Power, Indices & Surds — Notes, Rules and Shortcuts for SSC CGL

This chapter is pure marks. Almost every SSC paper carries two to four questions from it, and none of them need long calculation — they need the right rule applied in the right order. Work through these notes once, then practise the chapter tests linked at the bottom.

1. Base, index and the eight laws

In aⁿ, the number a is the base and n is the index (also called the power or exponent). It means a multiplied by itself n times.

#	Law	Example
1	a^m × aⁿ = a^m+n	2³ × 2⁴ = 2⁷
2	a^m ÷ aⁿ = a^m−n	5⁷ ÷ 5³ = 5⁴
3	(a^m)ⁿ = a^mn	(3²)⁴ = 3⁸
4	(ab)ⁿ = aⁿbⁿ	6³ = 2³ × 3³
5	(a/b)ⁿ = aⁿ/bⁿ	(2/3)⁴ = 16/81
6	a⁰ = 1 (a ≠ 0)	(97 − 12)⁰ = 1
7	a⁻ⁿ = 1/aⁿ	2⁻³ = 1/8
8	a^m = aⁿ ⇒ m = n	equal bases, equal indices

Memory hook Multiply → add the indices. Divide → subtract. Bracket → multiply. That one line covers laws 1 to 3, which between them solve most of the chapter.

Why law 8 works For a positive base other than 1, the power keeps growing (or keeps shrinking) as the index grows, so no two different indices can ever give the same value. That is why a = 1 and a = 0 must be excluded — 1⁵ = 1⁹ tells you nothing about 5 and 9.

2. Zero and negative indices

Two rules cause more mistakes than any others in this chapter, because they look similar and are not.

Non-zero base, zero index: a⁰ = 1. It does not matter how ugly the base is. (3⁷ − 2¹¹ + 5/9)⁰ = 1 — never compute the bracket.
Zero base, positive index: 0^k = 0. So 0⁸ = 0, not 1.
Negative index flips: a⁻ⁿ = 1/aⁿ and (a/b)⁻ⁿ = (b/a)ⁿ.

Example Evaluate (2⁻¹ + 3⁻¹)⁻¹.
Wrong: 2 + 3 = 5. An index never distributes over a sum.
Right: 1/2 + 1/3 = 5/6, so the answer is 6/5.

Example 7 ÷ 7⁻¹ − 49⁰ × 7 = 7 × 7 − 1 × 7 = 49 − 7 = 42.

3. Fractional indices and roots

A fractional index is just a root written another way.

Definition a^1/n = ⁿ√a and a^p/q = (^q√a)^p. Read it in that order: root first, power second. Both orders give the same answer, but the root first keeps the numbers small.

32^2/5 = (⁵√32)² = 2² = 4 — far easier than finding 32² = 1024 first.
(27/125)^−2/3 = (125/27)^2/3 = (5/3)² = 25/9.
Root of a root: the orders multiply. ³√(√x) = x^1/6.

Perfect powers worth memorising Squares to 30; cubes to 15 (216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375); fourth powers 16, 81, 256, 625, 1296, 2401, 4096; and 1024 = 2¹⁰, 3125 = 5⁵, 4096 = 2¹² = 4⁶ = 8⁴ = 16³. Recognising 4096 as a fourth power is often the whole question.

4. Decimal indices in exams

SSC loves decimal indices because they look frightening and always cancel. The decimals are chosen so the total lands on a whole number or a simple fraction.

Example 625^0.16 × 625^0.09 = 625^0.25 = 5. The moment you see 0.16 + 0.09 = 0.25 = 1/4, take the fourth root.

Example 8^1.5 × 16^0.75 ÷ 4². On base 2 this is 2^4.5 × 2³ ÷ 2⁴ = 2^3.5. Remember to multiply each decimal by the base's own index — 8^1.5 is 2^4.5, not 2^1.5.

5. Solving for an unknown index

Method, every time:

Write every base as a power of the same prime.
Collect each side into a single power.
Equate the indices and solve the linear equation.

Example 5^2x × 25^x−1 = 125^x+1
Left: 5^2x × 5^2x−2 = 5^4x−2. Right: 5^3x+3.
4x − 2 = 3x + 3 ⇒ x = 5.

Two shapes worth learning by name

Sum of shifted powers: factor the smaller one. 3^x+2 − 3^x = 3^x(9 − 1) = 8 · 3^x. So 3^x+2 − 3^x = 72 ⇒ 3^x = 9 ⇒ x = 2.
Copies of one power: 3^x + 3^x + 3^x = 3 · 3^x = 3^x+1. Adding copies raises the index by one when the count equals the base.
Merging two primes: equal indices merge — 2¹¹ × 5¹¹ = 10¹¹, and 3^1.2 × 2^1.2 = 6^1.2.

