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School SEE Compulsory Mathematics

Indices

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Exponential equations are solved by making the bases equal on both sides; if bases are same, indices are equal (2^x = 8 → x = 3). When bases are different, express everything in terms of one common power (let 2^x = a) and form a simple or quadratic equation. Common types include direct form (4^x = 64), factored form (3^x + 3^x = 6 × 3^x), and substitution (2^x + 1/2^x = 5/2 → let a = 2^x). Complex equations become quadratic in a (like a² – 10a + 9 = 0) and give two values of x.

1. Laws of Indices (Quick Revision)

 

LawFormulaa^m × a^na^(m+n)a^m ÷ a^na^(m–n)(a^m)^na^(m n)a^01a^–n1/a^na^(1/n)nth root of a

2. Solving Exponential Equations – Golden Rule

If bases are SAME → indices equal a^f(x) = a^g(x) → f(x) = g(x)

Method

  1. Write both sides with same base
  2. Make indices equal
  3. Solve the simple equation

3. Types of Exponential Equations (All covered)

Type 1: Direct same base

  • Example: 2^x = 8 → 2^x = 2^3 → x = 3

Type 2: Change to same base

  • Example: 4^(x–2) = 1/16 → (2^2)^(x–2) = 2^–4 → 2^(2x–4) = 2^–4 → 2x – 4 = –4 → x = 0

Type 3: Factor out common term

  • Example: 3^(x+1) + 3^x = 4 × 3^x → 3^x × 3 + 3^x = 4 × 3^x → 3^x (3 + 1) = 4 × 3^x → 4 = 4 → true for all x (or cancel 3^x)

Type 4: Let common term = a (Most important)

  • Example: 2^x + 1/2^x = 5/2 Let 2^x = a → a + 1/a = 5/2 → 2a² – 5a + 2 = 0 → (2a – 1)(a – 2) = 0 → a = 2 or 1/2 → 2^x = 2 → x=1 or 2^x = 1/2 → x=–1

Type 5: Multiple terms

  • Example: 5 × 4^(x+1) – 16^x = 64 → 5 × 4 × 4^x – (4^x)² = 64 Let 4^x = a → 20a – a² = 64 → a² – 20a + 64 = 0 → (a–4)(a–16)=0 → a=4 → x=1; a=16 → x=2

4. 5 Most Common SEE Questions + Solutions

  1. 2^x + 2^(x–1) = 6 → 2^x + 2^x / 2 = 6 → (3/2)2^x = 6 → 2^x = 4 → x=2
  2. 5^(x+1) + 5^x = 150 → 5^x (5 + 1) = 150 → 6 × 5^x = 150 → 5^x = 25 → x=2
  3. 4^x + 1/4^x = 17/8 Let 4^x = a → a + 1/a = 17/8 → 8a² – 17a + 8 = 0 → a = 1 or 2 → x = 0 or 1/2
  4. 3^(2x) – 10 × 3^x + 9 = 0 Let 3^x = a → a² – 10a + 9 = 0 → (a–1)(a–9)=0 → a=9 → x=2; a=1 → x=0
  5. 2^(x+1) + 2^x + 2^(x–1) = 28 → 2^x (2 + 1 + 1/2) = 28 → 2^x × (7/2) = 28 → 2^x = 8 → x=3

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https://besidedegree.com/exam/s/academic

 

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Law of Indices Class 10 Mathematics Solution in Nepali || Class 10 Maths Chapter 10 || By: Gurubaa

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