INDICES IN MATHEMATICS

INDICES IN MATHEMATICS

Indices in mathematics are an important topic. Like the standard form, indices are key in mathematics. As a student, knowing this topic and the laws guiding its solution will develop your ability in solving complex mathematics problems. This article “Indices in mathematics” will introduce the meaning of indices, laws of indices with examples, and lots more.

INDICES MEANING

The plural of “power” and “index” are “power” and “indices” respectively. Since we treat several cases of powers or indices, we title this article “INDICES” and not “index”. Indices can apply to integers, fractions, positive or negative, or mixed numbers. Indices can be integral or fractional, positive or negative. We will treat other kinds of indices later in this article.

We use index form usually with 10 as the base and for very large numbers or very small numbers, for example, we write 10^{8} instead of 100,000,000. Indices play an important role in logarithm and vice-versa because the two concepts are related since the characteristics or integer of logarithms can be compared with the indices.

For example, in the notation 3^{2}, 2 is the Power or Index while 3 is called the Base. In mathematics, the words “Power” and “index” are used interchangeably, i.e. they are synonymous, they mean the same thing. So for 3^{2}, we say 3 raised to the power of 2 or 3 index 2.

Let us look at the various laws of indices, that way, we can be able to see principles that will guide us as we solve problems on indices.

LAWS OF INDICES

In order to solve problems involving indices more easily, the following laws governing the operations with indices are provided and are true for any numbers a,x and y.

  • Product (Multipication), a^{x}\times a^{y}

By definition, a^{x}\times a^{y}= a^{x+y}. Therefore, for multiplication, we add the indices.

Example: b^{3}\times b^{6}

= (b\times b\times b)\times (b\times b\times b\times b\times b\times b)

b^{3+6}=b^{9}

  • Division (Quotient), a^{x} ÷ a^{y}

By definition, a^{x} ÷ a^{y}=\frac{a\times a\times a\times a....to, x, places}{a\times a\times a\times .... to, y, places} = a^{x-y}. Therefore for division, we subtract the indices.

Note that if x> y, (x-y) will be positive, and if x< y, (x-y) will be negative. Finally, if x=y, (x-y) will be 0, and then we will have a^{0}=1.

Example: Simplify 6^{2n} ÷ 6^{n}.

= 6^{2n} ÷ 6^{n}=6^{2n-n}= 6^{n}

  • Power laws, (a^{x})^{y}

By definition, (a^{x})^{y}=a^{x}\times a^{x}\times a^{x}.... to y places. Therefore, (a^{x})^{y}=a^{xy}. So for power indices, we multiply the indices.

Example: Simplify (2^{4})^{2}

From the example above, (2^{4})^{2}=2^{4\times 2}=2^{8}

  • Fractional Indices (or Roots), a^{\frac{1}{x}}

We define the notation 2^{\frac{1}{2}} as \sqrt{2} , that is the square root of 2 ; 2^{\frac{1}{3}} as \sqrt[3]{2} that is the cube root of 2 ; 254^{\frac{1}{4}} as \sqrt[4]{254} , that is the fourth root of 254 and so on. In general,a^{\frac{1}{x}} = \sqrt[x]{a}

Which is the x^{th} root of a. Where x and a are real numbers and a is positive. In otherwords, fractional indices involves roots.

Example 1: Simplify 81^{\frac{1}{4}}

The above example will give us, 81^{\frac{1}{4}}=\sqrt[4]{81}

\sqrt[4]{81}=\sqrt[4]{3\times 3\times 3\times 3}

= (\sqrt[4]{3})^{4} ( from this value, 4 will cut the fourth root)

= 3

Or we can solve the above simply making 81 have the same power with the given root (which in this case is 4).

81^{\frac{1}{4}}=(3^{4})^{\frac{1}{4}}(3^{4})^{\frac{1}{4}}=3^{4\times \frac{1}{4}}= 3

Example 2: Simplify 1728^{\frac{1}{3}}

From the above, 1728^{\frac{1}{3}}=\sqrt[3]{1728}

\sqrt[3]{1728}=\sqrt[3]{(2\times 2\times 2)\times (2\times 2\times 2)\times (3\times 3\times 3)}

= \sqrt[3]{2^{3}\times 2^{3}\times 3^{3}}

\sqrt[3]{(2\times 2\times 3)^{3}}=\sqrt[3]{12^{3}}

= 12

Hence, for roots, we divide the indices, while for powers we multiply the indices. So we can say that a^{\frac{x}{y}}=\sqrt[y]{a^{x}} i.e. the y^{th} root of a^{x}. In fractional indices, the numerator is regarded as a power while the denominator is regarded as the root.

