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ILS1.1 Indices
July 22, 2021 - 3:34pm — AJ (e67821)
What is index notation? When a number such as 16 is written in the form 42 (which means 4 x 4) we say that it is written as an exponential, or in index notation.
There are laws about multiplying and dividing indices as well as how to deal with negative indices.
This module introduces rules for multiplying and dividing expressions using index notation. For example how to simplify expressions like \(4a^{3}b\times3ab^{5}\) or \(9a^{3}b^{2}c\div3ab^{5}\). We do not consider fractional indices which are covered in a different module. The plural of index is indices.
Here's a video showing how the rules work:
Hi, I’m Martin Lindsay from the Study and learning centre at RMIT University. This is a short movie on indices.
Start by looking at what we mean by a base and an index. Here we have two to the power of three, in other words the two is the base, the three is the index or the power. Notice also that students tend to confuse this with the same as two times three, which it is not. Two to the power of three means two multiplied by itself three times, in other words two times two times two, which gives us an answer of eight.
Let’s look at some examples; X to the power of four, well X is the base, four is the index, so it’s X multiplied by itself four times, X times X times X times X. Here we have A squared times B to the power of five. Notice here that there are two different bases, A and B, so multiplying them out, A square is A times A, and B to the power of five is B times B times B times B times B. Finally we have four times M cubed times N cubed, so we leave the four alone and we multiply out the M three times, M times M times M, and the N is multiplied out three times as well, N times N times N, again the bases are different.
We’ll now go through some laws of indices. Here’s the first law of indices which says that X to the power of A times X to the power of B is the same as X to the power of A plus B. Notice here that the base must be the same but also that when we multiply terms with indices we actually add the indices.
Let’s do an example to illustrate this. We have A square times A to the five, so using the law that’s the same as A to the two plus five. Notice here the base is the same so we can add the indices two plus five, so the answer is A to the seven. Similarly with numbers, three square times three cubed, the base are the same which means that we can add the indices, three to the two plus three which is three to the five, three to the five gives us an answer of 243.
And, finally, here’s a problem where we have a number and a fractional index, four times X to the half times X to the half. So we leave the four alone, the base is the same, so we can add the indices, a half plus a half which is one, so we have four X to the one. Notice another important thing here is that X to the power of one is exactly the same as just X so we can leave our answer as four X.
Now let’s look at the second law of indices; X to the A divided by X to the B is the same as X to the A minus B. A couple of things to notice here, the bases must be the same and when the bases are the same and we’re dividing terms we subtract the indices, so for example A to the five divided by A to the two, the base are the same therefore we can subtract the indices, five take away two is A cubed. Similarly with numbers, two to the seven divided by two to the four, the bases are the same therefore we can subtract the indices, two to the seven minus four is two to the three, which is two times two times two, which gives us an answer of eight.
And finally, here are two numbers, 10 and five, 10 to the X to 11 divided by five X to the six. Here we can divide the numbers, 10 divided by five gives us two, that’s one part of this problem. The other part is dividing the terms with indices, X to the 11 divided by X to the six, the bases are the same therefore we can subtract the indices, so it’s X to the 11 minus six which gives us five. In other words our answer is two X to the five.
The third law of indices says that X to the power of zero is equal to one. Now notice that any base with a power of zero will always give us an answer of one. Let’s look at the reason behind this. Let’s look at this problem, A cubed divided by A cubed, if I do this the long way I’ve got the fraction as shown, A times A times A on the top, A times A times A on the bottom, notice they all cancel giving us an answer of one. But there’s a quicker way of doing this using this law; A cubed divided by A cubed, same base therefore we can subtract the indices, three minus three is zero, that gives us A to the power of zero and the third law says that anything with a power of zero to it gives us an answer of one.
Now let’s look at the fourth law of indices. Notice here that we have brackets with an index inside and an index outside. So when we have this situation we multiply the indices, we don’t add them, we multiply them, so here X to the A all inside brackets with an index of B outside of the brackets gives us an answer of X to the A times B. So for example four cubed in brackets to the power of two would give us four times four times four multiplied by four times four times four, this is the long way of doing this, which gives us an answer of four to the six. But if we use the law we multiply out the indices, in other words four cubed to the power of two is four to the three times two which is four to the six, as you can see a much quicker way of working out this problem.
Notice also when we have a fraction as a base we do exactly the same as before. In other words the numerator is four, the denominator is five, so they both get indexed, in other words four cubed divided by five cubed and that gives us four times four times four which is 64 divided by five times five times five which is 125, so remember, with a fraction you use both the numerator and the denominator will have the base to it.
