Indices and Surds

The most direct question of this chapter is "find the square root of this perfect square". You do not need long division for it. Two clues are enough: the unit digit and the range.

Clue 1: the unit digit. The last digit of a square depends only on the last digit of the number. Look at the pattern:

So a number ending in 2, 3, 7 or 8 can never be a perfect square. This one fact often cuts options before you calculate anything.

[[fig:0]]

Clue 2: the range. Cover the last two digits of the number. Find the largest n whose square is less than or equal to what is left. That n is the first part of the root.

Radhika wants √5329. It ends in 9, so the root ends in 3 or 7. Cover "29": 53 is left, and 7² = 49 ≤ 53 < 64, so the root is 73 or 77. To choose, she multiplies 7 × 8 = 56. Since 53 < 56, she takes the smaller one: √5329 = 73. (Check: 73² = 5329.)

The choosing rule: if the leftover part is less than n(n + 1), take the smaller ending; otherwise take the larger one. It works because (10n + 5)² ends in 25 and its front part is exactly n(n + 1), so n(n + 1) is the dividing line between the two endings.

[[fig:1]]

Decimals. Make the number of decimal places even, then halve them. √0.0049 = 0.07 (4 places become 2) and √1.69 = 1.3. You can also convert to a fraction: √0.0049 = √4910000 = 7100.

Root equations. When a question says √x − 3 = 6 or √? + 14 = √2601, first work out every root you can, leave the unknown root alone on one side, and then square both sides once.

One warning before the questions: √a + b is not √a + √b. √9 + 16 = √25 = 5, while √9 + √16 = 7. Add inside first, then take the root.

[[fig:2]]

Cube roots. A cube root can be found with the same two clues. The unit digit of the cube decides the unit digit of the root, and now every digit has exactly one answer: 1 → 1, 2 → 8, 3 → 7, 4 → 4, 5 → 5, 6 → 6, 7 → 3, 8 → 2, 9 → 9, 0 → 0. Cut off the last three digits, and see between which two cubes the remaining part lies; that gives the tens digit.

[[fig:3]]

Radhika finds ∛5832. The last digit 2 means the root ends in 8. Cutting the last three digits leaves 5, which lies between 1³ = 1 and 2³ = 8, so the tens digit is 1. The root is 18.

The laws are the same as for square roots: ∛ab = ∛a × ∛b and ∛ab = ∛a∛b. Unlike a square root, the cube root of a negative number exists: ∛−27 = −3.

Every law of indices comes from one simple picture: a^m is a written m times. Once you see that, you never have to memorise the laws.

Multiplying the same base: add the powers. a³ × a² = (a × a × a) × (a × a) = a⁵. So a^m × aⁿ = a^{m + n}.

Dividing the same base: subtract the powers. a⁵a² = a × a × a × a × aa × a = a³. So a^m ÷ aⁿ = a^{m − n}.

[[fig:0]]

Power of a power: multiply the powers. (a²)³ = a² × a² × a² = a⁶. So (a^m)ⁿ = a^mn.

[[fig:1]]

Power of a product or a fraction: give the power to each factor. (ab)ⁿ = aⁿbⁿ and (ab)ⁿ = aⁿbⁿ.

The zero and negative powers are not new rules at all. They are an extension of the division law:

• a³a³ = 1, and the same thing is a^{3 − 3} = a⁰. So a⁰ = 1 (a ≠ 0).
• a²a⁵ = 1a³, and the same thing is a^{2 − 5} = a⁻³. So a⁻ⁿ = 1aⁿ, and (ab)⁻ⁿ = (ba)ⁿ.
• Fractional powers are roots: a¹ⁿ = ⁿ√a and a^mn = ⁿ√a^m = (ⁿ√a)^m.

[[fig:2]]

Say Radhika has to simplify 8² × 4³16². She brings everything to base 2: 2⁶ × 2⁶2⁸ = 2^{6 + 6 − 8} = 2⁴ = 16. The golden rule of this type: first convert every number into the smallest common prime base. 4, 8, 16, 32, 64 go to base 2; 9, 27, 81, 243 to base 3; 25, 125, 625 to base 5; 0.04 and 0.008 to base 0.2 (or 5).

Addition is different. There is no law for a^m + aⁿ, so you cannot add the powers when the terms are added. Take the smallest power common instead: 3^{n + 2} + 3ⁿ = 3ⁿ(3² + 1) = 10 × 3ⁿ. Big sums shrink the same way: 2¹⁰ + 2¹¹2⁸ = 2¹⁰(1 + 2)2⁸ = 4 × 3 = 12.

