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: