Algebra · Paper 1
Indices
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Section 1 of 13
What is an Index?
Indices = Powers = Exponentials. Three names for the same thing.
Find $2^{3}$
$2^{3} = 2 \times 2 \times 2$
$= 8$
Anatomy of an index
Key terms
$b^{p} = n$
$b$= base
$p$= power (also called index or exponential)
$n$= number (the value)
So in $2^{3} = 8$: base is $2$, power is $3$, value is $8$.
Section 2 of 13
Multiplying Same Base
$2^{2} \times 2^{3} = (2 \times 2)(2 \times 2 \times 2)$
$= 2 \times 2 \times 2 \times 2 \times 2$
$= 2^{5}$
The powers just add: $2 + 3 = 5$.
$3^{3} \times 3^{20} = 3^{23}$
Rule 1
$a^{p} \cdot a^{q} = a^{p+q}$
Different bases don't combine.
$2^{3} \times 3^{2} = 2 \times 2 \times 2 \times 3 \times 3$
Stays as $2^{3} \times 3^{2}$ — no shortcut.
You try
Simplify: $x^{4} \cdot x^{7}$
Same base — add the powers.
$x^{4} \cdot x^{7} = x^{4+7}$
$= x^{11}$
$x^{11}$
Section 3 of 13
Dividing Same Base
$\dfrac{2^{5}}{2^{2}} = \dfrac{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2}{2 \cdot 2}$
Cancel two $2$'s top and bottom.
$= 2^{3}$
The powers just subtract: $5 - 2 = 3$.
Rule 2
$\dfrac{a^{p}}{a^{q}} = a^{p-q}$
You try
Simplify: $\dfrac{x^{9}}{x^{3}}$
Same base — subtract the powers.
$\dfrac{x^{9}}{x^{3}} = x^{9-3}$
$= x^{6}$
$x^{6}$
Section 4 of 13
Zero and Negative Indices
Watch the pattern as the power steps down:
$\dfrac{3^{3}}{3^{1}} = \dfrac{27}{3} = 9 = 3^{2}$
$\dfrac{3^{2}}{3^{1}} = \dfrac{9}{3} = 3 = 3^{1}$
$\dfrac{3^{1}}{3^{1}} = \dfrac{3}{3} = 1 = 3^{0}$
Zero index
$a^{0} = 1$
(for any non-zero $a$)
Step down further:
$\dfrac{3^{0}}{3^{1}} = \dfrac{1}{3} = 3^{-1}$
$3^{-2} = \dfrac{1}{9} = \dfrac{1}{3^{2}}$
Negative index
$a^{-p} = \dfrac{1}{a^{p}}$
$5^{-3} = \dfrac{1}{5^{3}} = \dfrac{1}{125}$
Reciprocal flip
If the negative power is on the bottom, it flips up positive:
$\dfrac{1}{a^{-n}} = a^{n}$
$\dfrac{1}{2^{-2}} = 2^{2} = 4$
You try
Evaluate: $2^{-4}$
Flip the negative power: $a^{-p} = 1/a^{p}$.
$2^{-4} = \dfrac{1}{2^{4}}$
$= \dfrac{1}{16}$
$\dfrac{1}{16}$
Section 5 of 13
Power of a Power
$(2^{3})^{2} = 2^{3} \times 2^{3} = 2^{6}$
$(3^{3})^{3} = 3^{3} \times 3^{3} \times 3^{3} = 3^{9}$
The two powers multiply: $3 \times 2 = 6$, $3 \times 3 = 9$.
Rule 3
$\left(a^{p}\right)^{q} = a^{pq}$
You try
Simplify: $(x^{2})^{5}$
Powers multiply.
$(x^{2})^{5} = x^{2 \times 5}$
$= x^{10}$
$x^{10}$
Section 6 of 13
Fractional Indices
$9 = 3^{2}$, so what is $9^{\tfrac{1}{2}}$? Whatever it is, doing it twice must give $9$.
