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

- anonymous

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

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- 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

It's the second one!

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

But what would I do after that?

- 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

"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

Okay, and then what would happen?

- mathmate

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

- anonymous

|dw:1438826071903:dw|

- anonymous

What would be the Z and the M?

- mathmate

But that's a completely different problem! lol

- 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

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

- 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

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

- mathmate

You may post the integral problem to start! :)

- anonymous

|dw:1438826425496:dw|

- anonymous

|dw:1438826526116:dw|

- anonymous

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

- mathmate

|dw:1438827006677:dw|

- anonymous

Im trying to understand what you did at the bottom

- 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

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

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

- anonymous

OH

- anonymous

that makes so much sense, thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.