Law of exponents for real numbers — lesson. Mathematics State Board, Class 9.

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In the previous grade, we have learnt about exponent. Let us recall them.

We can write the number \(729\) as \(9 \times 9 \times 9 = 9^3\). Here, the number \(9\) is the base and \(3\) is the exponent. The exponent is also called as index.

Here, we have found the value of \(9^3\). Similarly, we can find the value of \(9^{-3}\) which is the multiplicative inverse of \(9^3\). That is, \(9^3 \times 9^{-3} = 9^{3 - 3} = 9^0 = 1\).

Hence, we can write \(9^{-3}\) as \(9^{-3} = \frac{1}{9^{3}}\)

In general, we can say that \(x^{-n} = \frac{1}{x^n}\)

Consider the numbers with some powers like

a^{m}, b^{n}, ...

Here \(a\) and \(b\) the base and \(m\) and \(n\) are its respective exponents.

The law of exponents are as follows:

\begin{array}{l} (i) a^{m} \cdot a^{n} = a^{m + n} \\ (ii) {(a^{m})}^{n} = a^{mn} \\ (iii) \frac{a^{m}}{a^{n}} = {(a)}^{m - n}, where m > n . \\ (iv) a^{m} b^{m} = {(ab)}^{m} \end{array}

Example:

1. Consider

3^{5} \cdot 3^{- 6}

Here the base(3) is the same but the exponents \(5\) and \(-6\) are different.

Comparing the property

a^{m} \cdot a^{n} = a^{m + n}

with the expression,

3^{5} \cdot 3^{- 6} = 3^{5 - 6} = 3^{- 1} = \frac{1}{3}

2. Let us take another expression

{(9^{3})}^{5}

Here the base (9) is the same, but the exponents \(3\) and \(5\) are different.

Comparing the property

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

with the expression,

{(9^{3})}^{5} = 9^{3 \times 5} = 9^{15}

3. Let the next expression be

\frac{2^{7}}{4^{3}}

Here the base values are \(2\) and \(4 = 2^2\), and the exponents are \(7\) and \(3\).

\frac{2^{7}}{4^{3}} = \frac{2^{7}}{{(2^{2})}^{3}}

Applying the property

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

\frac{2^{7}}{{(2^{2})}^{3}} = \frac{2^{7}}{2^{6}}

Now applying

\frac{a^{m}}{a^{n}} = {(a)}^{m - n}, where m > n .

\frac{2^{7}}{2^{6}} = 2^{7 - 6} = 2^{1} = 2

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1. Law of exponents for real numbers

Theory: