jayeamy81

06-22-2006, 12:52 PM

Can someone explain the advantage of using rational exponents over the radical and sign or at least show an example of the difference so i can see the advantage.

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jayeamy81

06-22-2006, 12:52 PM

Can someone explain the advantage of using rational exponents over the radical and sign or at least show an example of the difference so i can see the advantage.

jonboy

06-22-2006, 01:00 PM

Whenever you are wanting to know something like this always Google (http://www.google.com/search) it. I found this exact same question on this board. :shock:

Hopefully it will help: Link (http://www.freemathhelp.com/forum/viewtopic.php?p=42489&highlight=&sid=e8164aa0909dd0925a591680df777056)

Hopefully it will help: Link (http://www.freemathhelp.com/forum/viewtopic.php?p=42489&highlight=&sid=e8164aa0909dd0925a591680df777056)

Denis

06-22-2006, 02:24 PM

Can someone explain the advantage of using rational exponents over the radical and sign or at least show an example of the difference so i can see the advantage.

Your question makes no sense...can you rewrite it, with an example :idea:

Your question makes no sense...can you rewrite it, with an example :idea:

soroban

06-22-2006, 05:59 PM

Hello, jayeamy81!

Can someone explain the advantage of using rational exponents over the radical sign?

Or at least show an example of the difference, so i can see the advantage.

Simplify: \;\sqrt{x}\,\cdot\,^3\!\!\sqrt{x}

(a) It can be done with radicals . . . like this:

\;\;\sqrt{x}\;=\;^{^6}\!\sqrt{x^3}

\;\;^3\!\sqrt{x}\;=\;^{^6}\!\sqrt{x^2}

The problem becomes: \:^{^6}\!\sqrt{x^3}\,\cdot\,^{^6}\!\sqrt{x^2}\;=\; ^{^6}\!\sqrt{x^3\cdot x^2}\;=\;^{^6}\!\sqrt{x^5}

(b) With rational exponents, we have:

\;\;\sqrt{x}\,\cdot\,^3\!\sqrt{x}\;=\;x^{^{\frac{1 }{2}}}\cdot x^{^{\frac{1}{3}}}\;=\;x^{^{\frac{1}{2}+\frac{1}{3 }}}\;=\;x^{^{\frac{5}{6}}}

Can someone explain the advantage of using rational exponents over the radical sign?

Or at least show an example of the difference, so i can see the advantage.

Simplify: \;\sqrt{x}\,\cdot\,^3\!\!\sqrt{x}

(a) It can be done with radicals . . . like this:

\;\;\sqrt{x}\;=\;^{^6}\!\sqrt{x^3}

\;\;^3\!\sqrt{x}\;=\;^{^6}\!\sqrt{x^2}

The problem becomes: \:^{^6}\!\sqrt{x^3}\,\cdot\,^{^6}\!\sqrt{x^2}\;=\; ^{^6}\!\sqrt{x^3\cdot x^2}\;=\;^{^6}\!\sqrt{x^5}

(b) With rational exponents, we have:

\;\;\sqrt{x}\,\cdot\,^3\!\sqrt{x}\;=\;x^{^{\frac{1 }{2}}}\cdot x^{^{\frac{1}{3}}}\;=\;x^{^{\frac{1}{2}+\frac{1}{3 }}}\;=\;x^{^{\frac{5}{6}}}

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