# Negative Exponents

Let's learn about negative integer exponents.

When we talked about exponents and integers, we assumed that exponent is positive integer.

But what if we want to raise number to negative integer exponent?

Formula for raising number to negative integer power is

$\color{red}{a^{-b}=\frac{1}{a^b}}$

$b$ should be a positive integer.

This formula means that number raised to negative power is reciprocal of the number raised to the same, but positive power.

Let's go through a couple of examples.

Example 1. Find ${{4}}^{{-{3}}}$.

${{4}}^{{-{3}}}=\frac{{1}}{{{{4}}^{{4}}}}=\frac{{1}}{{{4}\cdot{4}\cdot{4}}}=\frac{{1}}{{64}}$.

Answer: $\frac{{1}}{{64}}$.

Next example.

Example 2. Find ${{3}}^{{-{4}}}$.

${{3}}^{{-{4}}}=\frac{{1}}{{{{3}}^{{4}}}}=\frac{{1}}{{{3}\cdot{3}\cdot{3}\cdot{3}}}=\frac{{1}}{{81}}$.

Answer: $\frac{{1}}{{81}}$.

Now, let's see how to deal with negative integers.

Example 3. Find ${{\left(-{3}\right)}}^{{-{2}}}$.

${{\left(-{3}\right)}}^{{-{2}}}=\frac{{1}}{{{{\left(-{3}\right)}}^{{2}}}}=\frac{{1}}{{{\left(-{3}\right)}\cdot{\left(-{3}\right)}}}=\frac{{1}}{{9}}$.

Answer: $\frac{{1}}{{9}}$.

Next example.

Example 4. Find ${{\left(-{5}\right)}}^{{-{3}}}$.

${{\left(-{5}\right)}}^{{-{3}}}=\frac{{1}}{{{{\left(-{5}\right)}}^{{3}}}}=\frac{{1}}{{{\left(-{5}\right)}\cdot{\left(-{5}\right)}\cdot{\left(-{5}\right)}}}=\frac{{1}}{{{25}\cdot{\left(-{5}\right)}}}=\frac{{1}}{{-{125}}}=-\frac{{1}}{{125}}$.

Answer: $-\frac{{1}}{{125}}$.

Last example.

Example 5. Find ${{\left(\frac{{2}}{{5}}\right)}}^{{-{3}}}$.

Wow! Till now we didn't see such examples.

Let's first get read of the negative exponent: ${{\left(\frac{{2}}{{5}}\right)}}^{{-{3}}}=\frac{{1}}{{{{\left(\frac{{2}}{{5}}\right)}}^{{3}}}}$.

Now, recall the definition of exponent (this works for fractions too): ${{\left(\frac{{2}}{{5}}\right)}}^{{3}}=\frac{{2}}{{5}}\cdot\frac{{2}}{{5}}\cdot\frac{{2}}{{5}}=\frac{{{2}\cdot{2}}}{{{5}\cdot{5}}}\cdot\frac{{2}}{{5}}=\frac{{4}}{{25}}\cdot\frac{{2}}{{5}}=\frac{{{4}\cdot{2}}}{{{25}\cdot{5}}}=\frac{{8}}{{125}}$.

We can easily see that fractions are nice things: numerator and denominator can be separately raised to power: ${{\left(\frac{{2}}{{5}}\right)}}^{{3}}=\frac{{{{2}}^{{3}}}}{{{{5}}^{{3}}}}=\frac{{8}}{{125}}$.

Now, we just divide fractions (and write 1 as $\frac{{1}}{{1}}$):

$\frac{{1}}{{{{\left(\frac{{2}}{{5}}\right)}}^{{3}}}}=\frac{{1}}{{\frac{{8}}{{125}}}}=\frac{{\frac{{\color{blue}{{{1}}}}}{{\color{red}{{{1}}}}}}}{{\frac{{\color{red}{{{8}}}}}{{\color{blue}{{{125}}}}}}}=\frac{{{1}\cdot{125}}}{{{1}\cdot{8}}}=\frac{{125}}{{8}}$.

Another way of evaluating it is to notice that $\frac{{1}}{{\frac{{8}}{{125}}}}$ is reciprocal of $\frac{{8}}{{125}}$, and reciprocal is just number that is turned "upside down": $\frac{{125}}{{8}}$.

Answer: $\frac{{125}}{{8}}={15}\frac{{5}}{{8}}$.

From the last example, we have following important result:

${\color{red}{{{{\left(\frac{{a}}{{b}}\right)}}^{{-{n}}}={{\left(\frac{{b}}{{a}}\right)}}^{{n}}}}}$

Now, take pen and paper and solve following problems.

Exercise 1. Find ${{3}}^{{-{2}}}$.

${{3}}^{{-{2}}}=\frac{{1}}{{{{3}}^{{2}}}}=\frac{{1}}{{9}}$.

Answer: $\frac{{1}}{{9}}$.

Next example.

Exercise 2. Find ${{1}}^{{-{15}}}$.

$\frac{{1}}{{{{1}}^{{-{15}}}}}=\frac{{1}}{{{{1}}^{{15}}}}=\frac{{1}}{{1}}={1}$.

Next exercise.

Exercise 3. Find ${{\left(-{3}\right)}}^{{-{3}}}$.

${{\left(-{3}\right)}}^{{-{3}}}=\frac{{1}}{{{{\left(-{3}\right)}}^{{3}}}}=\frac{{1}}{{-{27}}}=-\frac{{1}}{{27}}$.

Answer: $-\frac{{1}}{{27}}$.

Next exercise.

Exercise 4. Find ${{\left(\frac{{3}}{{2}}\right)}}^{{-{4}}}$.

Get rid of negative exponent: ${{\left(\frac{{3}}{{2}}\right)}}^{{-{4}}}=\frac{{1}}{{{{\left(\frac{{3}}{{2}}\right)}}^{{4}}}}$.

Now, get rid of positive exponent: ${{\left(\frac{{3}}{{2}}\right)}}^{{4}}=\frac{{{{3}}^{{4}}}}{{{{2}}^{{4}}}}=\frac{{81}}{{16}}$.

Last step: $\frac{{1}}{{{{\left(\frac{{3}}{{2}}\right)}}^{{4}}}}=\frac{{1}}{{\frac{{81}}{{16}}}}=\frac{{16}}{{81}}$.

Answer: $\frac{{16}}{{81}}$.

Last exercise.

Exercise 5. Find ${{\left(-\frac{{3}}{{7}}\right)}}^{{-{3}}}$.

Get rid of negative exponent: ${{\left(-\frac{{3}}{{7}}\right)}}^{{-{3}}}=\frac{{1}}{{{{\left(-\frac{{3}}{{7}}\right)}}^{{3}}}}$.

Now, get rid of positive exponent: ${{\left(-\frac{{3}}{{7}}\right)}}^{{{3}}}=\frac{{{{\left(-{3}\right)}}^{{3}}}}{{{{7}}^{{3}}}}=\frac{{27}}{{343}}$.

Last step: $\frac{{1}}{{{{\left(-\frac{{3}}{{7}}\right)}}^{{3}}}}=\frac{{1}}{{-\frac{{27}}{{343}}}}=-\frac{{343}}{{27}}$.

Answer: $-\frac{{343}}{{27}}=-{12}\frac{{19}}{{27}}$.