3.01c - Laws of indices

3.01c - Laws of indices

Can you simplify a product of powers without working out the large numbers? Learn why multiplying powers adds indices, dividing subtracts them and raising a power multiplies them, including zero and negative integer indices.

Multiplying powers

How could you write 23×242^3\times2^4 as one power without calculating either factor? Count the copies of two: the first power supplies three and the second supplies four, so altogether there are seven.

(2×2×2)×(2×2×2×2)=27(2\times2\times2)\times(2\times2\times2\times2)=2^7

The base is the repeated factor and the index tells you how it is used. For a positive integer index, it counts the factors. A letter in a rule stands for a number: aa is the base, while mm and nn stand for indices.

Multiplying powers with the same base

am×an=am+na^m\times a^n=a^{m+n}

For any nonzero base aa, this rule holds for integer indices mm and nn. It also holds for a zero base when both indices are positive. The bases must be the same.

Not on the 2027 formula sheets: learn it.

Negative indices represent reciprocals, so cancellation explains the same rule. For example, 23×2−1=2×2×22=222^3\times2^{-1}=\frac{2\times2\times2}{2}=2^2. Adding the indices gives 3+(−1)=23+(-1)=2, exactly as the cancellation predicts.

Worked example 1: joining factors

Question. Write 23×242^3\times2^4 as a single power of two.

Plan. Keep the common base and add the indices.

23×24=23+4count all factors of 2=27add the indices\begin{aligned} 2^3\times2^4&=2^{3+4} && \text{count all factors of 2} \\ &=2^7 && \text{add the indices} \end{aligned}

Answer. 272^7.

Check. 8×16=1288\times16=128, and 27=1282^7=128.

Change only the sign of the second index. Its factors now belong in a denominator.

Worked example 2: multiplying by a reciprocal power

Question. Write 23×2−42^3\times2^{-4} as a single power of two, then find its value.

Plan. Add the signed indices, then interpret the negative result.

23×2−4=23+(−4)same-base product=2−1add signed indices=12negative index means reciprocal\begin{aligned} 2^3\times2^{-4}&=2^{3+(-4)} && \text{same-base product} \\ &=2^{-1} && \text{add signed indices} \\ &=\frac12 && \text{negative index means reciprocal} \end{aligned}

Answer. 2−1=122^{-1}=\frac12.

Check. Directly, 8×116=816=128\times\frac1{16}=\frac8{16}=\frac12.

A calculation may ask for a single power rather than its numerical value. Match that requested form.

Watch out

23×24≠2122^3\times2^4\ne2^{12}: multiplying powers adds the indices. Also, 23+24=8+16=242^3+2^4=8+16=24; an addition between powers does not use this law.

Try the same sign change with a new base.

Dividing powers

Dividing powers with the same nonzero base cancels matching factors. For example, 5452\frac{5^4}{5^2} has four factors of five above the line and two below; cancelling two pairs leaves 525^2.

Dividing powers with the same base

am÷an=am−n(a≠0)a^m\div a^n=a^{m-n}\qquad(a\ne0)

The indices are integers. Subtract the denominator's index from the numerator's index, keeping that order.

Not on the 2027 formula sheets: learn it.

The nonzero condition is essential because the divisor must not be zero. If the indices are equal, am÷am=1a^m\div a^m=1. The index rule also gives am−m=a0a^{m-m}=a^0, explaining why a0=1a^0=1 for a≠0a\ne0. This argument does not define 000^0, because it would require division by zero.

Worked example 3: more factors in the numerator

Question. Write 56÷525^6\div5^2 as one power of five.

Plan. Subtract the index of the divisor from the index of the dividend.

56÷52=56−2cancel two matching factors=54four factors remain above\begin{aligned} 5^6\div5^2&=5^{6-2} && \text{cancel two matching factors} \\ &=5^4 && \text{four factors remain above} \end{aligned}

Answer. 545^4.