$3^{2} = 9 \quad\Rightarrow\quad 9^{\tfrac{1}{2}} = 3$
$2^{3} = 8 \quad\Rightarrow\quad 8^{\tfrac{1}{3}} = 2$
$5^{3} = 125 \quad\Rightarrow\quad 125^{\tfrac{1}{3}} = 5$
Fractional index = root
$a^{\tfrac{1}{q}} = \sqrt[q]{a}$
Must learn
$\sqrt{a} = a^{\tfrac{1}{2}}$
Power $p/q$ — combine root and power
Split $\tfrac{p}{q}$ as $p \times \tfrac{1}{q}$, then use power-of-a-power:
$a^{\tfrac{p}{q}} = a^{p \times \tfrac{1}{q}} = \left(a^{p}\right)^{\tfrac{1}{q}} = \sqrt[q]{a^{p}}$
Or do the root first:
$a^{\tfrac{p}{q}} = \left(a^{\tfrac{1}{q}}\right)^{p} = \left(\sqrt[q]{a}\right)^{p}$
Tip: root first, then power — keeps the numbers small.
$8^{\tfrac{2}{3}} = \left(\sqrt[3]{8}\right)^{2} = 2^{2} = 4$
$27^{\tfrac{2}{3}} = \left(\sqrt[3]{27}\right)^{2} = 3^{2} = 9$
You try
Evaluate: $16^{\tfrac{3}{4}}$
Fourth root first, then cube.
$16^{\tfrac{3}{4}} = \left(\sqrt[4]{16}\right)^{3}$
$= 2^{3}$
$= 8$
$8$
Section 7 of 13
Brackets and Reciprocal
Brackets distribute the power
$(2x)^{3} = 2 \cdot x \cdot 2 \cdot x \cdot 2 \cdot x = 8x^{3}$
$(9x)^{2} = 81x^{2}$
Rule 4
$(ab)^{p} = a^{p}\,b^{p}$
Fraction in brackets
$\left(\dfrac{2}{3}\right)^{2} = \dfrac{2}{3} \times \dfrac{2}{3} = \dfrac{4}{9}$
$\left(\dfrac{3}{5}\right)^{3} = \dfrac{27}{125}$
Rule 5
$\left(\dfrac{a}{b}\right)^{p} = \dfrac{a^{p}}{b^{p}}$
Negative bracket — flip the fraction
$\left(\dfrac{2}{3}\right)^{-2} = \dfrac{2^{-2}}{3^{-2}} = \dfrac{\;\tfrac{1}{2^{2}}\;}{\tfrac{1}{3^{2}}} = \dfrac{3^{2}}{2^{2}} = \dfrac{9}{4}$
Same answer if we flip first: $\left(\dfrac{3}{2}\right)^{2} = \dfrac{9}{4}$. ✓
$\left(\dfrac{3}{4}\right)^{-3} = \left(\dfrac{4}{3}\right)^{3} = \dfrac{64}{27}$
Must learn — flip the fraction
$\left(\dfrac{a}{b}\right)^{-p} = \left(\dfrac{b}{a}\right)^{p}$
You try
Evaluate: $\left(\dfrac{2}{5}\right)^{-3}$
Flip the fraction, then cube.
$\left(\dfrac{2}{5}\right)^{-3} = \left(\dfrac{5}{2}\right)^{3}$
$= \dfrac{5^{3}}{2^{3}}$
$= \dfrac{125}{8}$
$\dfrac{125}{8}$
Section 8 of 13
Important Notes
Warning — you can not split a power over a sum:
$(a + b)^{n}$ ≠ $a^{n} + b^{n}$
Check with $(2+3)^{2}$:
$(2+3)^{2} = 5^{2} = 25$
$2^{2} + 3^{2} = 4 + 9 = 13$
$25$ ≠ $13$ ✗
Warning — a coefficient on a power is not the same as multiplying the base:
$2(3^{x})$ ≠ $6^{x}$
Check with $x = 2$: $2(3^{2}) = 2(9) = 18$, but $6^{2} = 36$. ✗
Section 9 of 13
Type 1: Evaluations
Use the rules to find the value. Work the structure of the index first; the numbers fall out at the end.
