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What do the index laws really mean once the index can be zero, negative or a fraction, and how do you simplify index expressions, solve a simple index equation and use scientific notation without a slip?

Apply the index laws to expressions with rational indices: use zero, negative and fractional indices, simplify and evaluate index expressions, solve simple index equations, and write numbers in scientific notation

A focused answer to the Year 11 Maths Advanced groundwork on indices: the five index laws for the same base, zero and negative indices as reciprocals, fractional indices as roots and powers, the reciprocal-then-root-then-power order for messy indices, solving simple index equations by matching bases, and scientific notation, with worked examples and original practice questions.

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What this dot point is asking

You met the index laws in earlier years for whole-number powers. NESA now expects you to use them with full confidence when the index can be zero, negative or a fraction, to simplify and evaluate index expressions cleanly, to solve simple index equations such as 2x=322^x = 32 by matching the base, and to handle very large or very small numbers in scientific notation. This is the gateway page for the whole exponential and logarithmic module: a logarithm is just an index in disguise, the curve y=axy = a^x is built from these powers, and the derivative rules in calculus run on the power rule, which is the index laws wearing a hat. The mechanics are short, but they recur in almost every later topic, so the goal is speed without slips and, just as importantly, knowing why each rule holds so you can rebuild it under pressure.

The answer

Power, base and index

The expression ana^n is a power. The number aa is the base and nn is the index (also called the exponent; the words mean exactly the same thing). When nn is a positive whole number, the power is just repeated multiplication: an=a×a××aa^n = a \times a \times \dots \times a with nn factors, so 34=3×3×3×3=813^4 = 3 \times 3 \times 3 \times 3 = 81. Everything that follows extends this single idea to indices that are zero, negative or fractional, and the guiding principle is the one that makes the whole topic coherent: we define the new powers precisely so that the laws you already know keep working.

The five index laws (same base)

Five laws let you combine powers of the same base. The first three combine two powers; the last two push a power across a product or quotient:

  • Multiplying adds indices: am×an=am+na^m \times a^n = a^{m+n}.
  • Dividing subtracts indices: aman=amn\dfrac{a^m}{a^n} = a^{m-n}.
  • Power of a power multiplies indices: (am)n=amn\left(a^m\right)^n = a^{mn}.
  • Power of a product distributes: (ab)n=anbn(ab)^n = a^n b^n.
  • Power of a quotient distributes: (ab)n=anbn\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}.

The first law is obvious once you count factors: a2×a3=(a×a)(a×a×a)=a5a^2 \times a^3 = (a \times a)(a \times a \times a) = a^5, and 2+3=52 + 3 = 5. The division law is the same count in reverse, and the power-of-a-power law is repeated use of the first. None of these laws lets you combine different bases: a3×b2a^3 \times b^2 does not simplify, because the bases differ. When an expression has several factors, work through it systematically: the number coefficients first, then each pronumeral in turn. A common and costly slip is to add the coefficients when the law only adds the indices: in 3x4×4x3=12x73x^4 \times 4x^3 = 12 x^7 the indices add (4+34 + 3) but the numbers multiply (3×4=123 \times 4 = 12).

Zero and negative indices: where the definitions come from

Why is a0=1a^0 = 1, and why does a negative index mean "take the reciprocal"? Because those are the only values that keep the division law true. Consider a3a3\dfrac{a^3}{a^3}. By ordinary cancelling it equals 11. By the division law it equals a33=a0a^{3-3} = a^0. For both to agree we are forced to define a0=1a^0 = 1. The same argument fixes negative indices: a2a3=1a\dfrac{a^2}{a^3} = \dfrac{1}{a} by cancelling, but a23=a1a^{2-3} = a^{-1} by the law, so we must define a1=1aa^{-1} = \dfrac{1}{a}, and in general an=1ana^{-n} = \dfrac{1}{a^n}. Both definitions need a0a \neq 0, because they came from dividing by a power of aa.