Check. Multiply back: 54×52=565^4\times5^2=5^6, the original dividend.

Reverse the two powers. The same factors cancel, but the unmatched factors are now below the fraction line.

Worked example 4: more factors in the denominator

Question. Write 52÷565^2\div5^6 as one power of five and as a fraction.

Plan. Subtract in the written order and use the reciprocal meaning.

52÷56=52−6numerator index minus denominator index=5−4four factors remain below=1625since 54=625\begin{aligned} 5^2\div5^6&=5^{2-6} && \text{numerator index minus denominator index} \\ &=5^{-4} && \text{four factors remain below} \\ &=\frac1{625} && \text{since }5^4=625 \end{aligned}

Answer. 5−4=16255^{-4}=\frac1{625}.

Check. 1625×15625=25\frac1{625}\times15625=25, so multiplying by the original divisor restores 525^2.

The pattern is continuous: dividing by the base reduces an index by one, through positive indices, zero and negative indices. The value at index zero is 11, not 00.

Watch out

Do not change 52−65^{2-6} into 56−25^{6-2} to avoid a negative index. Division order matters; a smaller positive dividend divided by a larger one gives a value between zero and one.

Use the equal-index boundary case.

Raising a power to a power

In (32)3(3^2)^3, the outer index repeats the whole bracket three times: 32×32×323^2\times3^2\times3^2. There are three groups of two factors of three, making six factors altogether. This time the indices multiply.

A power raised to a power

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

For a nonzero base aa, this rule holds for integer indices mm and nn. A zero base is also allowed when both indices are positive. This lesson uses integer indices only.

Not on the 2027 formula sheets: learn it.

For a positive outer index, add the inner index once for each bracket. A negative outer index takes a reciprocal: (23)−2=1(23)2=126=2−6(2^3)^{-2}=\frac1{(2^3)^2}=\frac1{2^6}=2^{-6}. That agrees with multiplying 33 by −2-2.

Worked example 5: groups of factors

Question. Write (32)3(3^2)^3 as one power of three.

Plan. Multiply the inner and outer indices.

(32)3=32×3three groups of two factors=36six factors altogether\begin{aligned} (3^2)^3&=3^{2\times3} && \text{three groups of two factors} \\ &=3^6 && \text{six factors altogether} \end{aligned}

Answer. 363^6.

Check. 93=7299^3=729, and 36=7293^6=729.

Now change only the sign of the inner index. Each repeated bracket is a reciprocal.

Worked example 6: a reciprocal raised to a power

Question. Write (3−2)3(3^{-2})^3 as one power of three and as a fraction.

Plan. Multiply the signed indices, then evaluate the reciprocal.

(3−2)3=3(−2)×3multiply indices=3−6negative times positive=1729take the reciprocal of 36\begin{aligned} (3^{-2})^3&=3^{(-2)\times3} && \text{multiply indices} \\ &=3^{-6} && \text{negative times positive} \\ &=\frac1{729} && \text{take the reciprocal of }3^6 \end{aligned}

Answer. 3−6=17293^{-6}=\frac1{729}.

Check. Each bracket is 19\frac19, and 19×19×19=1729\frac19\times\frac19\times\frac19=\frac1{729}.

The opening shortcut is now explained: counting factors, including cancelled factors, lets you handle large powers without writing large numbers. Later it will make standard-form calculations much shorter.

Watch out

Distinguish 32×33=353^2\times3^3=3^5 from (32)3=36(3^2)^3=3^6. A product of powers adds indices; a power of a power multiplies them.

Keep the negative sign attached to the inner index.

Quick Recap

  • With the same base, multiplication adds indices and division subtracts them.
  • Raising a power to a power multiplies indices.
  • For nonzero bases, index zero gives 11 and a negative index gives a reciprocal.
  • The rules describe products, quotients and powers; they do not combine an ordinary sum of powers.

You can simplify the opening product immediately: 23×24=272^3\times2^4=2^7, because it contains seven factors of two.