(i) $3^{2}$
$3^{2} = 9$
(ii) $5^{-3}$
$5^{-3} = \dfrac{1}{5^{3}}$
$= \dfrac{1}{125}$
(iii) $25^{\tfrac{1}{2}}$
$25^{\tfrac{1}{2}} = \sqrt{25}$
$= 5$
(iv) $9^{-\tfrac{1}{2}}$
$9^{-\tfrac{1}{2}} = \dfrac{1}{9^{\tfrac{1}{2}}} = \dfrac{1}{\sqrt{9}}$
$= \dfrac{1}{3}$
(v) $8^{-\tfrac{1}{3}}$
$8^{-\tfrac{1}{3}} = \dfrac{1}{8^{\tfrac{1}{3}}} = \dfrac{1}{\sqrt[3]{8}}$
$= \dfrac{1}{2}$
(vi) $\dfrac{1}{2^{-2}}$
$\dfrac{1}{2^{-2}} = 2^{2}$
$= 4$
(vii) $125^{-\tfrac{1}{3}}$
$125^{-\tfrac{1}{3}} = \dfrac{1}{\sqrt[3]{125}}$
$= \dfrac{1}{5}$
(viii) $27^{\tfrac{2}{3}}$
$27^{\tfrac{2}{3}} = \left(\sqrt[3]{27}\right)^{2}$
$= 3^{2}$
$= 9$
(ix) $\left(\dfrac{2}{3}\right)^{2}$
$\left(\dfrac{2}{3}\right)^{2} = \dfrac{2^{2}}{3^{2}}$
$= \dfrac{4}{9}$
(x) $\left(\dfrac{3}{4}\right)^{-2}$
Negative bracket — flip the fraction.
$\left(\dfrac{3}{4}\right)^{-2} = \left(\dfrac{4}{3}\right)^{2} = \dfrac{16}{9}$
(xi) $27^{-\tfrac{2}{3}}$
$27^{-\tfrac{2}{3}} = \dfrac{1}{27^{\tfrac{2}{3}}} = \dfrac{1}{\left(\sqrt[3]{27}\right)^{2}}$
$= \dfrac{1}{3^{2}}$
$= \dfrac{1}{9}$
Simplify with algebra
$x^{-\tfrac{1}{2}} = \dfrac{1}{x^{\tfrac{1}{2}}} = \dfrac{1}{\sqrt{x}}$
$x^{-\tfrac{3}{2}} = \dfrac{1}{x^{\tfrac{3}{2}}} = \dfrac{1}{\left(\sqrt{x}\right)^{3}} = \dfrac{1}{x\sqrt{x}}$
You try
Evaluate: $4^{\tfrac{3}{2}}$
Square root first ($\sqrt{4} = 2$), then cube.
$4^{\tfrac{3}{2}} = \left(\sqrt{4}\right)^{3}$
$= 2^{3}$
$= 8$
$8$
You try
Evaluate: $\left(\dfrac{8}{27}\right)^{-\tfrac{2}{3}}$
Flip the fraction (negative power), then root, then square.
$\left(\dfrac{8}{27}\right)^{-\tfrac{2}{3}} = \left(\dfrac{27}{8}\right)^{\tfrac{2}{3}}$
$= \left(\sqrt[3]{\dfrac{27}{8}}\right)^{2} = \left(\dfrac{3}{2}\right)^{2}$
$= \dfrac{9}{4}$
$\dfrac{9}{4}$
Section 10 of 13
Type 2: $x$ in the Power
Must learn
One base = one base ⟹ powers are equal.
(i) $5^{x} = 25$
$5^{x} = 5^{2}$
$x = 2$
(ii) $2^{x} = 8$
$2^{x} = 2^{3}$
$x = 3$
(iii) $2^{x} = \dfrac{8}{\sqrt{2}}$
Get everything to base $2$ first.
$2^{x} = \dfrac{2^{3}}{2^{\tfrac{1}{2}}}$
$2^{x} = 2^{3 - \tfrac{1}{2}}$
$2^{x} = 2^{\tfrac{5}{2}}$
$x = \dfrac{5}{2}$
(iv) $9^{x+1} = \dfrac{27}{\sqrt{3}}$
Bring both sides to base $3$.