Finally, let’s look at square roots and cubed roots as indices. Notice a couple of things here; that the index inside the root is the numerator of our new index and the index outside the root is the denominator of our new index. Let’s illustrate this with an example, here we have the square root of X which is the same as X to the power of one inside the square roots and we could put a two outside the square root as well and using the rule above that will give us an answer of X to the one over two, or X to the half, so the square root of X is the same as X to the half.
Similarly here we have the cubed root of X squared, so the square is on the top of our new index, the cubed root is on the bottom of our new index, so that gives us an answer of X to the two over three or X to the two thirds.
Now try some problems for yourself. The answers to these questions are on the next slide. Thanks for watching this short movie.
Index Notation
Consider the following examples:
\[\begin{align*} 3^{4} & =3\times3\times3\times3=81\\ 5^{3} & =5\times5\times5=125\\ 2^{7} & =2\times2\times2\times2\times2\times2\times2=128 \end{align*}\]
In general: \[ a^{n}=\underbrace{a\times a\times a\times\ldots\times a}_{n\textrm{ factors}} \] The letter \(n\) in \(a^{n}\) is referred to in one of three ways:
\(n\) is the index in \(a^{n}\) with \(a\) known as the base.
\(n\) is the exponent or power to which the base \(a\) is raised.
\(n\) is the logarithm, with \(a\) as the base. (see the Logarithms module)
When a number such as \(125\) is written in the form \(5^{3}\) we say it is written as an exponential or in index notation. Multiplication and division of numbers or expressions written in index notation is achieved using index laws.
Index Laws
This section states and gives examples of universal index laws.
First Index Law
To multiply index expressions you add the indices. For example: \[\begin{align*} 2^{3}\times2^{2} & =\left(2\times2\times2\right)\times\left(2\times2\right)\\ & =2\times2\times2\times2\times2\\ & =2^{5} \end{align*}\] Therefore \(2^{3}\times2^{2}=2^{3+2}=2^{5}\). In general:
First Index Law: \[ a^{m}\times a^{n}=a^{m+n} \]
Second Index Law
To divide expressions subtract the indices. For example: \[\begin{align*} 3^{5}\div3^{3} & =\frac{3^{5}}{3^{3}}\\ & =\frac{3\times3\times3\times3\times3}{3\times3\times3}\\ & =\frac{3\times3}{1}\quad\textrm{cancelling three lots of 3}\\ & =3^{2} \end{align*}\] Therefore \(3^{5}\div3^{3}=\frac{3^{5}}{3^{3}}=3^{5-3}=3^{2}.\) In general:
Second Index Law: \[ a^{m}\div a^{n}=\frac{a^{m}}{a^{n}}=a^{m-n},\quad a\neq0 \]
Note that expressions in index form can only be multiplied or divided if they have the same base.
Third Index Law
To raise an expression in index form to a power, multiply the indices. For example: \[\begin{align*} \left(5^{2}\right)^{3} & =5^{2}\times5^{2}\times5^{2}\\ & =5^{2+2+2}\quad\textrm{using the first index law}\\ & =5^{6} \end{align*}\] Therefore \(\left(5^{2}\right)^{3}=5^{2\times3}=5^{6}.\) In general:
Third Index Law: \[ \left(a^{m}\right)^{n}=a^{m\times n} \]
This also leads to the expression: \[\begin{align*} (a^{m}b^{p})^{n} & =a^{mn}b^{pn} \end{align*}\]
Be careful as this is true for multiplication and division only , not addition or subtraction , so that \((a+b)^{n}\neq a^{n}+b^{n}\)
Examples of Index Laws
Simplify \(x^{5}\times x^{6}\).
Solution: \[\begin{align*} x^{5}\times x^{6} & =x^{5+6}\quad\textrm{by the first law}\\ & =x^{11} \end{align*}\]Simplify \(a^{5}\div a^{3}.\)
Solution: \[\begin{align*} a^{5}\div a^{3} & =\frac{a^{5}}{a^{3}}=a^{5-3}\quad\textrm{by the second law}\\ & =a^{2} \end{align*}\]Simplify \(\left(c^{3}\right)^{4}\).
Solution: \[\begin{align*} \left(c^{3}\right)^{4} & =c^{3\times4}\quad\textrm{by the third law}\\ & =c^{12} \end{align*}\]Simplify \(\left(2x^{2}\right)^{3}\).
Solution: \[\begin{align*} \left(2x^{2}\right)^{3} & =2^{3}\left(x^{2}\right)^{3}\\ & =8x^{2\times3}\quad\textrm{by the third law}\\ & =8x^{6} \end{align*}\]
Note that terms with different bases must be considered seperately when using the index laws , such as \((2a^{3}b^{2})^{4}=2^{4}a^{12}b^{8}\)
Exercise 1 provides practice for these laws.
