Law of exponents for real numbers — lesson. Maths CBSE Mentoring, Class 9 Mentoring product.

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 \(4 = 2²\), 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

Did you find an error?

1. Law of exponents for real numbers