## Fractional Exponents Examples

In my last post, I ended with

${x}^{\frac{m}{n}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[n]{{x}^{m}}\hspace{0.33em}{=}\hspace{0.33em}{\left({\sqrt[n]{x}}\right)}^{m}$

If I rewrite this, replacing the radical symbols with their exponent equivalents, I get

${x}^{\frac{m}{n}}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}^{m}{)}^{\frac{1}{n}}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}^{\frac{1}{n}}{)}^{m}$. Note that ${m}\hspace{0.33em}\times\hspace{0.33em}\frac{1}{n}\hspace{0.33em}{=}\hspace{0.33em}\frac{1}{n}\hspace{0.33em}\times\hspace{0.33em}{m}\hspace{0.33em}{=}\hspace{0.33em}\frac{m}{n}$.

This seems to imply that if you have a base raised to a power and that is again raised to a power, you can simplify this by just multiplying the exponents. This is, in fact, true. You can show this for yourself by expanding

${(}{2}^{3}{)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{2}^{{3}\times{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{6}$.

In general,

${(}{x}^{m}{)}^{n}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{m}\times{n}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{mn}$.

By the way, ${x}^{0}\hspace{0.33em}{=}\hspace{0.33em}{1}$ no matter how small or large x is, except for x = 0. ${0}^{0}$ is not defined and if you try to do it, like dividing by 0, the math police will be knocking on your door.

Two other exponent rules, which you can demonstrate for yourself with simple examples are:

${(}{xy}{)}^{m}\hspace{0.33em}{=}\hspace{0.33em}{x}^{m}{y}^{m}{,}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\left({\frac{x}{y}}\right)}^{m}\hspace{0.33em}{=}\hspace{0.33em}\frac{{x}^{m}}{{y}^{m}}$

With these rules and the ones covered previously, let’s do some examples:

1. ${(}{x}^{2}{y}{)(}{x}^{3}{y}^{4}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{2}{+}{3}}{y}^{{1}{+}{4}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{5}{y}^{5}$
2. $\frac{{x}^{2}{y}^{5}}{{xy}^{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{2}{-}{1}}{y}^{{5}{-}{3}}\hspace{0.33em}{=}\hspace{0.33em}{xy}^{2}$
3. ${\left({7{a}^{3}{b}^{{-}{1}}}\right)}^{0}\hspace{0.33em}{=}\hspace{0.33em}{1}$
4. ${\left({{z}^{5}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{z}^{{5}\times{2}}\hspace{0.33em}{=}\hspace{0.33em}{z}^{10}$
5. ${\left({\frac{2x}{3{y}^{2}}}\right)}^{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}^{3}{x}^{3}}{{3}^{3}{y}^{{2}\times{3}}}\hspace{0.33em}{=}\hspace{0.33em}\frac{8{x}^{3}}{27{y}^{6}}$  (This problem uses several rules at the same time)
6. ${\left({{-}{2}{x}^{2}{y}^{{-}{4}}}\right)}^{{-}{2}}\hspace{0.33em}{=}\hspace{0.33em}\left({{-}{2}^{{-}{2}}{x}^{{2}\times{(}{-}{2}{)}}{y}^{{-}{4}\times{(}{-}{2}{)}}}\right)\hspace{0.33em}{=}\hspace{0.33em}\left({{-}{2}^{{-}{2}}{x}^{{-}{4}}{y}^{8}}\right)\hspace{0.33em}{=}\hspace{0.33em}\frac{{y}^{8}}{4{x}^{4}}$

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## Exponents and Roots

Well you might suspect that there is a connection between the roots of a number and exponents given my last couple of posts, and you would be correct. Now in my post on roots, I limited myself to using numbers that are perfect roots. For example, $\sqrt{8}\hspace{0.33em}{=}\hspace{0.33em}{2}$. But many roots do not have such a simple answer. You just have to represent it exactly as a root. For example, $\sqrt{2}$ has no simple integer solution. In fact, its decimal equivalent cannot be written down exactly as it is a non-repeating decimal. The answer you get on a calculator is just the first few decimals. $\sqrt{2}$ is an example of something called an irrational number. That doesn’t mean you can’t reason with it. It just means you can’t write it down exactly using decimals. You can only exactly represent it as $\sqrt{2}$.

But it is still true that $\sqrt{2}\hspace{0.33em}\times\hspace{0.33em}\sqrt{2}\hspace{0.33em}{=}\hspace{0.33em}{2}$ (ignoring the negative solution for now).

