Consider the exponential term \({\left( { - 2} \right)^{\frac{1}{2}}}\). This is an entity which when squared should generate\(- 2\). But we know that whenever we *square any real number*, we always *end up with a non-negative real number*. Thus, there can be no real number which when squared can generate\(- 2\). This means that this exponential term is a *non-real* quantity. In fact, it is a *complex number*.

Now, consider the exponential term \({\left( { - 8} \right)^{\frac{1}{3}}}\). This is an entity which when cubed should generate \( - 8\). It turns out that this is a real number, \(- 2\), because\({\left( { - 8} \right)^{\frac{1}{3}}} = - 2\).

From these two simple examples, we can conclude that if the base is negative, the exponential term may or may not be defined. Can you figure out the cases in which it will be defined?

To keep matters simple, therefore, we restrict the base to have only positive values. We even exclude 0 from the possible base values. Can you understand why? Think what would happen if we were to raise 0 to a negative integer power.

From now on, whenever we consider an exponential term of the form \({b^p}\), we will assume that the base *b* is a *positive real number*, while the exponent *p* is an *arbitrary real number*.