# Powers with negative bases

To calculate the power of a negative number we proceed as before, multiplying by itself so many times as indicated by the exponent. But in this case, it is necessary to bear in mind the hierarchy of the signs in the product:

• The product of two positives is positive. $(+)\cdot (+)=+$

For example: $(+3)\cdot (+3)=3\cdot 3=9=+9$

• The product of a positive (multiplied) by a negative is negative. $(+)\cdot (-)=-$

For example: $(+5)\cdot(-3)=5\cdot(-3)=-15$

and also

$(-9)\cdot (+2)=(-9)\cdot 2=-18$

• The product of two negatives is positive. $(-)\cdot (-) =+$

For example: $(-2)\cdot(-2)=+4$

This way, to calculate powers of a negative base, these rules have to be used.

Let's illustrate this with the following calculations:

$$\begin{array}{rcl}(-3)^3&=&(-3)\cdot (-3)\cdot (-3)=9\cdot (-3)=-27 \\\\ (-5)^2&=&(-5)\cdot (-5)=25 \\\\ (-1)^4&=&(-1)\cdot (-1) \cdot (-1) \cdot (-1)=1 \cdot (-1) \cdot (-1)=1 \cdot 1=1 \\\\ (-1)^3&=& (-1)\cdot (-1)\cdot (-1)=1 \cdot (-1)=-1\end{array}$$

In general, raising a negative number to an even exponent has a positive result. While if it is raised to the power of an odd exponent it will be negative.

When the negative is not the power but the exponent, for instance $8^{-3}$ we have to consider the following fact:

$\displaystyle 8^{-3}=\frac{1}{8^3}$

which, therefore, is the inverse of the power with negative exponent.

For example: $$\displaystyle \begin{array}{rcl} 4^{-2}&=&\frac{1}{4^2} \\\\ (3^7)^{-2}&=&\frac{1}{(3^7)^2} \\\\ 6^{-8}\cdot 6^3&=& 6^3\cdot \frac{1}{6^8}=\frac{6^3}{6^8}\end{array}$$