C1.7; E1.7 - Integer index laws

C1.7; E1.7 - Integer index laws

Powers can make a calculation much shorter, even when an index is zero or negative. You will use repeated factors to explain the index laws, then combine them to find exact numerical values. This lesson is for Core and Extended.

Multiplying and dividing powers

How can 23÷252^3\div2^5 have a value when there seem to be more factors underneath than on top? We will begin with the easier case, where some factors remain on top, then follow the same pattern further.

In 343^4, the base is 33 and the index is 44: it means four factors of 33, not 3×43\times4. Multiplying 343^4 by 323^2 joins a group of four factors to a group of two. There are six factors altogether. Dividing cancels matching nonzero factors instead.

Product and quotient laws

For the same nonzero base aa and integer indices mm and nn,

am×an=am+n,am÷an=am−na^m\times a^n=a^{m+n},\qquad a^m\div a^n=a^{m-n}

Here the letters are labels: aa stands for the base number, and m,nm,n for the two indices. The nonzero condition prevents division by zero and allows zero and negative indices later. Multiplication with positive indices also works for base zero.

Not given in the List of formulas: learn these laws.

These are laws for powers with the same base. In 23×322^3\times3^2, the factors are different, so adding the indices does not work.

Worked example 1: joining factors

Question. Work out 34×323^4\times3^2.

Plan. Add the indices because the base is the same, then evaluate.

34×32=34+2join four and two factors=36six factors of three=729evaluate the power\begin{aligned} 3^4\times3^2&=3^{4+2} && \text{join four and two factors}\\ &=3^6 && \text{six factors of three}\\ &=729 && \text{evaluate the power} \end{aligned}

Answer. 729729.

Check. Separately, 34=813^4=81 and 32=93^2=9; 81×9=72981\times9=729.

Change only multiplication to division. The factors now cancel instead of joining.

Worked example 2: cancelling factors

Question. Work out 34÷323^4\div3^2.

Plan. Subtract the divisor's index from the first index.

34÷32=3×3×3×33×3write repeated factors=34−2=32cancel two matching factors=9evaluate\begin{aligned} 3^4\div3^2&=\frac{3\times3\times3\times3}{3\times3} && \text{write repeated factors}\\ &=3^{4-2}=3^2 && \text{cancel two matching factors}\\ &=9 && \text{evaluate} \end{aligned}

Answer. 99.

Check. 9×32=9×9=81=349\times3^2=9\times9=81=3^4, undoing the division.

Watch out

34×32=383^4\times3^2=3^8 is wrong: multiplying the indices would describe a power of a power. Here you add the numbers of factors, so 34×32=363^4\times3^2=3^6.

Try the division method with a different base.

Zero and negative indices

Reducing an index by one divides the value by the base. For powers of 22, the pattern continues through index zero:

Index33221100−1-1−2-2
Power of 228844221112\frac1214\frac14

Each step right divides by 22. Zero as an index therefore gives 11, and a negative index gives a reciprocal, not a negative answer.

Zero and negative integer indices

For a nonzero base aa and a positive integer nn,

a0=1,a−n=1ana^0=1,\qquad a^{-n}=\frac{1}{a^n}

For example, 70=17^0=1 and 7−2=1497^{-2}=\frac1{49}. We do not assign a value to 000^0 here; 0−n0^{-n} is undefined because it would divide by zero.

Not given in the List of formulas: learn these rules.

Why must the zero rule work? Any nonzero number divided by itself gives 11. For instance, 53÷53=15^3\div5^3=1. The quotient law writes the same calculation as 53−3=505^{3-3}=5^0. The two forms must agree.

Worked example 3: reaching index zero

Question. Work out 23÷232^3\div2^3.

Plan. Use the quotient law and the zero-index rule.

23÷23=23−3subtract the indices=20=1nonzero base to index zero\begin{aligned} 2^3\div2^3&=2^{3-3} && \text{subtract the indices}\\ &=2^0=1 && \text{nonzero base to index zero} \end{aligned}

Answer. 11.

Check. The original calculation is 8÷8=18\div8=1.

Increase only the divisor's index from 33 to 55. Two uncancelled factors now remain in the denominator.

