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anonymous
 one year ago
Need Help! How do you solve for (3/7)^7/3 to be in fraction form?
anonymous
 one year ago
Need Help! How do you solve for (3/7)^7/3 to be in fraction form?

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mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Is 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
 one year ago
Best ResponseYou've already chosen the best response.0It's the second one!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So if I were using the first method, it would be something like: 3^7/3 all over 7^7/3 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But what would I do after that?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.0"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
 one year ago
Best ResponseYou've already chosen the best response.0Okay, and then what would happen?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0Then you would use the second rule to make it a simple fraction.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438826071903:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What would be the Z and the M?

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0But that's a completely different problem! lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh, 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
 one year ago
Best ResponseYou've already chosen the best response.0But I guess now that the air is clear, do you mind helping me with the problem I drew? lol

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0I don't understand the second problem. \(\frac{3}{7}+\frac{4}{9}=\frac{55}{63}\) but not with the exponents that you included!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thats really weird... btw this came from me trying to solve an integral problem

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0You may post the integral problem to start! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438826425496:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438826526116:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0.... and then I dont know what to do

mathmate
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438827006677:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im trying to understand what you did at the bottom

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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
 one year ago
Best ResponseYou've already chosen the best response.0note that 1^(7/3)=1, and 1^(9/4)=1, 0^(7/3)=0, and 0^(9/4)=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that makes so much sense, thanks!
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