Zero Index
So far we have only considered expressions in which each index is a positive whole number1 Whole numbers are called integers and positive whole numbers are called the positive integers.. The index laws also apply if the index is zero, negative or a fraction (fractional indices will be dealt with in another module).
Consider \(2^{3}\div2^{3}=\frac{2^{3}}{2^{3}}=\frac{2\times2\times2}{2\times2\times2}=\frac{8}{8}=8\div8=1\). Using the second law, \(2^{3}\div2^{3}=2^{3-3}=2^{0}\)
therefore \(1=2^{0}\). In general any expression with a zero index is equal to 1. Also note that \(0^{0}\) is ambiguous and so we don’t allow \(a=0\) in this law.
Zero law of indices: \[ a^{0}=1,\qquad a\neq0 \]
Examples of the Zero Index Law
\[\begin{array}{lllllll} 7^{0}=1 & & \left(xy\right)^{0}=1 & & \left(\frac{1}{2}\right)^{0}=1 & & \left(28x^{2}\right)^{0}=1\\ \end{array}\]
Negative Indices
Consider \(2^{0}\div2^{4}\). \[\begin{align*} 2^{0}\div2^{4} & =\frac{2^{0}}{2^{4}}\quad\textrm{remember that $2^{0}=1$ }\\ & =\frac{1}{2^{4}}. \end{align*}\] But \[\begin{align*} 2^{0}\div2^{4} & =2^{0-4}\quad\textrm{using the second law}\\ & =2^{-4}. \end{align*}\] So \(2^{0}\div2^{4}=2^{-4}\) and \(2^{0}\div2^{4}=\frac{1}{2^{4}}\)
therefore \(2^{-4}=\frac{1}{2^{4}}\). In general2 Note that \(1/0\) is undefined and so \(a\neq0\) in the law below.,
\[ a^{-n}=\frac{1}{a^{n}}\textrm{ which also leads to $\frac{1}{a^{-n}}=a^{n},\qquad a\neq0$ } \]
Examples of Negative Indices
\[\begin{array}{lllll} 1. \ \ 2^{-3}=\frac{1}{8} & & 2. \ \ \frac{1}{x}=x^{-1} & & 3. \ \ 2y^{-1}=\frac{2}{y}\\ & & & & \\ 4. \ \ \frac{1}{3x^{-2}}=\frac{x^{2}}{3} & & 5. \ \ \frac{1}{\left(-2a\right)^{-3}}=\left(-2a\right)^{3} & & 6. \ \ 5ab^{-4}=\frac{5a}{b^{4}}\\ \end{array}\]
Exercise 2 provides practice on zero and negative indices.
Summary of Index Laws
The following laws should be remembered.
Summary of Index Laws
\(\ a^{m}\times a^{n}=a^{m+n}\)
\(\ a^{m}\div a^{n}=a^{m-n}\) , \(\qquad a\neq0\)
\(\ \left(a^{m}\right)^{n}=a^{mn}\)
\(\ a^{0}=1\) , \(\qquad a\neq0\)
\(\ a^{-n}=\frac{1}{a^{n}}\) , \(\qquad a\neq0\)
Combing of Index Laws
Index laws may be used to simplify complex expressions.
Examples
Simplify \(\left(4a^{2}b\right)^{3}\div b^{2}.\)
Solution: \[\begin{align*} \left(4a^{2}b\right)^{3}\div b^{2} & =\left(4^{3}a^{6}b^{3}\right)\div b^{2}\quad\textrm{using law 3}\\ & =4^{3}a^{6}b^{1}\quad\textrm{using law 2}\\ & =4^{3}a^{6}b\\ & =64a^{6}b \end{align*}\]Simplify \(\left(\frac{3a^{3}b}{c^{2}}\right)^{2}\div\left(\frac{ab}{3c^{-2}}\right)^{-3}\)
remember that \(a\div\frac{b}{c}=a\times\frac{c}{b}\)
Solution: \[\begin{align*} \left(\frac{3a^{3}b}{c^{2}}\right)^{2}\div\left(\frac{ab}{3c^{-2}}\right)^{-3} & =\frac{3^{2}a^{6}b^{2}}{c^{4}}\div\frac{a^{-3}b^{-3}}{3^{-3}c^{6}}\quad\textrm{by law 3}\\ & =\frac{3^{2}a^{6}b^{2}}{c^{4}}\times\frac{3^{-3}c^{6}}{a^{-3}b^{-3}}\quad\textrm{inverting the last term and multiplying}\\ & =\frac{3^{2-3}a^{6}b^{2}c^{6}}{c^{4}a^{-3}b^{-3}}\quad\textrm{by law 1}\\ & =3^{-1}a^{6-\left(-3\right)}b^{2-\left(-3\right)}c^{6-4}\quad\textrm{by law 2}\\ & =3^{-1}a^{9}b^{5}c^{2}\quad\textrm{simplifying}\\ & =\frac{a^{9}b^{5}c^{2}}{3}\quad\textrm{by negative index law} \end{align*}\]
- Write \(x^{-1}+x^{2}\) as a single fraction.
