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:
- \[
{(}{x}^{2}{y}{)(}{x}^{3}{y}^{4}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{2}{+}{3}}{y}^{{1}{+}{4}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{5}{y}^{5}
\] - \[
\frac{{x}^{2}{y}^{5}}{{xy}^{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}^{{2}{-}{1}}{y}^{{5}{-}{3}}\hspace{0.33em}{=}\hspace{0.33em}{xy}^{2}
\] - \[
{\left({7{a}^{3}{b}^{{-}{1}}}\right)}^{0}\hspace{0.33em}{=}\hspace{0.33em}{1}
\] - \[
{\left({{z}^{5}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{z}^{{5}\times{2}}\hspace{0.33em}{=}\hspace{0.33em}{z}^{10}
\] - \[
{\left({\frac{2x}{3{y}^{2}}}\right)}^{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{{2}^{3}{x}^{3}}{{3}^{3}{y}^{{2}\times{3}}}\hspace{0.33em}{=}\hspace{0.33em}\frac{8{x}^{3}}{27{y}^{6}}
\] (This problem uses several rules at the same time) - \[
{\left({{-}{2}{x}^{2}{y}^{{-}{4}}}\right)}^{{-}{2}}\hspace{0.33em}{=}\hspace{0.33em}\left({{-}{2}^{{-}{2}}{x}^{{2}\times{(}{-}{2}{)}}{y}^{{-}{4}\times{(}{-}{2}{)}}}\right)\hspace{0.33em}{=}\hspace{0.33em}\left({{-}{2}^{{-}{2}}{x}^{{-}{4}}{y}^{8}}\right)\hspace{0.33em}{=}\hspace{0.33em}\frac{{y}^{8}}{4{x}^{4}}
\]
If you have questions regarding any of these, send me an email via the Contact page, leave a comment via the Comments page, or post a comment on my facebook page.