# Multiplying and Dividing Exponential Terms

Suppose that we multiply two exponential terms with the same base, as follows: \({2^3} \times {2^4}\). The first exponential term gives 3 twos, while the second gives 4 twos, and so we can say that

\[{2^3} \times {2^4} = {2^{\left( {3 + 4} \right)}} = {2^7}\]

Carefully observe how the exponents have added. This is a general behavior of exponential terms: *whenever you multiply two exponential terms with the same base, the exponents add*, because in a sense, the *contributions* of the base from both the terms are adding up. It is easy to see that this property will hold for the multiplication of any number of exponential terms with the same base:

\[{a^{{p_1}}} \times {a^{{p_2}}} \times {a^{{p_3}}} \times ... = {a^{\left\{ {{p_1} + {p_2} + {p_3} + ...} \right\}}}\]

Note that in the expression above, the exponents are all integers (we are yet to see how to interpret non-integer exponents).

Now, suppose that we divide two exponential terms which have the same base: \({3^7} \div {3^4}\). The first exponential term has 7 threes, while the second has 4 threes; dividing the first by the second will *take away* 4 threes from the first term. Alternatively, you can say that you have to contribute 3 more threes to the second term to be able to generate the first term. Thus,

\[{3^7} \div {3^4} = {3^{\left( {7 - 4} \right)}} = {3^3}\]

Generalizing: *whenever you divide two exponential terms with the same base, the exponents subtract*, because in a sense, the divisor *takes away* some bases from the dividend.

Can the exponent be 0? As we have mentioned already, it can be, and any base raised to the power 0 will yield 1. A simple proof (in the following, take *p* to be any integer):

\[\begin{align} & {b^0} = {b^{\left\{ { - p + p} \right\}}} = {b^{ - p}} \times {b^p}\;\;\left( {{\rm{how}}?} \right)\\\;\;\;\; & = \frac{1}{{{b^p}}} \times {b^p} = 1\end{align}\]