Consider the following division:
$$ \large \dfrac{3^2}{3^3} = 3^{2-3} = 3^{-1}$$
Now, if we attempt to calculate the value of this division:
$$ \large \dfrac{3^2}{3^3} = \dfrac{9}{27} = \dfrac{1}{3}$$
From this conclusion, we can say that $3^{-1} = \dfrac{1}{3}$.
This conclusion can be generalised:
$$ \large a^{-1} = \dfrac{1}{a}$$
Example 1
Write $4^{-1}$ in fractional form.
$4^{-1} = \dfrac{1}{4}$
Example 2
Write $\dfrac{1}{x}$ in exponent (index) form.
$\dfrac{1}{x} = x^{-1}$
This rule can be extended for negative exponents (negative indices) other than $-1$.
Consider the expression $\dfrac{a^2}{a^5}$.
Using the exponent laws $\dfrac{a^2}{a^5} = a^{2-5} = a^{-3}$ and simplifying the fraction, we obtain
\( \begin{align} \require{AMSsymbols} \displaystyle \require{cancel}
\dfrac{a^2}{a^5} &= \dfrac{\bcancel{a} \times \bcancel{a}}{\bcancel{a} \times \bcancel{a} \times a \times a \times a} = \dfrac{1}{a^3}.
\end{align} \)
Thus $a^{-3} = \dfrac{1}{a^3}$.
In general, we can say:
$$ \large a^{-n} = \dfrac{1}{a^n}$$
Example 3
Write $5^{-2}$ in fractional form.
$5^{-2} = \dfrac{1}{5^2} = \dfrac{1}{25}$
Example 4
Write $\dfrac{1}{a^3}$ using negative exponents (negative indices).
$\dfrac{1}{a^3} = a^{-3}$
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