Solved Examples

Example 1: Simplify (2³ × 2⁵) ÷ 2⁴. = 2^(3+5−4) = 2⁴ = 16

Example 2: Simplify √50 + √18. - √50 = √(25×2) = 5√2 - √18 = √(9×2) = 3√2 - Sum = 5√2 + 3√2 = 8√2

Example 3: Rationalise 1/(√5 − √3). - Multiply by conjugate (√5 + √3): = (√5+√3) / [(√5)² − (√3)²] = (√5+√3)/(5−3) = (√5+√3)/2

Example 4: If x = 2^(1/3) + 2^(−1/3), find x³ (a typical SSC-CGL indices problem, using the identity (a+b)³ = a³+b³+3ab(a+b)). - Let a = 2^(1/3), b = 2^(−1/3); ab = 2^(1/3 − 1/3) = 2⁰ = 1 - a³ = 2, b³ = 1/2 - x³ = a³ + b³ + 3ab(a+b) = 2 + 0.5 + 3(1)(x) = 2.5 + 3x - So x³ − 3x − 2.5 = 0 (this type is generally solved by back-substituting the numeric value of x on a calculator-free exam using approximation, or is set up so the options make the arithmetic clean — always check whether the exam wants x³ in terms of x, as shown here, or a fully numeric answer via a cleaner given expression)

Example 5: If 5^(x−3) × 3^(2x−8) = 225, find x. - 225 = 15² = (3×5)² = 3² × 5² - So 5^(x−3) × 3^(2x−8) = 3² × 5² - Matching powers of 5: x − 3 = 2 → x = 5 - Matching powers of 3: 2x − 8 = 2 → 2x = 10 → x = 5 (consistent with the first equation) Answer: x = 5

Example 6 (Negative-exponent edge case): Simplify (4^(−2) × 2³) ÷ 2^(−4). - 4^(−2) = (2²)^(−2) = 2^(−4), so the numerator = 2^(−4) × 2³ = 2^(−4+3) = 2^(−1) - Dividing by 2^(−4): 2^(−1) ÷ 2^(−4) = 2^(−1−(−4)) = 2^(−1+4) = 2³ = 8 Answer: 8 (Direct check: 4^(−2) = 1/16, 2³ = 8, so numerator = 8/16 = 0.5; 2^(−4) = 1/16; 0.5 ÷ (1/16) = 0.5 × 16 = 8 ✓ — matches.)

Example 7 (Fractional/negative indices together): Simplify 27^(−2/3) × 9^(3/2). - 27 = 3³, so 27^(−2/3) = 3^(3 × (−2/3)) = 3^(−2) = 1/9 - 9 = 3², so 9^(3/2) = 3^(2 × 3/2) = 3³ = 27 - Product = (1/9) × 27 = 3 Answer: 3