# Negative Exponents

## Key Questions

• I suppose you mean the fact that a number to the zero exponent is always equal to one, for example:

${3}^{0} = 1$

The intuitive explanation can be found remembering that:
1) dividing two equal numbers gives 1;
ex. $\frac{4}{4} = 1$
2) The fraction of two equal numbers a to the power of m and n gives:
${a}^{m} / {a}^{n} = {a}^{m - n}$

Now:

• Negative exponents are an extension of the initial exponent concept.

To understand negative exponents ,
first review what we mean by positive (integer) exponents

What do we mean when we write something like:
${n}^{p}$ (for now, assume that $p$ is a positive integer.

One definition would be that
${n}^{p}$ is $1$ multiplied by $n$, $p$ times.

Note that using this definition
${n}^{0}$ is $1$ multiplied by $n$, $0$ times
i.e. ${n}^{0} = 1$ (for any value of $n$)

Suppose you know the value of ${n}^{p}$ for some particular values of $n$ and $p$
but you would like to know the value of ${n}^{q}$ for a value $q$ less than $p$

For example suppose you knew that
${2}^{10} = 1024$ but you wanted to know what ${2}^{9}$ was equal to.
Is there a faster way than multiplying $1$ by $2$, $9$ times?
Yes.
If we note that ${2}^{9} = \frac{{2}^{10}}{2}$
we can simply divide $1024$ by $2$ (giving 512) to obtain ${2}^{9}$

In general if we know that the value of ${n}^{p}$ is $k$
and we want to know the value of ${n}^{q}$ when $q < p$
we can simply divide k by n^(p-q)

With this in mind what is the value of
${n}^{- t}$ ?
We know that ${n}^{0} = 1$
so ${n}^{- t}$ must be $1$ divided by $n$, $\left(0 - \left(- t\right)\right)$ times

That is ${n}^{- t} = \frac{1}{n} ^ t$

As a final example consider the descending powers of 3 in the following, noting that with each line down the result is decreased by dividing the current value by 3

${3}^{4} = 81$
${3}^{3} = 27$
${3}^{2} = 9$
${3}^{1} = 3$
${3}^{0} = 1$
${3}^{- 1} = \frac{1}{3}$
${3}^{- 2} = \frac{1}{9}$
${3}^{- 3} = \frac{1}{27}$

• Raising to the -1 power is equivalent to taking the reciprocal, so we have

${\left(\frac{a}{b}\right)}^{- 1} = \frac{b}{a}$

I hope that this was helpful.

• ${x}^{- n} = \frac{1}{{x}^{n}}$

Maybe you were asking for something more than this (???)

• You can start by rewriting in the following way:

${b}^{- x} = \frac{1}{b} ^ x$

I hope that this was helpful.