Cyclic powers. A question like (x^ax^b)^{a + b} × (x^bx^c)^{b + c} × (x^cx^a)^{c + a} looks frightening, but the total power is (a² − b²) + (b² − c²) + (c² − a²) = 0, so the whole thing is x⁰ = 1. Whenever the letters go round in a cycle, expect the powers to cancel.

An exponential equation is one where the unknown sits in the power, like 2^x = 32 or 81^a = 3²⁰. You cannot "divide by 2" to free x here. We use one fact instead:

If a^x = a^y (with a > 0 and a ≠ 1), then x = y.

Why must a not be 1? Because 1⁵ = 1⁷, yet 5 ≠ 7. With any other positive base, different powers always give different values, so equal values force equal powers.

[[fig:0]]

So the method always has the same three steps:

• Write both sides as powers of the same base (the smallest prime base is the safest).
• Using the laws of Type 2, turn each side into one single power.
• Equate the powers and solve the ordinary equation that is left.

[[fig:1]]

Radhika has to solve 27^x = 9^{x + 1}. In base 3 it becomes 3^3x = 3^{2x + 2}, so 3x = 2x + 2 and x = 2. Check: 27² = 729 and 9³ = 729.

When the equation has a sum, take the smallest power common: 2^{x + 2} + 2^x = 80 ⇒ 2^x(4 + 1) = 80 ⇒ 2^x = 16 ⇒ x = 4.

Two more useful facts.

• If a^x = b^x with a ≠ b (both positive), then x = 0, because only the zero power makes two different bases equal (both become 1).
• If a^x = b^y = c^z = k (k ≠ 1), write a = k^1x, b = k^1y and c = k^1z. Now any relation among a, b, c turns into a relation among the powers. For example, if c = ab, then k^1z = k^{1x + 1y}, so 1z = 1x + 1y. Putting everything equal to k is how the tougher questions of this type are opened up.

Exponential equations that turn into a quadratic. Sometimes both a^2x and a^x appear in the equation, as in 4^x − 5 × 2^x + 4 = 0. Write 4^x = (2^x)² and put y = 2^x. The equation becomes y² − 5y + 4 = 0, so y = 1 or y = 4. Then 2^x = 1 gives x = 0 and 2^x = 4 gives x = 2.

[[fig:2]]

One check is needed at the end. A power a^x is always positive, so any negative value of y must be rejected. For 4^x − 3 × 2^x − 4 = 0, y² − 3y − 4 = 0 gives y = 4 or y = −1. Only y = 4 is allowed, so x = 2 is the single solution.

Equations like 2^x + 2^{3 − x} = 9 are of the same family: write 2^{3 − x} = 82^x, put y = 2^x, and you get y + 8y = 9, that is y² − 9y + 8 = 0.

Before you add or subtract surds, always ask one question: can this surd be made smaller? √72, √32 and √8 look like three different surds, but all three are secretly multiples of √2.

Simplest form. Break the number under the root into (largest perfect square) × (rest), and take the square out.
√50 = √25 × 2 = 5√2
√75 = √25 × 3 = 5√3
√108 = √36 × 3 = 6√3
√175 = √25 × 7 = 5√7

To find the largest square factor quickly, try dividing by 4, 9, 16, 25, 36, 49, 64, 81 … in turn and keep the biggest one that divides exactly.

[[fig:0]]

Adding and subtracting. After simplifying, combine only like surds, exactly as you combine like terms in algebra: 5√2 + 3√2 − 2√2 = (5 + 3 − 2)√2 = 6√2. Unlike surds such as √2 and √3 stay separate.

[[fig:1]]

Multiplying and dividing. Here the surds need not be alike. Surds of the same order multiply and divide under one root:
√a × √b = √ab, √a√b = √ab, and √a × √a = a.
So √6 × √24 = √144 = 12, and √72√2 = √36 = 6.

[[fig:2]]

Mixed to pure. To compare or multiply, you can push a non-negative number inside the root: a√b = √a²b (for a ≥ 0). Going the other way, √a²b = |a|√b: with a = −3 and b = 2 the result is 3√2, not −3√2. For example, 3√6 = √54.

Here is Radhika's sum: √28 + √63. She writes √28 = 2√7 and √63 = 3√7, so the sum is 5√7. If the question then asks for the square of this sum, it is 25 × 7 = 175. She needs no expansion at all, because she combined the surds first.

Equations in √a and √b. If √a + √b = p and √a − √b = q, treat √a and √b as two unknowns. Add the equations to get √a = p + q2, subtract them to get √b = p − q2, and then square.

Using a given value. When a question gives √3 = 1.732 (or a sum that hides it), first reduce every surd to a multiple of √3, and multiply by the value only once, at the end.

When a surd is joined to a number, like 3 + √2 or √5 − √3, we call it a binomial surd. Squaring it, or multiplying two of them, is ordinary algebra: use (x + y)² = x² + y² + 2xy and (x + y)(x − y) = x² − y², and remember that (√b)² = b.

