Question

Can someone please help me solve this problem? http://i158.photobucket.com/albums/t120/brinazarski2/q_zps284189ee.png

Answer 1

\[7x^5=11y^{13}\]

\[y ^{13}=\frac{7}{11}x^5\]
So one of the factors of x has to be 11

Answer 2

Why 11?

Answer 3

And shouldn't x^5 be divided by 11 as well?

Answer 4

I divided both sides of the equation by 11

Answer 5

Yes... but how does that make one of the factors 11?

Answer 6

the problem says that y and x are both integers. That means that the powers of x and y are also integers and the multiples of x and y are integers. So y^13 is an integer. but y^13 = 7/11 x^5. So if y^13 is going to be an integer, 11 must divide into x^5 since it will not divide into 7

Answer 7

Okay... but since we're looking for x, shouldn't it be the other way around?

Answer 8

I thought you wanted the factors of x. sorry.

Answer 9

I do. I'm bad at math, I have no idea what I'm doing lol

Answer 10

That's why I need help

Answer 11

Tushara, tbansal, are you two still here and do you know how to do this? >.<

Answer 12

hey... m still here, im not quite sure how to do it, ill let u know if i figure it out

Answer 13

okay, thank you!!! I'm so lost ._.

Answer 14

D 33

Answer 15

How did you get that answer?

Answer 16

We have seen that 11 has to be one of the factors. Similarly, 7 has to be one of the factors. So 7 (11^?)(7^?)=11(11^?)(7^?)
Put the smallest exponents on those necessary to make them equal. You will see that those numbers are 13 and 12. So 13 + 12 + 7 + 11 = 33

Answer 17

Dang... how'd you come up with 12 and 13? Though I see it works, did you just plug it in or was there some way you found them? And thank you!

Answer 18

I found the smallest numbers that would make the exponents on both sides the same.

Answer 19

You had to start with 13 because one side was raised to the 13th power. Since there was already a 7, the factor 7 only had to be raised to the 12th power.

Answer 20

Ah, that makes sense... okay, thank you so much! :)