Indices and Equations: Revision Notes

Condensed for the final weeks. For worked examples, use the Indices and Equations study guide.

Index laws

a^m x a^n  =  a^(m+n)        a^0     =  1
a^m / a^n  =  a^(m-n)        a^-n    =  1 / a^n
(a^m)^n    =  a^(mn)         a^(1/n) =  nth root of a
                             a^(m/n) =  (nth root of a)^m

Worked: 8^(2/3) = (∛8)² = 2² = 4. Take the root first, then the power — the numbers stay small.

Equations with the unknown in the index, e.g. 5^(x+1) = 25^x: rewrite both sides with a matching base first, then equate the indices.

5^(x+1) = (5^2)^x = 5^(2x)
so  x + 1 = 2x  ->  x = 1

Matching bases, then equating indices, is the standard technique — logarithms are not needed at this level.

Solving linear equations

Do the same to both sides; unwind the operations in reverse order. With fractions, multiply every term by the common denominator first.

Worked example (fractional equation). Solve x/(x + 2) = 3/(x − 6).

x(x - 6) = 3(x + 2)
x^2 - 6x = 3x + 6
x^2 - 9x - 6 = 0

This reduces to a quadratic — solve with the formula below, and always check the answer doesn’t make an original denominator zero.

Simultaneous equations

Method	Use when
Elimination	Both equations are linear
Substitution	One equation is already in the form y = … (or can easily be rearranged into it)

Construct the two equations from the wording first if a question describes a situation rather than giving the equations directly — this is where marks are lost before any algebra even begins.

3x + 2y = 16
 x -  y =  2   ->  x = y + 2

Substitute:  3(y + 2) + 2y = 16
             3y + 6 + 2y = 16
             5y = 10  ->  y = 2,  x = 4

Always check in the other equation — it catches almost every arithmetic slip.

Quadratic equations — three methods

1. Factorising — try first, since it’s fastest when it works. Two numbers multiplying to ac, adding to b.
2. Completing the square — also gives the turning point.
3. Formula — always works, whatever the numbers, even when the expression won’t factorise neatly:

x = [ -b +/- sqrt(b^2 - 4ac) ] / 2a

Rearrange to = 0 before doing anything else.

Discriminant b² − 4ac: positive → two roots · zero → one repeated root · negative → no real roots.

Worked example (surd form). Solve x² − 4x − 3 = 0, giving answers in surd form.

x = (4 +/- sqrt(16 + 12)) / 2 = (4 +/- sqrt(28)) / 2 = 2 +/- sqrt(7)

Since 28 has no integer square root, leave the answer as a surd rather than approximating it — that’s exactly what “surd form” is asking for.

Changing the subject

Treat it as solving for a letter. Unwind operations in reverse; if the required letter appears twice, collect those terms on one side and factorise.

Make r the subject:  A = pi r^2
                     r^2 = A / pi
                     r = sqrt(A / pi)

Worked example (subject inside a root). Make x the subject of y = √(x + 3) − 2.

y + 2 = sqrt(x + 3)
(y + 2)^2 = x + 3
x = (y + 2)^2 - 3

When the subject is under a root, isolate the root first, then square both sides to remove it — squaring is always the last step, once the root stands alone. This rearrangement is only valid when y + 2 ≥ 0, since a square root cannot itself be negative; squaring both sides must always be checked for reversibility like this.

Exam traps

• a^-n is a reciprocal, not a negative number: 2⁻³ = 1/8, not −8.
• a^0 = 1 for any non-zero a.
• Rearrange a quadratic to = 0 before factorising or using the formula.
• In the formula, −b means the opposite sign of b — if b = −5, then −b = +5.
• When the subject appears twice, you must factorise; you cannot just divide.
• Multiply every term when clearing fractions, including those without a denominator.
• In a fractional equation, always reject any solution that would make an original denominator zero.
• Approximating a surd answer when a question explicitly asks for surd form, or vice versa.
• Squaring too early, before the root is isolated on its own, when the subject sits under a root.

Self-test

1. Simplify (2x³)⁴.
2. Evaluate 27^(2/3) and 5⁻².
3. Solve simultaneously: 2x + y = 11, x − y = 1.
4. Solve x² − 5x + 6 = 0.
5. Make h the subject of V = πr²h.
6. Solve 5^(x+1) = 25^x.
7. Make x the subject of y = √(x + 3) − 2.

Answers: 1. 2⁴ × x¹² = 16x¹². 2. 27^(2/3) = (∛27)² = 3² = 9; 5⁻² = 1/25 = 0.04. 3. Adding: 3x = 12 → x = 4, y = 3. 4. (x − 2)(x − 3) = 0 → x = 2 or 3. 5. h = V / (πr²). 6. Rewrite 25 as 5²: x + 1 = 2x → x = 1. 7. x = (y + 2)² − 3, valid provided y + 2 ≥ 0.

For worked examples with full explanations, see the Indices and Equations study guide; for exam-style practice with full mark schemes, see the Indices and Equations practice questions.