Exponents or Indices are used to tell how many times a factor must be multiplied by itself. The factor may be a number (constant) or a variable. Consider $9^{2}$. The factor is 9 and is called as the base and the exponent or the index is 2. It means 9 must be multiplied 2 times 9 × 9.

We can also multiply the factors in an exponential notation. Multiplication of variables or constants with exponents is simple and the process is the same for both numbers and variables. For example,

($2^{2}$)($2^{2}$) = $2 \times 2 \times 2 \times 2$

= $2^{4}$

($x^{2}$)($x^{2}$) = $x \times x \times x \times x$

= $x^{4}$

If m is a positive integer and a in the Real Number Set and a ? 0, then a × a × a × ... m times is $a^{m}$ and a × a × a ×.... n times is $a^{n}$ and the product of $a^{m}$ and $a^{n}$ is

$a^{m}$ × $a^{n}$ = (a × a × a × ... m times) (a × a × a ×.... n times)

$a^{m}$ × $a^{n}$ = $a^{m+n}$

### Multiplication of Indices

To multiply indices, we have to add the exponents!

**Note:** This rule is applicable only when the bases are the same.

We can use this rule for multiplying positive integer exponents, like

$2^{2}$ × $2^{3}$ = $2^{2+3}$= $2^{5}$

We can use the same Product rule to multiply the negative integer exponents

$2^{-2}$ × $2^{3}$ = $2^{-2+3}$ = 2

For any variable x and m, n in the Rational Number set

$x^{m}$ × $x^{n}$ = $x^{m+n}$

The product of $x^{1/2}$ and $x^{3/4}$ is

($x^{1/2}$ )($x^{3/4}$ ) = $x^{(1/2 + 3/4)}$ = $x^{5/2}$

Hence, the rule is applicable to fractional exponents as well! We can also have a combination of constants and variables with exponents.

### Examples on Multiplying Exponents

Given below are some examples that explain the steps to perform multiplication of exponents.

**Example 1:**

Simplify the following algebraic expression with exponents

($x^{3}y^{-2}$)(3x$y^{4}$)

**Solution:**$x^{3}y^{-2}$ = $1x^{3}y^{-2}$

Hence ($x^{3}y^{-2}$)(3x$y^{4}$) = ($1x^{3}y^{-2}$)(3x$y^{4}$)

Multiply the constants and group the x terms and y terms together.

($1x^{3}y^{-2}$)(3x$y^{4}$) = $3(x \times x^{3})(y^{-2} \times y^{4})$

Simplifying using the Product Rule,

$3(x \times x^{3})(y^{-2} \times y^{4})$ = $3x^{1+3} y^{-2+4}$

Simplifying further,

$3x^{1+3} y^{-2+4}$ = $3x^{4}y^{2}$

Hence, $(x^{3}y^{-2})(3xy^{4})$ = $3x^{4}y^{2}$

**Example 2:**

Simplify $(2^{n} \times 8^{2n})16^{3n}$

**Solution:**

We cannot use the product rule of indices to simplify as the bases are not same. But if we observe carefully, it is possible to rewrite each term with base 2.

$8 = 2^{3}$ and $16 = 2^{4}$

$(2^{n} \times 8^{2n})16^{3n}$ = $(2^{n} \times (2^{3^{2n}}))(2^{4^{3n}})$

According to the order of operations, start simplifying the innermost brackets. According to the law of Indices, $a^{m^{n}}$ = $a^{mn}$. Hence $2^{3^{2n}}$ = $2^{3\times2n}$ and $2^{4^{3n}}$ = $2^{4\times3n}$

$(2^{n} \times (2^{3^{2n}}))(2^{4^{3n}})$ = $(2^{n} \times 2^{3\times2n})(2^{4\times3n})$

= $(2^{n} \times 2^{6n})(2^{12n})$

Using the Product Rule, we can simplify further as the bases are the same. The base is 2.

$(2^{n} \times 2^{6n})(2^{12n})$ = $(2^{n+6n})(2^{12n})$

= $(2^{7n})(2^{12n})$

= $2^{7n+12n}$

= $2^{19n}$

Hence, $(2^{n} \times 8^{2n})16^{3n}$ = $2^{19n}$