So the negative sign in an index says, "take the reciprocal" - and the cleanest habit is to do that first. For a fraction this is especially neat, because the reciprocal just turns the fraction upside down: (23)3=(32)3=278\left(\dfrac{2}{3}\right)^{-3} = \left(\dfrac{3}{2}\right)^{3} = \dfrac{27}{8}. Write the reciprocal of ab\dfrac{a}{b} as ba\dfrac{b}{a}, not as 1a/b\dfrac{1}{a/b}, and the rest is just a positive power.

Fractional indices: roots and powers

A fractional index brings in roots, and again the definition is chosen to keep the power-of-a-power law alive. Take a1/2a^{1/2}. The law says (a1/2)2=a(1/2)×2=a1=a\left(a^{1/2}\right)^2 = a^{(1/2) \times 2} = a^1 = a, so a1/2a^{1/2} is the number that squares to aa: it is a\sqrt{a}. The same reasoning gives a1/3=a3a^{1/3} = \sqrt[3]{a} and in general a1/n=ana^{1/n} = \sqrt[n]{a}, the nn-th root. Because roots are involved, the base must satisfy a0a \ge 0 for an even root.

A general fraction mn\dfrac{m}{n} in the index combines a root and a power, and the power-of-a-power law tells you the order does not affect the answer:

am/n=(an)m=amn.a^{m/n} = \left(\sqrt[n]{a}\right)^m = \sqrt[n]{a^m}.

In practice take the root first, because it keeps the numbers small. To evaluate 272/327^{2/3}, take the cube root before squaring: (273)2=32=9\left(\sqrt[3]{27}\right)^2 = 3^2 = 9. Doing it the other way (272=72927^2 = 729, then 7293=9\sqrt[3]{729} = 9) gives the same value but with far heavier arithmetic.

The order for a messy index: reciprocal, then root, then power

When an index is both negative and fractional, deal with the pieces in a fixed order and each step stays simple. This single routine handles every numerical index question in the course, and it is the kind of explicit method the textbook leaves you to assemble yourself.

Solving simple index equations

An equation such as 2x=322^x = 32 asks "what power of 22 gives 3232?". The reliable method at this level is to write both sides as powers of the same base and then equate the indices, because a power is completely determined by its index once the base is fixed. Since 32=2532 = 2^5, the equation 2x=252^x = 2^5 forces x=5x = 5. The same trick handles equations where the two sides use different-looking bases that are secretly powers of one common base: for 8x=48^x = 4, write 8=238 = 2^3 and 4=224 = 2^2, so (23)x=22\left(2^3\right)^x = 2^2 gives 23x=222^{3x} = 2^2 and hence 3x=23x = 2, that is x=23x = \dfrac{2}{3}. (When the two sides cannot be matched to a common base, you need logarithms - the very next page in this module.)

Scientific notation

Scientific notation writes a number as a×10ka \times 10^k, where 1a<101 \le a < 10 (one non-zero digit before the decimal point) and kk is an integer. It is the natural home for the index laws, because multiplying and dividing such numbers just means handling the number parts and the powers of ten separately. To multiply, multiply the aa values and add the indices of 1010; to divide, divide the aa values and subtract the indices. After combining, you may need to renormalise so the front number again lies between 11 and 1010: for instance 12×108=1.2×101×108=1.2×10912 \times 10^8 = 1.2 \times 10^1 \times 10^8 = 1.2 \times 10^9.

The growth curve below is the simplest exponential, y=2xy = 2^x, the same doubling that runs through index equations like 2x=322^x = 32. Reading values off it shows the index laws at work: each step of 11 to the right doubles yy (multiplying by 212^1), each step left halves it (the negative-index reciprocal), and the value at x=0x = 0 is 20=12^0 = 1.

y equals 2 to the x growth curveThe exponential curve y equals 2 to the x rising from near zero on the left, through zero one, doubling at each unit step to one two, two four and three eight, with the x axis as a horizontal asymptote.xy1248-2-1123y = 2ˣ(0, 1)(1, 2)(2, 4)(3, 8)As x decreases the curve approaches y = 0 but never touches it.