Square of a binomial surd.
(a + √b)² = a² + b + 2a√b
(√a + √b)² = a + b + 2√ab
(√a − √b)² = a + b − 2√ab
See the pattern: the rational part is the sum of the two squares, and the surd part is twice the product. So (√7 + √2)² = 9 + 2√14, in one look.

[[fig:0]]

Conjugate pair. a + √b and a − √b are called conjugates. The product of two conjugates has no surd at all:
(a + √b)(a − √b) = a² − b, and (√a + √b)(√a − √b) = a − b.
This is the most useful fact of the whole chapter, because rationalising in the next type rests on it.

[[fig:1]]

Sum and difference of the two squares. Call the two parts x and y:
(x + y)² + (x − y)² = 2(x² + y²)
(x + y)² − (x − y)² = 4xy
So (√6 + √3)² − (√6 − √3)² = 4√18 = 12√2, without expanding anything.

Conjugate values a, b. If a = p + √q and b = p − √q, then a + b = 2p and ab = p² − q are clean numbers. Every symmetric expression then becomes easy: a² + b² = 2(p² + q).

Suppose Radhika's x is 2 + √3. Then x² = 4 + 3 + 4√3 = 7 + 4√3. If a question asks for x² − 4√3, the surd part cancels and the answer is 7.

Equating parts. If p + q√r = s + t√r, where p, q, s, t are rational and √r is irrational, then p = s and q = t. A rational number can never equal a surd, so the two parts must match separately. That is how you find K in (5 + 3√3)² = 52 + K√3.

[[fig:2]]

1√2 and √22 are the same number, but the second one is far easier to use. Dividing 1 by 1.414 is difficult; halving 1.414 is quick. Rationalising means removing the surd from the denominator by multiplying the top and the bottom by a suitable factor. Since we multiply by something like √2√2 = 1, the value does not change.

Case 1: a single surd below. Multiply by the same surd.
a√b = a√bb. For example, 6√3 = 6√33 = 2√3.

[[fig:0]]

Case 2: a binomial surd below. Multiply by its conjugate (change the sign in the middle). The denominator becomes a difference of squares and the surd disappears:
1a + √b = a − √ba² − b, 1√a − √b = √a + √ba − b.

[[fig:1]]

Watch Radhika rationalise 4√5 − 1. She multiplies the top and the bottom by √5 + 1 and gets 4(√5 + 1)5 − 1 = √5 + 1.

When the top and the bottom are conjugates, the result is a perfect square over a number:
√a + √b√a − √b = (√a + √b)²a − b = a + b + 2√aba − b.

The reciprocal pairs. If the product of two conjugates is 1, each one is the reciprocal of the other. (3 + 2√2)(3 − 2√2) = 9 − 8 = 1, so 13 + 2√2 = 3 − 2√2. The same is true for 2 ± √3, √2 ± 1, 5 ± 2√6 and √3 ± √2. Keep these pairs in mind; Algebra uses them again and again.

[[fig:2]]

Using a given value. If a question gives √6 = 2.45 and asks for √23, rationalise in a way that makes √6 appear: √23 = √2√3 × √3√3 = √63.

After rationalising, if the question says "= a + b√c", compare the rational parts and the surd parts separately, as in Type 5.

Square root of a surd. In Type 5 we saw that (√x + √y)² = (x + y) + 2√xy. Read it backwards and you get the square root of a binomial surd:

√a + 2√b = √x + √y, where x + y = a and xy = b.

So the job is simple: write the surd part as 2√b, then find two numbers whose sum is a and whose product is b.

Radhika wants √7 + 4√3. She writes 4√3 = 2√12. Two numbers with sum 7 and product 12 are 4 and 3. So √7 + 4√3 = √4 + √3 = 2 + √3. Check: (2 + √3)² = 7 + 4√3.

[[fig:0]]

With a minus sign, √a − 2√b = √x − √y with x > y, so that the answer stays positive. For example, √5 − 2√6 = √3 − √2, not √2 − √3.

If the surd part has some other coefficient, bring it to the form 2√b first: 4√7 = 2√28, 10√3 = 2√75. For √3 + √5 write √5 = 2√54: the numbers with sum 3 and product 54 are 52 and 12, so the root is √10 + √22. Integer pairs are the quick special case, not a condition for simplifying.

Infinite nested roots. Let x = √a + √a + √a + …. The part under the first root, after "a +", is x itself again. So x = √a + x, that is x² − x − a = 0, which gives x = 1 + √1 + 4a2.

