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

1. 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.

2. anonymous

It's the second one!

3. anonymous

So if I were using the first method, it would be something like: 3^7/3 all over 7^7/3 ?

4. anonymous

But what would I do after that?

5. 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.

6. 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.

7. anonymous

Okay, and then what would happen?

8. mathmate

Then you would use the second rule to make it a simple fraction.

9. anonymous

|dw:1438826071903:dw|

10. anonymous

What would be the Z and the M?

11. mathmate

But that's a completely different problem! lol

12. 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

13. anonymous

But I guess now that the air is clear, do you mind helping me with the problem I drew? lol

14. 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!

15. anonymous

Thats really weird... btw this came from me trying to solve an integral problem

16. mathmate

You may post the integral problem to start! :)

17. anonymous

|dw:1438826425496:dw|

18. anonymous

|dw:1438826526116:dw|

19. anonymous

.... and then I dont know what to do

20. mathmate

|dw:1438827006677:dw|

21. anonymous

Im trying to understand what you did at the bottom

22. 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

23. mathmate

Yes, but it's the same as evaluating term by term, $$[(3/7)*x^{7/3}+(4/9)x^{9/4}]_0^1=((3/7)*1^{7/3}+(4/9)1^{9/4})-((3/7)*^{7/3}+(4/9)0^{9/4}])$$ $$=(3/7)+(4/9)-(0+0)=3/7+4/9=55/63$$

24. mathmate

note that 1^(7/3)=1, and 1^(9/4)=1, 0^(7/3)=0, and 0^(9/4)=0

25. anonymous

OH

26. anonymous

that makes so much sense, thanks!