Johan14th
  • Johan14th
can someone help me find the power form of the reciprocal radical form 1/(3 sqrt.three(x^2))
Mathematics
chestercat
  • chestercat
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Johan14th
  • Johan14th
\[\frac{ 1 }{ 3\sqrt[3]{x ^{2}}}\]
anonymous
  • anonymous
you have a simple formula for radicals \[\sqrt[a]{x^b}=x^\frac{b}{a}\]|dw:1444074783490:dw|
anonymous
  • anonymous
Also \[x^n=\frac{1}{x^{-n}} \space \space \space ; \space \space \space \frac{1}{x^n}=x^{-n}\] Just add a minus sign to the exponent when u move the number from numerator to denominator or vice versa

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anonymous
  • anonymous
|dw:1444075056599:dw|
anonymous
  • anonymous
@Johan14th r u there?
Johan14th
  • Johan14th
yes so then it would be x^ 2/3
Johan14th
  • Johan14th
\[3x^{\frac{ 2 }{ 3 }}\]
anonymous
  • anonymous
no! you're not done yet!!
anonymous
  • anonymous
\[\frac{1}{3x^{\frac{2}{3}}}\]
anonymous
  • anonymous
We need to bring the number to the top now!! |dw:1444075590776:dw|
Johan14th
  • Johan14th
oh okay so then it is \[3x^{\frac{- 3 }{ 2 }}\]
Johan14th
  • Johan14th
and then i could apply the power rule right?
anonymous
  • anonymous
Hmm, you only change the sign, you leave the fraction unchanged, also the number 3 will stay in the bottom, if you want to bring the 3 to the top, you must write -1 to it's power \[\frac{1}{3x^{\frac{2}{3}}}=\frac{x^{-\frac{2}{3}}}{3}=3^{-1}x^{-\frac{2}{3}}\]
Johan14th
  • Johan14th
okay so basically adding the (-) to the power makes it go on to the top of the fraction, numerator. if i want to do the same to the 3 (too get rid of the fraction) i add the (-) to the power
anonymous
  • anonymous
Yes, anytime you bring a number or letter from top to bottom, or from bottom to top, you change the sign of it's power, very important, remember this rule
anonymous
  • anonymous
for example if we had \[\frac{1}{3^{-1}}\] If we bring it to the top it becomes \[\frac{1}{3^{-1}}=3^{1}=3\]
Johan14th
  • Johan14th
okay that just cleared everything out. thank you. i was trying too do some weird multiplication. you saved me!
anonymous
  • anonymous
you're absolutely welcome

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