Worked example 4: continuing below zero

Question. Work out 23÷252^3\div2^5.

Plan. Subtract the indices, then turn the negative power into a reciprocal.

23÷25=23−5the same quotient law=2−2=122take the reciprocal=14evaluate the denominator\begin{aligned} 2^3\div2^5&=2^{3-5} && \text{the same quotient law}\\ &=2^{-2}=\frac1{2^2} && \text{take the reciprocal}\\ &=\frac14 && \text{evaluate the denominator} \end{aligned}

Answer. 14\frac14.

Check. Directly, 23÷25=832=142^3\div2^5=\frac8{32}=\frac14. Multiplying the answer by 3232 gives 88.

The rules also apply to negative nonzero bases, with brackets: (−3)−2=1(−3)2=19(-3)^{-2}=\frac1{(-3)^2}=\frac19, while (−3)−3=−127(-3)^{-3}=-\frac1{27}. The base's sign and whether its positive power is odd or even determine the answer's sign.

Watch out

4−2=−164^{-2}=-16 is wrong. The minus sign belongs to the index and means reciprocal: 4−2=1164^{-2}=\frac1{16}. By contrast, −42=−(42)=−16-4^2=-(4^2)=-16; brackets would be needed to make −4-4 the base.

Apply the reciprocal rule after subtraction.

Powers of powers and combined laws

A power outside brackets repeats the whole power inside. For example, (23)2=(23)×(23)(2^3)^2=(2^3)\times(2^3). There are two groups of three factors of 22, making six factors altogether.

Power of a power

For a nonzero base aa and integer indices m,nm,n,

(am)n=amn(a^m)^n=a^{mn}

Multiply the indices. If all indices are positive, base zero is also allowed.

Not given in the List of formulas: learn this law.

When laws occur together, deal with bracketed powers first, then multiplication or division. Keep the value as a single power until the end; this avoids calculating large numbers unnecessarily.

Worked example 5: a bracketed power and a quotient

Question. Work out (23)224\dfrac{(2^3)^2}{2^4}.

Plan. Multiply the indices in the numerator, then subtract the denominator's index.

(23)224=23×224power of a power=26−4=22quotient with the same base=4evaluate last\begin{aligned} \frac{(2^3)^2}{2^4}&=\frac{2^{3\times2}}{2^4} && \text{power of a power}\\ &=2^{6-4}=2^2 && \text{quotient with the same base}\\ &=4 && \text{evaluate last} \end{aligned}

Answer. 44.

Check. 23=82^3=8, so the original is 82÷16=64÷16=48^2\div16=64\div16=4.

Change only the inner index from 33 to −3-3. The same laws work, but the answer is now a reciprocal.

Worked example 6: a negative inner index

Question. Work out (2−3)224\dfrac{(2^{-3})^2}{2^4}.

Plan. Combine the indices before interpreting the negative result.

(2−3)224=2−624multiply −3 by 2=2−6−4=2−10subtract the divisor’s index=11024reciprocal of 210\begin{aligned} \frac{(2^{-3})^2}{2^4}&=\frac{2^{-6}}{2^4} && \text{multiply }-3\text{ by }2\\ &=2^{-6-4}=2^{-10} && \text{subtract the divisor's index}\\ &=\frac1{1024} && \text{reciprocal of }2^{10} \end{aligned}

Answer. 11024\frac1{1024}.

Check. Squaring 2−3=182^{-3}=\frac18 gives 164\frac1{64}; dividing this by 1616 gives 11024\frac1{1024}.

Watch out

(23)2=25(2^3)^2=2^5 is wrong. Adding indices is for a product of powers, such as 23×222^3\times2^2. A power of a power repeats each factor, so (23)2=26(2^3)^2=2^6.

You can now solve the opening puzzle without inventing a new division rule: extra factors below the line become a reciprocal.

Quick Recap

  • For the same nonzero base, multiply powers by adding indices and divide by subtracting indices.
  • A nonzero number to index zero is 11. A negative index means reciprocal.
  • For a power of a power, multiply the indices.
  • Keep brackets around negative bases; a negative base and a negative index have different jobs.

The opening calculation is 23÷25=2−2=142^3\div2^5=2^{-2}=\frac14: two factors of 22 remain underneath.