anonymous
  • anonymous
Need Help! How do you solve for (3/7)^7/3 to be in fraction form?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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mathmate
  • mathmate
Is the question \(\Large \frac{(\frac{3}{7})^7}{3}\) or is it \(\Large (\frac{3}{7})^\frac{7}{3}\) ? In any case, use \((\frac{A}{B})^X=\frac{A^X}{B^X}\) and \(M^\frac{x}{y}=\sqrt[y]{M^x}\) to help you.
anonymous
  • anonymous
It's the second one!
anonymous
  • anonymous
So if I were using the first method, it would be something like: 3^7/3 all over 7^7/3 ?

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anonymous
  • anonymous
But what would I do after that?
mathmate
  • mathmate
@mary_am Please remember the priority of operations require that exponentiation is before multiplication and division, so use brackets as required. \(\Large (\frac{3}{7})^\frac{7}{3}\) is written (3/7)^(7/3) to avoid further confusion.
mathmate
  • mathmate
"So if I were using the first method, it would be something like: 3^7/3 all over 7^7/3 ? " probably meant = ( 3^(7/3) ) / ( 7^(7/3) ) Yes written that way would be correct.
anonymous
  • anonymous
Okay, and then what would happen?
mathmate
  • mathmate
Then you would use the second rule to make it a simple fraction.
anonymous
  • anonymous
|dw:1438826071903:dw|
anonymous
  • anonymous
What would be the Z and the M?
mathmate
  • mathmate
But that's a completely different problem! lol
anonymous
  • anonymous
Oh, I was just going to figure out how to find (3/7)^(7/3) in its fraction form and try to apply the same method for the other value
anonymous
  • anonymous
But I guess now that the air is clear, do you mind helping me with the problem I drew? lol
mathmate
  • mathmate
I don't understand the second problem. \(\frac{3}{7}+\frac{4}{9}=\frac{55}{63}\) but not with the exponents that you included!
anonymous
  • anonymous
Thats really weird... btw this came from me trying to solve an integral problem
mathmate
  • mathmate
You may post the integral problem to start! :)
anonymous
  • anonymous
|dw:1438826425496:dw|
anonymous
  • anonymous
|dw:1438826526116:dw|
anonymous
  • anonymous
.... and then I dont know what to do
mathmate
  • mathmate
|dw:1438827006677:dw|
anonymous
  • anonymous
Im trying to understand what you did at the bottom
anonymous
  • anonymous
I learned to evaluate the integral by plugging in the top bound and bottom bound in the x values and then subtracting the top and bottom bound
mathmate
  • mathmate
Yes, but it's the same as evaluating term by term, \([(3/7)*x^{7/3}+(4/9)x^{9/4}]_0^1=((3/7)*1^{7/3}+(4/9)1^{9/4})-((3/7)*^{7/3}+(4/9)0^{9/4}])\) \(=(3/7)+(4/9)-(0+0)=3/7+4/9=55/63\)
mathmate
  • mathmate
note that 1^(7/3)=1, and 1^(9/4)=1, 0^(7/3)=0, and 0^(9/4)=0
anonymous
  • anonymous
OH
anonymous
  • anonymous
that makes so much sense, thanks!

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