How exam questions ask about indices

A handful of command words tell you exactly which skill is wanted:

  • "Simplify" an index expression means combine the powers using the laws and present the answer in index form. Unless told otherwise, leave no negative or zero index in the final form, and remember to multiply the number coefficients.
  • "Simplify, leaving your answer with positive indices" is the explicit version of the same instruction: any ana^{-n} must be rewritten as 1an\dfrac{1}{a^n}.
  • "Evaluate" or "find the value of" a numerical power means produce an exact number. For a fractional index, take the root first; for a negative index, take the reciprocal first.
  • "Solve" an index equation such as 2x=322^x = 32 means write both sides as powers of one base and equate the indices.
  • "Express in scientific notation" signals an answer of the form a×10ka \times 10^k with 1a<101 \le a < 10; renormalise the front number if a calculation leaves it outside that range.
  • "Express in index form" or "using a fractional index" asks you to rewrite a root, such as x23=x2/3\sqrt[3]{x^2} = x^{2/3}, ready for the index laws or for differentiation later.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksEvaluate (a) 50+325^0 + 3^{-2} and (b) (23)1\left(\dfrac{2}{3}\right)^{-1}, leaving each answer as an exact fraction.
Show worked solution →

(a) Apply the zero and negative index meanings. Any non-zero base to the power 00 is 11, so 50=15^0 = 1. A negative index says "take the reciprocal", so 32=132=193^{-2} = \dfrac{1}{3^2} = \dfrac{1}{9}. Add them over a common denominator:

50+32=1+19=99+19=109.5^0 + 3^{-2} = 1 + \frac{1}{9} = \frac{9}{9} + \frac{1}{9} = \frac{10}{9}.

(b) A negative index on a fraction flips the fraction. Taking the reciprocal of 23\dfrac{2}{3} gives

(23)1=32.\left(\frac{2}{3}\right)^{-1} = \frac{3}{2}.

Check numerically. 1+1/9=1.1111=10/91 + 1/9 = 1.1111 = 10/9, and 3/2=1.53/2 = 1.5, matching the reciprocal of 2/30.66672/3 \approx 0.6667.

foundation2 marksSimplify 12a74a3\dfrac{12 a^7}{4 a^3} and (2x4)3\left(2 x^4\right)^3, leaving your answers in index form.
Show worked solution →

Divide powers of the same base by subtracting indices. Divide the number coefficients (12÷4=312 \div 4 = 3) and subtract the indices of aa (73=47 - 3 = 4):

12a74a3=3a73=3a4.\frac{12 a^7}{4 a^3} = 3 a^{7-3} = 3 a^4.

Raise a product to a power by raising each factor. The power 33 applies to both the 22 and the x4x^4, and a power of a power multiplies indices (4×3=124 \times 3 = 12):

(2x4)3=23×x4×3=8x12.\left(2 x^4\right)^3 = 2^3 \times x^{4 \times 3} = 8 x^{12}.

The number coefficients are handled separately from the indices: you divide 1212 by 44, but you cube the 22.

core2 marksEvaluate 272/327^{2/3} and 163/416^{-3/4} without a calculator, leaving each as an exact value.
Show worked solution →

Read the fractional index as root then power. The denominator is the root, the numerator is the power. For 272/327^{2/3}, take the cube root first, then square:

272/3=(273)2=32=9.27^{2/3} = \left(\sqrt[3]{27}\right)^2 = 3^2 = 9.

Taking the root first keeps the numbers small (32=93^2 = 9) rather than cubing 2727 first (272=72927^2 = 729, then 7293=9\sqrt[3]{729} = 9): same answer, easier arithmetic.

For the negative fractional index, take the reciprocal first. Deal with the negative sign, then the root, then the power:

163/4=1163/4=1(164)3=123=18.16^{-3/4} = \frac{1}{16^{3/4}} = \frac{1}{\left(\sqrt[4]{16}\right)^3} = \frac{1}{2^3} = \frac{1}{8}.

