Learning Lab

Index Laws

Remember the basic index laws:

1. \(\quad\quad\) \(a^{m}\times a^{n}=a^{m+n}\)

2. \(\quad\quad\) \(a^{m}\div a^{n}=a^{m-n}\)

3. \(\quad\quad\) \(\left(a^{m}\right)^{n}=a^{mn}\)

4. \(\quad\quad\) \(a^{0}=1\)

5. \(\quad\quad\) \(a^{-n}=\frac{1}{a^{n}}\)

These hold when \(m\) and \(n\) are whole numbers or fractions. In what follows, we will refer to these as the first index law, the second index law etc.

Fractional Indices

What meaning can we give to \(3^{1/2}.\) Using the first index law, we find: \[\begin{align*} 3^{1/2}\times3^{1/2} & =3^{1/2+1/2}\\ & =3^{1}\\ & =3. \end{align*}\] So \(3^{1/2}\) is the number that when multiplied by itself, gives three. But the square root of \(3\) is the number that when multiplied by itself gives \(3.\) That is: \[\begin{align*} \sqrt{3}\times\sqrt{3} & =3. \end{align*}\] Since \(3^{1/2}\) behaves like \(\sqrt{3}\) we say \[ 3^{1/2}=\sqrt{3}. \]

Consider now \(2^{1/3}\). Using the first index law, we can write: \[\begin{align*} 2^{1/3}\times2^{1/3}\times2^{1/3} & =2^{1/3+1/3+1/3}\\ & =2^{1}\\ & =2. \end{align*}\] So \(2^{1/3}\) behave like the cube root of \(2.\) That is, \[\begin{align*} 2^{1/3} & =\sqrt[3]{2} \end{align*}\]

In general we say that \(a^{1/n}\) is the \(n^{th}\) root of \(a.\) That is \[ a^{1/n}=\sqrt[n]{a} \] for any positive integer \(n\) (whole numbers like \(1,2,\ldots\) ).

Exercise 1

Evaluate the following expressions. If the answer is not exact, use a calculator to give a decimal approximation to two decimal places.

a)\(\;64^{1/2}\) \(\quad\quad\) b) \(\;125^{1/3}\) \(\quad\quad\)c) \(\;36^{1/4}\) \(\quad\quad\)d) \(\;81^{-1/2}\) \(\quad\quad\)e) \(\;128^{-1/7}\) \(\quad\quad\)f) \(\;250^{1/5}\)

More Complicated Fractional Indices

In this section, we introduce terms like \(a^{2/3}\).

Using the third index law11 The third law is: \(\left(a^{m}\right)^{n}=a^{mn}\) we have: \[\begin{align*} a^{2/3} & =\left(a^{2}\right)^{1/3}\\ & =\sqrt[3]{a^{2}}\quad\textrm{using the definition of indices as roots.} \end{align*}\]

Again using the third index law:22 We use the commutative law: \[\begin{align*} mn & =m\times n\\ & =n\times m\\ & =nm. \end{align*}\] In our case, we are using the fact that \(2\times\left(1/3\right)=\left(1/3\right)\times2.\) \[\begin{align*} a^{2/3} & =a^{2\times\left(1/3\right)}\\ & =a^{\left(1/3\right)\times2}\\ & =\left(\sqrt[3]{a}\right)^{2}. \end{align*}\]

In general, \[\begin{align*} a^{p/q} & =\sqrt[q]{a^{p}} \end{align*}\] or \[\begin{align*} a^{p/q} & =\left(\sqrt[q]{a}\right)^{p} \end{align*}\] where \(p\) and \(q\) are integers with \(q\neq0.\)

Simplification of Expressions

The index laws may be used to simplify algebraic expressions as shown in the following examples. Note that you don’t have to put in the lines of text explaining each step.

1. Simplify \(\left(x^{3}y^{9}\right)^{2/3}.\)
   Solution:

Open imageThe third index law is: \[ \left(a^{m}\right)^{n}=a^{mn} \]

\[\begin{align*} \left(x^{3}y^{9}\right)^{2/3} & =x^{3\times\left(2/3\right)}y^{9\times\left(2/3\right)}\quad\textrm{using the third law,}\\ & =x^{2}y^{6}. \end{align*}\]

2. Simplify \(3^{1/3}\div3^{4/3}.\)
   Solution:

Open imageThe second index law is: \[ a^{m}\div a^{n}=a^{m-n}. \]

\[\begin{align*} 3^{1/3}\div3^{4/3} & =3^{\left(1/3\right)-\left(4/3\right)}\quad\textrm{using the second index law,}\\ & =3^{-\left(3/3\right)}\\ & =3^{-1}\\ & =\frac{1}{3}. \end{align*}\]

3. Simplify \(32^{3/5}\).
   Solution:

Open imageFor this example, you need to know the fifth root of \(32\) is \(2\) . In symbols \(\sqrt[5]{32}=2.\) That is \(2\times2\times2\times2\times2=32.\)

\[\begin{align*} 32^{3/5} & =\left(\sqrt[5]{32}\right)^{3}\\ & =2^{3}\\ & =8. \end{align*}\]

4. Simplify \(25^{-1/2}.\)
   Solution:

Open imageFor this example you need to know the square root of 25 is 5. In symbols \(\sqrt{25}=5.\) That is \(5\times5=25\). You also need to use index law 5: \[ a^{-n}=\frac{1}{a^{n}}. \]

\[\begin{align*} 25^{-1/2} & =\frac{1}{\sqrt{25}}\quad\textrm{using index law 5,}\\ & =\frac{1}{5} \end{align*}\]

5. Simplify \(\left(a^{2}b^{5}\right)^{1/3}\times a^{1/3}b^{-2/3}.\)
   Solution: \[\begin{align*} \left(a^{2}b^{5}\right)^{1/3}\times a^{1/3}b^{-2/3} & =a^{2/3}b^{5/3}\times a^{1/3}b^{-2/3}\quad\textrm{using the third index law,}\\ & =a^{\left(2/3\right)+\left(1/3\right)}b^{\left(5/3\right)-\left(2/3\right)}\\ & =ab. \end{align*}\]

Please do Exercise 2 for some practice.

Exercise 2

Simplify the following expressions. Give your answers in index notation with positive indices.

a)\(\left(\frac{8}{27}\right)^{2/3}\quad\) b)\(\sqrt{5}\times\sqrt[3]{5}\times\sqrt[6]{5}\quad\) c) \(\sqrt{a}\times\sqrt[4]{a}\times\sqrt[3]{a^{2}}\quad\) d) \(\left(125a^{6}b\right)^{1/3}\times b^{2/3}\quad\) e) \(\frac{\left(2xy^{3}\right)^{1/2}}{2}\times\left(\frac{x^{3/2}}{y^{2}}\right)^{4}\quad\) f) \(2^{5/2}-2^{3/2}\)