Solution: \[\begin{align*} x^{-1}+x^{2} & =\frac{1}{x}+x^{2}\quad\textrm{by negative index law}\\ & =\frac{1}{x}+\frac{xx^{2}}{x}\\ & =\frac{1+xx^{2}}{x}\quad\textrm{using a common denominator}\\ & =\frac{1+x^{3}}{x}\quad\textrm{using law 1} \end{align*}\]
Exercise 1
Simplify the following:
\[\begin{array}{lllll} a). \ c^{5}\times c^{3}\times c^{7} & & b). \ 3\times2^{2}\times2^{3} & & c). \ a^{3}\times a^{2}b^{3}\times ab^{4}\\ & & & & \\ d). \ 3^{6}\div3^{4} & & e). \ a^{8}\div a^{3} & & f). \ x^{4}y^{6}\div x^{2}y^{3}\\ & & & & \\ g). \ \left(x^{3}\right)^{4} & & h). \ \left(x^{m}y^{n}\right)^{5} & & \\ \end{array}\]
\[\begin{array}{llll} a)\:c^{15} & b)\;3\times2^{5}=96 & c)\;a^{6}b^{7} & d)\;3^{2}=9\\ e)\;a^{5} & f)\;x^{2}y^{3} & g)x^{12} & h)\;x^{5m}y^{5n} \end{array}\]
Exercise 2
Write with positive indices and evaluate if possible:
\[\begin{array}{lllllll} a). \ x^{-6} & & b). \ 250^{0} & & c). \ 3ab^{-5} & & d). \ \left(pq\right)^{-2}\\ & & & & & & \\ e). \ \left(5xy\right)^{-3} & & f). \ \frac{2y}{z^{-5}} & & g). \ 2^{-5} & & h).\ (-2)^{-3}\\ & & & & & & \\ i).\ -(3^{-2}) & & j).\ 2\times(-5)^{-2} & & & & \\ \end{array}\]
\[\begin{array}{lllll} a)\;\frac{1}{x^{6}} & b)\;1 & c)\;\frac{3a}{b^{5}}. & d)\;\frac{1}{(pq)^{2}} & e)\;\frac{1}{(5xy)^{3}}=\frac{1}{125x^{3}y^{3}}\\ \\ f)\;2yz^{5} & g)\;\frac{1}{32} & h)\;-\frac{1}{8} & i)-\frac{1}{9} & j)\;\frac{2}{25} \end{array}\]
Exercise 3
Simplify the following:
\[\begin{array}{lllll} a). \ 2a^{3}b^{2}\times a^{-1}\times b^{3} & & b). \ (5x^{-2}y)^{-3} & & c). \ (3x^{3}y^{-1})^{5}\\ & & & & \\ d). \ (a^{-4}b^{-5})^{-2} & & e). \frac{a^{2}b^{3}c^{-4}}{a^{4}bc^{5}} & & f). \frac{a^{7}\times a^{8}\times a^{3}}{a^{2}\times a^{5}}\\ & & & & \\ g). \ x\left(x-x^{-1}\right) & & h). \ \frac{(2^{4})^{n}}{2^{3}} & & i).\ \frac{15a^{2}b}{3a^{4}b}\times\frac{4a^{5}b^{2}}{5a^{3}b^{4}}\\ & & & & \\ j).\ 2^{4}-2^{3} & & & & \\ \end{array}\]
\[\begin{array}{lllllc} a)\;2a^{2}b^{5} & b)\;\frac{x^{6}}{5^{3}y^{3}} & c)\;\frac{3^{5}x^{15}}{y^{5}} & d)\;a^{8}b^{10} & e)\;\frac{b^{2}}{a^{2}c^{9}}\\ \\ f)\;a^{11} & g)\;x^{2}-1 & h)\;2^{4n-3} & i)\;\frac{4a^{7}b^{3}}{a^{7}b^{5}}=4b^{-2}=\frac{4}{b^{2}} & j)\;2^{3}(2^{1}-2^{0})=2^{3}=8 \end{array}\]
What's next... ILS1.2 Fractional indices