  • Zero Index Law, a^{0}

We know that a^{x} ÷ a^{x}=1, since any number dividing itself has the result as 1. But from the division law 2 above, a^{x} ÷ a^{x}=a^{x-x}=a^{0}. Equating the two results, we see that a^{0}=1 (i.e. any number a raised to zero power is equal to 1).

For example, 5^{0}=1, 100^{0}=1,-2^{0}=1 etc.

  • Negative Indices law, (a^{-x})

This law states that a^{-x}=\frac{1}{a^{x}}.

Proof: Consider that, a^{0} ÷ a^{x}=a^{0-x}=a^{-x}

But a^{0} ÷ a^{x}=\frac{a^{0}}{a^{x}}=1 ÷ a^{-x}=\frac{1}{a^{x}} . Therefore, a^{-x}=\frac{1}{a^{x}}.

Example 1: Simplify 4^{-4}

From the example above, \frac{1}{4^{4}}=\frac{1}{4\times 4\times 4\times 4}

=\frac{1}{256}Example 2: Simplify 3(2^{-2})

From the example above, 3(2^{-2})=3(\frac{1}{2^{2}})

3(\frac{1}{2^{2}})=3(\frac{1}{4})=\frac{3}{4}

Therefore a number raised to a negative power or index is equal to the reciprocal of that number raised to that positive index or power.

INDICES EXAMPLES

Let us consider some series of examples that involves the combination of all the laws we have discussed so far.

Example 1: Simplify 64^{\frac{5}{6}}

In solving the above question, we will use two methods. You can choose the most easiest out of the two methods.

From fractional indices, 64^{\frac{5}{6}}=\sqrt[6]{64^{5}}

\sqrt[6]{64^{5}}=\sqrt[6]{(2^{6})^{5}}=\sqrt[6]{2^{(6\times 5)}}

= 2^{\frac{6\times 5}{6}}=2^{5}

=32

Alternatively, 64^{\frac{5}{6}}=(2^{6})^{\frac{5}{6}}

=(2)^{6\times \frac{5}{6}}=2^{5}=32

Example 2: Simplify 40a^{4}b^{6}c^{3} ÷ -8a^{4}b^{3}c^{2}

Using the 2nd law we discussed above, the division law, we can say that  \frac{40}{-8}a^{4-4}\times b^{6-3}\times c^{3-2}

=-5\times a^{0}\times b^{3}\times c^{1}

=-5\times 1\times b^{3}\times c^{1}=-5b^{3}c^{1}

Example 3: Simplify (5^{-\frac{1}{2}})^{4}

From the above, (5^{-\frac{1}{2}})^{4}=5^{-\frac{1}{2}\times 4}

5^{-\frac{1}{2}\times 4}=5^{-2}=\frac{1}{5^{2}}
=\frac{1}{25}
INDICIAL EQUATIONS

Let us consider some examples of indicial equations which involves the application of the laws of indices.

Solve the following equations:

  1. 3^{3x}=81^{\frac{3}{4}}
  2. 4^{x}=\frac{1}{64}
  • From the first example, 3^{3x}=81^{\frac{3}{4}}.

Recall the fractional law of indices, 81^{\frac{3}{4}}=(3^{4})^{\frac{3}{4}}=3^{3}

So 3^{3x}=3^{3} ( since they have same base, the 3 will clear each other)

∴ 3x=3

x=\frac{3}{3}

Therefore, x=1

  • For the second example, 4^{x}=\frac{1}{64}

Applying the laws of indices, \frac{1}{64}=\frac{1}{2^{6}}=2^{-6}

So 4^{x}=2^{-6} ( making both sides have the same base)

We get (2^{2})^{x}=2^{-6}

2^{2x}=2^{-6}

∴ 2x=-6 and x=\frac{-6}{2}

x=-3

Solving indices can be very interesting, when you know the various laws and how to apply them. Click HERE to download Indices Worksheet containing questions on indices. From the above examples, We are convinced that you can solve numerous questions on indices.

Download: Mathematics Textbooks with Past Questions and Answers. Also watch and learn from Mathematics videos in our YouTube Channel “NEW TRACK MATHEMATICS VIDEOS.

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September 2, 2021
MATH ARTICLES

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