Now from my post on exponents, you saw that

${2}^{3}\hspace{0.33em}\times\hspace{0.33em}{2}^{2}\hspace{0.33em}{=}\hspace{0.33em}{2}^{{3}{+}{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{5}$

Now bear with me, but consider

${2}^{\frac{1}{2}}\hspace{0.33em}\times\hspace{0.33em}{2}^{\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{\frac{1}{2}{+}\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{1}\hspace{0.33em}{=}\hspace{0.33em}{2}$.

But remember that

$\sqrt{2}\hspace{0.33em}\times\hspace{0.33em}\sqrt{2}\hspace{0.33em}{=}\hspace{0.33em}{2}$.

Perhaps

${2}^{\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{2}$.

This is in fact true and the rules of exponents apply here as well. In general,

${x}^{\frac{1}{n}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[n]{x}$.

Now consider

$\sqrt{2}\hspace{0.33em}\times\hspace{0.33em}\sqrt{2}\hspace{0.33em}\times\hspace{0.33em}\sqrt{2}\hspace{0.33em}{=}\hspace{0.33em}{2}^{\frac{1}{2}}\hspace{0.33em}\times\hspace{0.33em}{2}^{\frac{1}{2}}\hspace{0.33em}\times\hspace{0.33em}{2}^{\frac{1}{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}^{\frac{3}{2}}$.

Looks like a number raised to a fraction does have meaning. Now in decimals, this number is 2.82842712475… , which actually goes on forever. But on a calculator, you can get this answer by first raising 2 to the “3” power then taking the square root of the result, or first take the square root of 2 then raise the result to the “3” power. In general:

${x}^{\frac{m}{n}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt[n]{{x}^{m}}\hspace{0.33em}{=}\hspace{0.33em}{\left({\sqrt[n]{x}}\right)}^{m}$.

In my next post, I will show several examples of fractional exponents.

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## Back to our Roots

I am tired of fractions, you too? Well let’s switch gears and talk about the roots of numbers. This is preparing you for a post or posts on the rules of exponents.

So before, I introduced the concept of the square root and how it is the reverse operation of squaring a number:

$\sqrt{25}\hspace{0.33em}{=}\hspace{0.33em}\pm{5}$ because ${5}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}$. That is $\sqrt{{5}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{5}$. Or in general, $\sqrt{{x}^{2}}\hspace{0.33em}{=}\hspace{0.33em}\pm{x}$. Remember that when taking the square root, there are two solutions since ${\left({{-}{5}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}$ as well.

Well, what about the opposite operation to ${x}^{3}$? Well there is one:

$\sqrt{{x}^{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}$

Notice a few things here. First, there is only one solution. There are no plus/minus solutions because the index (the “3”) of the root is odd and using the rules of signs, the sign of the odd root of a number will be the same as the number in the radical symbol. That is:

$\sqrt{125}\hspace{0.33em}{=}\hspace{0.33em}{5}{,}\hspace{0.33em}\hspace{0.33em}\sqrt{{-}{125}}\hspace{0.33em}{=}\hspace{0.33em}{-}{5}$

The other thing to notice is that if there is no index shown, then a “2” is assumed to be there. If the index is a “3”, it is called a cube root. After that, you use ordinal numbers, that is fourth root, fifth root, etc.

The last thing to notice is if the index is even, then you do get two plus/minus solutions. If the index is odd, you only get one solution.

So in general,

$\sqrt[n]{{x}^{n}}\hspace{0.33em}{=}\hspace{0.33em}\pm{x}$ if n is even and

$\sqrt[n]{{x}^{n}}\hspace{0.33em}{=}\hspace{0.33em}{x}$ if n is odd.

Examples:

$\begin{array}{l} {\sqrt{16}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{{(}\pm{2}{)}^{4}}\hspace{0.33em}{=}\hspace{0.33em}\pm{2}}\\ {\sqrt{{-}{8}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{{(}{-}{2}{)}^{3}}\hspace{0.33em}{=}\hspace{0.33em}{-}{8}}\\ {\sqrt{32}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{{2}^{5}}\hspace{0.33em}{=}\hspace{0.33em}{2}}\\ {\sqrt{{-}{32}}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{{(}{-}{2}{)}^{5}}\hspace{0.33em}{=}\hspace{0.33em}{-}{2}}\\ {\sqrt{81}\hspace{0.33em}{=}\hspace{0.33em}\sqrt{{(}\pm{3}{)}^{4}}\hspace{0.33em}{=}\hspace{0.33em}\pm{3}} \end{array}$

In my next post, I’ll introduce some rules regarding exponents.

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