6. Power chains and common-value questions

Closed chains

If a^x = b, b^y = c, c^z = a, then substituting round the loop gives a = a^xyz, so xyz = 1. If the loop closes on a⁸ instead, the product is 8; if it closes on 1/a, the product is −1. The rule is simply: the product of all the indices equals the final index of a.

Common-value questions

When a^x = b^y = c^z = k, write a = k^1/x, b = k^1/y, c = k^1/z. Then any multiplicative relation among the bases becomes an additive relation among the reciprocals.

Relation among bases	Result
c = ab	1/z = 1/x + 1/y
c = a²b	1/z = 2/x + 1/y
c = a^pb^q	1/z = p/x + q/y
a^x = b^y = (ab)^−z	1/x + 1/y + 1/z = 0

Match each index to its own base In c = 2³ × 5² with 2^x = 5^y = c^z, the 3 goes with x (base 2) and the 2 with y (base 5): 1/z = 3/x + 2/y. Swapping them is the commonest error here.

7. Surds: pure, mixed and like

Definition A surd is a root of a rational number that cannot itself be written as a rational number — √2, ³√9. If the root comes out exactly, as in ³√64 = 4, it is not a surd.

Pure surd: nothing outside the root — √50.
Mixed surd: a number outside — 5√2.
Like surds: same root part after simplifying — 3√5 and 4√5 are like; √12 and √18 are not. Only like surds add or subtract.

The four conversions

Job	Rule	Example
Pure → mixed	ⁿ√(mⁿ·a) = m·ⁿ√a	³√108 = 3³√4
Mixed → pure	k·ⁿ√a = ⁿ√(kⁿa)	2·³√5 = ³√40
Raise the order	ⁿ√a = ^nk√(a^k)	³√5 = ⁶√25
Lower the order	^nk√(a^k) = ⁿ√a	⁶√125 = √5

The commonest slip When taking a number inside a root, raise it to the order of that root. Inside a cube root the 2 becomes 8, not 4.

8. Rationalising the denominator

Denominator	Multiply top and bottom by	Result
√b	√b	a√b / b
p + q	p − q	p² − q²
p + q + r	(p + q) − r, then the conjugate again	rational after two stages

Standard results to keep ready

1/(√(n+1) + √n) = √(n+1) − √n — the denominator becomes 1. This is what makes telescoping chains collapse.
(p+q)/(p−q) + (p−q)/(p+q) = 2(p²+q²)/(p²−q²) — a rational number.
(p+q)/(p−q) − (p−q)/(p+q) = 4pq/(p²−q²).
If x = a + √b and a² − b = 1, then 1/x = a − √b exactly. So x + 1/x = 2a and x − 1/x = 2√b.

Telescoping example 1/(√2+1) + 1/(√3+√2) + … + 1/(√16+√15)
Each term becomes √(n+1) − √n. Everything in the middle cancels, leaving √16 − 1 = 3.

Watch the numerator If the gap under the roots is 3 and the numerator is also 3, each term is still √m − √n — the numerator has already cancelled the gap. Do not divide again.

9. Compound surds — √(a ± 2√b)

The splitting rule √(a + 2√b) = √m + √n where m + n = a and mn = b, with m > n. For the minus sign the answer is √m − √n.

The 2 in front of the inner root is essential. If it is missing, create it:

√(9 − 4√5): write 4√5 = 2√20, so m + n = 9, mn = 20 ⇒ 5 and 4 ⇒ √5 − 2.
√(4 + √15): double inside and divide by √2 — √(8 + 2√15)/√2 = (√5 + √3)/√2.

Two results that appear again and again √(a+2√b) + √(a−2√b) = 2√m (the smaller surd cancels)
√(a+2√b) − √(a−2√b) = 2√n (the larger surd cancels)

Three-term surds

(√a + √b − √c)(√a + √b + √c): treat √a + √b as one term and use the difference of squares, giving (a + b + 2√ab) − c. One pass usually leaves a surd; a second conjugate finishes the rationalisation.

10. Nested radicals

Finite nests — work inside out

√(45 − √(67 + √196)) = √(45 − √81) = √36 = 6. Always start at the innermost root and check the order written on each root sign — a small 3 or 4 changes everything.

Infinite nests — set up an equation

Form	Equation	Value
√(a + √(a + …))	x² = a + x	(1 + √(1+4a))/2; if a = n(n+1), value = n+1
√(a − √(a − …))	x² = a − x	if a = n(n+1), value = n
√(a√(a√a …))	x² = ax	a

Example √(30 + √(30 + …)): since 30 = 5 × 6, the answer is 6. The minus version of the same nest would be 5.

Inverse framing If the nest is given as equal to u, then a = u² − u for the plus form and a = u² + u for the minus form.

11. Comparing surds and large powers

Different root orders

Convert everything to the LCM order and compare the numbers inside.

Example √2, ³√3, ⁶√6 → order 6 → ⁶√8, ⁶√9, ⁶√6 → so ³√3 > √2 > ⁶√6. Note how close the first two are; never judge these by two-decimal estimates.