$\left(3^{2}\right)^{x+1} = \dfrac{3^{3}}{3^{\tfrac{1}{2}}}$
$3^{2x+2} = 3^{3 - \tfrac{1}{2}}$
$3^{2x+2} = 3^{\tfrac{5}{2}}$
$2x + 2 = \dfrac{5}{2}$
$2x = \dfrac{1}{2}$
$x = \dfrac{1}{4}$
(v) $3^{x} = \dfrac{\sqrt{3}}{27}$
Both sides to base $3$. Watch the signs.
$3^{x} = \dfrac{3^{\tfrac{1}{2}}}{3^{3}}$
$3^{x} = 3^{\tfrac{1}{2} - 3}$
$3^{x} = 3^{-\tfrac{5}{2}}$
$x = -\dfrac{5}{2}$
(vi) $8^{x-3} = \left(\dfrac{4}{\sqrt{2}}\right)^{3}$
Everything to base $2$.
$\left(2^{3}\right)^{x-3} = \left(\dfrac{2^{2}}{2^{\tfrac{1}{2}}}\right)^{3}$
$2^{3x-9} = \left(2^{2 - \tfrac{1}{2}}\right)^{3}$
$2^{3x-9} = \left(2^{\tfrac{3}{2}}\right)^{3} = 2^{\tfrac{9}{2}}$
$3x - 9 = \dfrac{9}{2}$
$3x = \dfrac{27}{2}$
$x = \dfrac{9}{2}$
(vii) $\dfrac{1}{25^{x-1}} = \dfrac{\sqrt{5}}{5}$
Everything to base $5$. The $\tfrac{1}{\cdots}$ on the left becomes a negative power.
$\dfrac{1}{\left(5^{2}\right)^{x-1}} = \dfrac{5^{\tfrac{1}{2}}}{5^{1}}$
$5^{-(2x-2)} = 5^{\tfrac{1}{2} - 1}$
$5^{2 - 2x} = 5^{-\tfrac{1}{2}}$
$2 - 2x = -\dfrac{1}{2}$
$-2x = -\dfrac{5}{2}$
$x = \dfrac{5}{4}$
Impossible cases
$3^{x} = -9$
Try $x = 2$: $3^{2} = 9$ (positive) ✗
Try $x = -2$: $3^{-2} = \dfrac{1}{9}$ (still positive) ✗
You cannot go from a positive base to a negative answer.
$\therefore\;\;$ Impossible
You try
Solve: $4^{x} = \dfrac{1}{8}$
Get both sides to base $2$. $4 = 2^{2}$ and $\tfrac{1}{8} = 2^{-3}$.
$\left(2^{2}\right)^{x} = 2^{-3}$
$2^{2x} = 2^{-3}$
$2x = -3$
$x = -\dfrac{3}{2}$
$x = -\dfrac{3}{2}$
You try
Solve: $2^{x} = -16$
Look at the sign of the right-hand side.
$2^{x}$ is positive for every real $x$.
A positive base cannot give a negative answer.
Impossible
Impossible
Section 11 of 13
Type 3: Write in Terms of $y$
Given $y = 3^{x}$, write each in terms of $y$.
(i) $3^{x+1}$
$3^{x+1} = 3^{x} \cdot 3^{1}$
$= y \cdot 3$
$= 3y$
(ii) $3^{x+2}$
$3^{x+2} = 3^{x} \cdot 3^{2} = 9 \cdot 3^{x}$
$= 9y$
(iii) $3^{x-3}$
$3^{x-3} = \dfrac{3^{x}}{3^{3}}$
$= \dfrac{y}{27}$
(iv) $3^{2x}$
$3^{2x} = \left(3^{x}\right)^{2}$
$= y^{2}$
(v) $3^{2x+1}$
$3^{2x+1} = 3 \cdot 3^{2x} = 3 \cdot \left(3^{x}\right)^{2}$
$= 3y^{2}$
You try
Given $t = 5^{x}$, write in terms of $t$: (a) $5^{x+3}$ (b) $5^{2x}$
(a) Split off the $5^{3}$ as a coefficient. (b) Use power-of-a-power.