Check numerically. 272/39.000027^{2/3} \approx 9.0000 and 163/4=0.125=1/816^{-3/4} = 0.125 = 1/8. Both agree.

core3 marksSimplify (2a2b)3×ab44a3b2\dfrac{\left(2 a^2 b\right)^3 \times a b^4}{4 a^3 b^2}, where aa and bb are non-zero, leaving your answer with positive indices.
Show worked solution →

Raise the bracket to the power first. A power of a product raises each factor, and a power of a power multiplies indices: (2a2b)3=23a2×3b1×3=8a6b3\left(2 a^2 b\right)^3 = 2^3 a^{2 \times 3} b^{1 \times 3} = 8 a^6 b^3.

Multiply out the numerator. Multiply 8a6b38 a^6 b^3 by ab4a b^4: multiply the numbers (8×1=88 \times 1 = 8) and add the indices of each letter (aa: 6+1=76 + 1 = 7; bb: 3+4=73 + 4 = 7):

(2a2b)3×ab4=8a7b7.\left(2 a^2 b\right)^3 \times a b^4 = 8 a^7 b^7.

Divide by the denominator. Divide the numbers (8÷4=28 \div 4 = 2) and subtract indices (aa: 73=47 - 3 = 4; bb: 72=57 - 2 = 5):

8a7b74a3b2=2a73b72=2a4b5.\frac{8 a^7 b^7}{4 a^3 b^2} = 2 a^{7-3} b^{7-2} = 2 a^4 b^5.

Every index is already positive, so the simplified expression is 2a4b52 a^4 b^5.

exam3 marksSolve for xx: (a) 8x=48^x = 4 and (b) 9x=279^x = 27. Give each answer as an exact fraction.
Show worked solution →

(a) Write both sides as powers of the same base. Both 88 and 44 are powers of 22: 8=238 = 2^3 and 4=224 = 2^2. So 8x=(23)x=23x8^x = \left(2^3\right)^x = 2^{3x}, and the equation becomes

23x=22.2^{3x} = 2^2.

Equate the indices (a power is determined by its index once the base is fixed): 3x=23x = 2, so

x=23.x = \frac{2}{3}.

(b) Use base 33. Here 9=329 = 3^2 and 27=3327 = 3^3, so 9x=(32)x=32x9^x = \left(3^2\right)^x = 3^{2x} and the equation is 32x=333^{2x} = 3^3. Equating indices, 2x=32x = 3, so

x=32.x = \frac{3}{2}.

Check numerically. 82/3=48^{2/3} = 4 and 93/2=279^{3/2} = 27, confirming both solutions.

exam4 marksA single bacterium in a Petri dish at a Sydney lab divides so that the number of cells doubles every hour, giving N=2tN = 2^t cells after tt hours. (a) Write NN as an index expression and find the number of cells after 55 hours. (b) The dish is full when it holds about 1.05×1061.05 \times 10^6 cells, which is 2202^{20} cells. After how many whole hours does the dish first become full? (c) The lab also stores 3×1043 \times 10^4 dishes, each able to hold 2×1062 \times 10^6 cells. Find the total capacity in scientific notation.
Show worked solution →

(a) Substitute t=5t = 5 into N=2tN = 2^t. This is a single power of 22:

N=25=32 cells after 5 hours.N = 2^5 = 32 \text{ cells after } 5 \text{ hours}.

(b) Set the count equal to the full value and match the powers of 22. The dish is full when N=220N = 2^{20}, so 2t=2202^t = 2^{20}, giving t=20t = 20. The dish first becomes full after

t=20 hours.t = 20 \text{ hours}.

(As a check, 220=10485761.05×1062^{20} = 1\,048\,576 \approx 1.05 \times 10^6, matching the stated capacity.)

(c) Multiply the two quantities in scientific notation. Multiply the number parts and use the index law on the powers of ten (add the indices 4+6=104 + 6 = 10):

(3×104)×(2×106)=(3×2)×104+6=6×1010.\left(3 \times 10^4\right) \times \left(2 \times 10^6\right) = (3 \times 2) \times 10^{4+6} = 6 \times 10^{10}.

Since 66 already has one non-zero digit before an implied decimal point, 6×10106 \times 10^{10} is in standard scientific form.

Answers. (a) 3232 cells; (b) 2020 hours; (c) 6×10106 \times 10^{10} cells.

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