Different fractional indices

Raise all of them to the LCM of the denominators. For 3^1/2, 5^1/3, 7^1/4, raise to the 12th power: 3⁶ = 729, 5⁴ = 625, 7³ = 343. So the order is exactly as written.

Large powers

Divide every index by their common factor and compare the reduced bases. 2³⁶, 3²⁴, 5¹⁸, 7¹² all have index-factor 6, giving bases 64, 81, 125, 49 — so 5¹⁸ wins.

Sums and differences of surds

Sums with the same total inside: square them. (√p + √q)² = p + q + 2√(pq), so the more balanced pair is larger. With p + q = 20, √10 + √10 beats √17 + √3.
Differences with the same gap: √p − √q = (p−q)/(√p + √q), so the smaller numbers give the larger difference. √8 − √4 > √35 − √31.

Power towers

a^{b^c} is read from the top down: 2^3² = 2⁹ = 512. But (2³)² = 2⁶ = 64. The brackets change the answer completely.

12. Approximation questions

The paper gives you one or two root values and asks for a messy-looking expression. The work is always: simplify first, substitute once.

Collect like surds: 3√20 − 2√45 + √125 = 6√5 − 6√5 + 5√5 = 5√5. One multiplication left.
Rationalise first: 2/(√3 − 1) = √3 + 1 = 2.732 — far safer than dividing decimals.
Use the shift rule: two decimal places under a square root become one place outside. Given √1369 = 37: √13.69 = 3.7, √0.1369 = 0.37, √136900 = 370. For cube roots the step is three places inside, one outside.
Look for a hidden factor: (√6 + √2)/(√3 + 1) = √2(√3 + 1)/(√3 + 1) = √2.

Values worth memorising √2 = 1.414 · √3 = 1.732 · √5 = 2.236 · √6 = 2.449 · √7 = 2.646 · √10 = 3.162

13. Ten traps that cost marks

Indices over a sum. (2⁻¹ + 3⁻¹)⁻¹ is not 2 + 3. Finish the bracket first.
0⁸ is 0, not 1. Only a non-zero base gives 1 at the zero index.
Forgetting the base's own index. 8^1.5 = 2^4.5, because 8 is 2³.
Double negatives when subtracting indices. b⁻³ ÷ b⁻⁴ = b¹, not b⁻⁷.
Squaring a coefficient. (3√2)² = 18, not 6.
Order of a root. Taking a factor inside a cube root cubes it; pulling one out of a fourth root needs a perfect fourth power.
Sign of an odd root. ³√(−27) = −3. But an even power afterwards kills the sign: (−8)^2/3 = 4.
Reciprocals reverse an inequality. Larger surd, smaller reciprocal. Order the surds first, then flip.
Reading the question. Many questions solve for x but ask for x², 2x or x − 5. The option list always includes the un-finished value.
Bare comparison by eye. ³√3 and √2 differ by 0.03. Always go to a common order.

14. Quick revision sheet

Situation	Do this
Different bases	Write all as powers of one prime
Decimal indices	Multiply by the base's index; the total will be clean
Unknown in the index	Same base both sides, then equate indices
a^x+k ± a^x = N	Factor out a^x
Closed power chain	Product of indices = final index of the base
a^x = b^y = c^z	Reciprocals of the indices add the way the bases multiply
Surd in a denominator	Multiply by the conjugate
√(a ± 2√b)	Find m, n with m + n = a, mn = b
Infinite plus-nest	x² = a + x; if a = n(n+1), answer n+1
Compare different orders	Convert to the LCM order
Compare big powers	Divide indices by their common factor
Approximation	Simplify fully, substitute once at the end

15. Frequently asked questions

How many questions come from this chapter in SSC CGL?

Typically two to four in Tier 1 and three to five in Tier 2, counting simplification questions that are really index or surd questions in disguise. It is one of the highest-return chapters per hour of study.

Is ³√64 a surd?

No. It equals 4 exactly, so it is a rational number. A root is a surd only when the number inside is not a perfect power of that order.

Why is a⁰ equal to 1?

From the division law, aⁿ ÷ aⁿ = a⁰. But any non-zero number divided by itself is 1. So a⁰ = 1. With a = 0 the division is not allowed, which is why the base must be non-zero.

Can I compare surds using a calculator-style decimal estimate?

Not safely. Exam options are often chosen so that two of them differ in the third decimal place. Converting to a common order takes ten seconds and never fails.

Do I need logarithms for this chapter?

No. Every SSC question in this chapter is solvable with index laws, prime factorisation and the surd rules above.

Practise this chapter

Chapter tests300 questions on Power, Indices & Surds, in two levels CGL previous yearsReal questions, shift by shift CHSL previous yearsSame chapter, easier calibration Full mock testsPractise it under time pressure