(a) $5^{x+3} = 5^{x} \cdot 5^{3} = t \cdot 125 = 125t$
(b) $5^{2x} = (5^{x})^{2} = t^{2}$
$125t \;$ and $\; t^{2}$
(a) $125t$ · (b) $t^{2}$
Section 12 of 13
Quadratic in Indices
When you see two index terms with the same base — typically $b^{2x+c}$ and $b^{x}$ — substitute $t = b^{x}$ to turn the equation into a normal quadratic in $t$.
Solve $3^{2x+1} - 28\left(3^{x}\right) + 9 = 0$
Step 1: rewrite the awkward term.
$3^{2x+1} = 3 \cdot 3^{2x} = 3\left(3^{x}\right)^{2}$
Step 2: let $t = 3^{x}$.
$3t^{2} - 28t + 9 = 0$
Step 3: factorise (guide number $+27 \cdot 1$, add $-28$).
$3t^{2} - 27t - t + 9 = 0$
$3t(t - 9) - 1(t - 9) = 0$
$(3t - 1)(t - 9) = 0$
$t = \dfrac{1}{3} \quad$ or $\quad t = 9$
Step 4: back-substitute $t = 3^{x}$.
$3^{x} = \dfrac{1}{3} = 3^{-1} \;\;\Rightarrow\;\; x = -1$
$3^{x} = 9 = 3^{2} \;\;\Rightarrow\;\; x = 2$
$x = -1 \;$ or $\; x = 2$
Solve $2^{2x+1} - 9\left(2^{x}\right) + 4 = 0$
$2^{2x+1} = 2 \cdot 2^{2x} = 2\left(2^{x}\right)^{2}$
Let $t = 2^{x}$:
$2t^{2} - 9t + 4 = 0$
$(2t - 1)(t - 4) = 0$
$t = \dfrac{1}{2} \quad$ or $\quad t = 4$
$2^{x} = 2^{-1} \;\;\Rightarrow\;\; x = -1$
$2^{x} = 2^{2} \;\;\Rightarrow\;\; x = 2$
$x = -1 \;$ or $\; x = 2$
Solve $2^{2x} - 2^{x+1} - 3 = 0$
Rewrite the two power-of-$2$ terms in terms of $2^{x}$.
$2^{2x} = \left(2^{x}\right)^{2}$
$2^{x+1} = 2 \cdot 2^{x}$
Let $t = 2^{x}$:
$t^{2} - 2t - 3 = 0$
$(t - 3)(t + 1) = 0$
$t = 3 \quad$ or $\quad t = -1$
Back-substitute and check each:
$2^{x} = -1$ — a positive base can never give a negative. Impossible.
$2^{x} = 3$ — a real solution exists, but $3$ is not a power of $2$.
We'll come back to $2^{x} = 3$ when we learn logs — that's how we get $x$ out of the power exactly.
You try
Solve: $3^{2x+1} - 82\left(3^{x}\right) + 27 = 0$
$3^{2x+1} = 3(3^{x})^{2}$. Let $t = 3^{x}$. Factor — guide number is $+81 \cdot 1$, sum $-82$.
$3(3^{x})^{2} - 82(3^{x}) + 27 = 0$
Let $t = 3^{x}$:
$3t^{2} - 82t + 27 = 0$
$3t^{2} - 81t - t + 27 = 0$
$3t(t-27) - 1(t-27) = 0$
$(3t - 1)(t - 27) = 0$
$t = \tfrac{1}{3} \;$ or $\; t = 27$
$3^{x} = 3^{-1} \;\Rightarrow\; x = -1$
$3^{x} = 3^{3} \;\Rightarrow\; x = 3$
$x = -1 \;$ or $\; x = 3$
$x = -1 \;$ or $\; x = 3$
Section 13 of 13
Sequences with Indices
$U_{n}$ is the $n$th term of a sequence. When the term contains an index, the rules let you write $U_{n+1}$, $U_{n+2}$ in terms of $U_{n}$.
