View Full Version : rational exponents and radical signs
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)
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:
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}}}
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