⁻ⁿ = 1/aⁿ (negative index → reciprocal)Example: 2³ × 2⁻⁵ = 2⁻² = 1/4.

🧮 Fractional indices

The bottom number is the root, the top number is the power. Take the root first to keep numbers small: 8^(2/3) = (∛8)² = 2² = 4, and 27^(−1/3) = 1/∛27 = 1/3.

a^(m/n) = (ⁿ√a)ᵐ

⚠️ Power of a power: multiply, don't add

O/L examiners found many students writing (a⁻²)³ = a¹ by adding the indices, or a⁶ by losing the sign. The rule is (aᵐ)ⁿ = aᵐⁿ, so (a⁻²)³ = a^(−2 × 3) = a⁻⁶ = 1/a⁶. Add only when the same base is multiplied, as in a⁻² × a³ = a.

Solving index equations

Key idea: if aˣ = aʸ, then x = y (same base, so the indices must be equal).So: 1. write both sides as powers of the same base, 2. equate the indices, 3. solve.2ˣ = 1024 → 2ˣ = 2¹⁰ → x = 10. In round 10 the Vesak notice reaches 1024 phones.

✏️ Making the bases match

Solve (i) 3^(x + 1) = 81 (ii) 9ˣ = 27

1(i) 81 = 3⁴, so 3^(x + 1) = 3⁴.
2Equate indices: x + 1 = 4, so x = 3.
3(ii) 9 and 27 are both powers of 3: (3²)ˣ = 3³.
4So 3²ˣ = 3³, giving 2x = 3 and x = 3/2.

✅ (i) x = 3 (ii) x = 3/2

✏️ When the answer is negative

Solve (i) 2ˣ = 1/64 (ii) 8ˣ = 1/16

1(i) 64 = 2⁶, so 1/64 = 2⁻⁶.
22ˣ = 2⁻⁶, so x = −6.
3(ii) Use base 2: 8ˣ = (2³)ˣ = 2³ˣ and 1/16 = 2⁻⁴.
43x = −4, so x = −4/3.

✅ (i) x = −6 (ii) x = −4/3

⚠️ Ignoring the reciprocal

In an O/L paper only about a third of candidates solved an equation like 2ˣ = 1/64. Many answered x = 6. But 2⁶ = 64, not 1/64. A fraction 1 over a power means a negative index: 1/64 = 2⁻⁶, so x = −6. Always substitute back to check.

Logarithms: the other side of indices

aˣ = N is the same statement as logₐ N = x.2⁵ = 32 ⇔ log₂ 32 = 510² = 100 ⇔ lg 100 = 2 (lg means base 10)Laws:• log(mn) = log m + log n• log(m/n) = log m − log n• log mⁿ = n log m• logₐ a = 1, logₐ 1 = 0

⚠️ lg 8 + lg 25 is not lg 33

A common slip in term tests: adding the numbers inside the logs. Adding logs multiplies the numbers: lg 8 + lg 25 = lg 200. Subtracting logs divides. Also, lg(x + 3) can't be split into lg x + lg 3: there is no law for the log of a sum.

📌 Remember

• Multiply → add indices; power of a power → multiply indices.• a^(m/n) = (ⁿ√a)ᵐ and a⁻ⁿ = 1/aⁿ.• Index equation: same base on both sides, then equate the indices.• aˣ = N ⇔ logₐ N = x.• Log equation: combine into one log on each side, then remove the logs or change to index form. Check x > 0 inside every log.

🧾 Quick revision sheet

(aᵐ)ⁿ = ?	aᵐⁿ (multiply the indices)
a⁻ⁿ = ?	1/aⁿ
a^(m/n) = ?	(ⁿ√a)ᵐ: root n, power m
If aˣ = aʸ, then…	x = y (equate the indices)
log₂ 32 = 5 in index form	2⁵ = 32
lg m + lg n = ?	lg mn
lg 20 in terms of lg 2	lg 10 + lg 2 = 1 + lg 2
logₐ a and logₐ 1	1 and 0