$U_{n} = 2^{n}$ — find $U_{3}$ and $U_{n+1}$
$U_{3} = 2^{3} = 8$
$U_{n+1} = 2^{n+1} = 2^{n} \cdot 2$
$= 2\left(2^{n}\right)$
$U_{n} = 5^{n}$ — find $U_{2}$ and $U_{n+2}$
$U_{2} = 5^{2} = 25$
$U_{n+2} = 5^{n+2} = 5^{n} \cdot 5^{2}$
$= 25\left(5^{n}\right)$
$U_{n} = 3^{n}$ — find $U_{n-1}$
$U_{n-1} = 3^{n-1} = \dfrac{3^{n}}{3}$
Proofs
Technique
Write out $U_{n}$, $U_{n+1}$, $U_{n+2}$ using the rules. Let $a$ and $b$ stand in for the index terms. Substitute into the LHS — every coefficient should cancel to give $0$.
$U_{n} = 2^{n} + 3^{n}$ — prove $U_{n+2} - 5U_{n+1} + 6U_{n} = 0$
Step 1: write the terms out.
$U_{n+1} = 2^{n+1} + 3^{n+1} = 2\left(2^{n}\right) + 3\left(3^{n}\right)$
$U_{n+2} = 2^{n+2} + 3^{n+2} = 4\left(2^{n}\right) + 9\left(3^{n}\right)$
Step 2: let $a = 2^{n}$ and $b = 3^{n}$.
$U_{n} = a + b$
$U_{n+1} = 2a + 3b$
$U_{n+2} = 4a + 9b$
Step 3: substitute into the LHS.
$U_{n+2} - 5U_{n+1} + 6U_{n}$
$= (4a + 9b) - 5(2a + 3b) + 6(a + b)$
$= 4a + 9b - 10a - 15b + 6a + 6b$
$= (4 - 10 + 6)a + (9 - 15 + 6)b$
$= 0 \quad\checkmark$
Spot the pattern: for $U_{n} = a^{n} + b^{n}$, the identity $U_{n+2} - (a+b)U_{n+1} + ab \cdot U_{n} = 0$ always works. Here $a + b = 2 + 3 = 5$ and $ab = 2 \times 3 = 6$.
You try
Given $U_{n} = 5^{n} + 7^{n}$, prove $U_{n+2} - 12\,U_{n+1} + 35\,U_{n} = 0$.
$5+7 = 12$ and $5 \times 7 = 35$ — the pattern fits. Let $a = 5^{n}$, $b = 7^{n}$.
$U_{n+1} = 5(5^{n}) + 7(7^{n}) = 5a + 7b$
$U_{n+2} = 25(5^{n}) + 49(7^{n}) = 25a + 49b$
$U_{n+2} - 12U_{n+1} + 35U_{n}$
$= (25a + 49b) - 12(5a + 7b) + 35(a + b)$
$= 25a + 49b - 60a - 84b + 35a + 35b$
$= (25 - 60 + 35)a + (49 - 84 + 35)b$
$= 0 \quad\checkmark$
Both coefficients sum to $0$. Proved.
You try
Given $U_{n} = 3^{n} + 8^{n}$, prove $U_{n+2} - 11\,U_{n+1} + 24\,U_{n} = 0$.
$3 + 8 = 11$ and $3 \times 8 = 24$. Let $a = 3^{n}$, $b = 8^{n}$.
$U_{n+1} = 3(3^{n}) + 8(8^{n}) = 3a + 8b$
$U_{n+2} = 9(3^{n}) + 64(8^{n}) = 9a + 64b$
$U_{n+2} - 11U_{n+1} + 24U_{n}$
$= (9a + 64b) - 11(3a + 8b) + 24(a + b)$
$= 9a + 64b - 33a - 88b + 24a + 24b$
$= (9 - 33 + 24)a + (64 - 88 + 24)b$
$= 0 \quad\checkmark$
Both coefficients sum to $0$. Proved.
That's Indices.
Next up on Mathslive.ie: Logs — the inverse of indices.