[[fig:1]]

In exam questions a is usually a product of two consecutive numbers, a = n(n + 1). Then the answer is the bigger of the two: √12 + √12 + … = 4 (12 = 3 × 4) and √30 + √30 + … = 6 (30 = 5 × 6). With minus signs, √a − √a − … is the smaller one: √12 − √12 − … = 3.

[[fig:2]]

For products, √a√a√a…∞ = a, because x = √ax gives x² = ax. A finite chain of n roots gives a^{1 − 12n}, so √a√a√a = a⁷⁸.

Square root of a three-term surd. The same reading-backwards works with three surds. Since

(√a + √b + √c)² = (a + b + c) + 2√ab + 2√bc + 2√ca,

an expression of the form (a + b + c) + 2√ab + 2√bc + 2√ca has the square root √a + √b + √c.

[[fig:3]]

Take √15 + √60 + √84 + √140. First write each surd as 2√…: √60 = 2√15, √84 = 2√21, √140 = 2√35. The products 15, 21 and 35 split as 3 × 5, 3 × 7 and 5 × 7, and 3 + 5 + 7 = 15 matches the rational part. So the root is √3 + √5 + √7.

The trick is to look for three numbers whose pairwise products are the three numbers under the roots and whose sum is the rational part. If a coefficient is not 2, make it 2 first: 6√2 = 2√18.

To compare surds, never find their decimal values one by one. First make them comparable, then compare.

1. Same order: compare what is under the root. Push every coefficient inside: 3√3 = √27, 2√7 = √28. Now √26, √27 and √28 can be compared directly. The bigger radicand gives the bigger surd.

[[fig:0]]

2. Different orders: make the orders the same. Take the LCM of the orders and raise every surd to that power. For √3, ∛4 and ⁶√10, the LCM of 2, 3 and 6 is 6.
(√3)⁶ = 3³ = 27, (∛4)⁶ = 4² = 16, (⁶√10)⁶ = 10.
So √3 > ∛4 > ⁶√10. The surds keep the same order, because raising positive numbers to the same power does not change their order.

[[fig:1]]

3. Comparing with a decimal: square both. Is √15 < 3.9? Square both: 15 < 15.21. Yes.

4. Sums with the same total. Compare √10 + √2 with √7 + √5. Inside, both have the same total: 10 + 2 = 7 + 5 = 12. Square both sums: 12 + 2√20 against 12 + 2√35. The one with the bigger product wins, so √7 + √5 is bigger. When the totals are equal, the pair of numbers closer to each other gives the bigger sum.

[[fig:2]]

5. Differences with the same gap. Compare √7 − √5 with √5 − √3. Rationalise: √a − √b = a − b√a + √b. Both tops are 2, so the one with the smaller bottom is bigger, and √5 − √3 wins. With the same gap, the smaller numbers give the bigger difference.

The same thinking handles big powers. Radhika has to compare 7²⁰⁰ and 9¹⁵⁰. The bases differ, but the powers have HCF 50: 7²⁰⁰ = (7⁴)⁵⁰ = 2401⁵⁰ and 9¹⁵⁰ = (9³)⁵⁰ = 729⁵⁰. So 7²⁰⁰ is bigger. Roots can be handled this way too, since they are only fractional powers.

What is a logarithm? log_a N = x means exactly the same as a^x = N. So log₂ 32 = 5 because 2⁵ = 32. The logarithm is the power: it answers the question "a raised to what power gives N?". The base a must be positive and not 1, and N must be positive, because a positive base never gives a zero or negative result.

[[fig:0]]

Start with the values that follow straight from the definition:

• log_a 1 = 0 and log_a a = 1.
• log₃ 81 = 4 (3⁴ = 81), log₁₀ 0.001 = −3 (10⁻³ = 0.001), log₁₂ 16 = −4.
• a^{log_a N} = N. For example 5^{log₅ 9} = 9.

The laws. They are the laws of indices in a different form. With the same base on every line:

[[fig:1]]

Two forms help in equations: log_a^q m^p = pq log_a m (so log₄ x = 12 log₂ x), and log_b a = 1log_a b. A chain such as log₂ 3 × log₃ 4 × log₄ 5 collapses to log₂ 5, because each base cancels the next number.

Common logarithm and digits. When no base is written, it is base 10. If log 2 = 0.3010 is given, the number of digits in 2³⁰ is found like this: log 2³⁰ = 30 × 0.3010 = 9.03. The whole-number part, 9, plus 1 gives 10 digits. In general, the number of digits in N is the integer part of log N, plus 1.

Logarithmic equations. Change to the power form, or combine the logs into one and remove them. Always check the answers at the end: every number inside a log must be positive. In log(x + 1) + log(x − 1) = log 24, combine to get (x + 1)(x − 1) = 24, so x² = 25 and x = ±5. Only x = 5 keeps both x + 1 and x − 1 positive, so x = −5 is rejected.