This is my first time using this web site. Could anyone help me with the solution to [3x^2/3y^1/4]^3 i dont know if i am just over complicating the simple or what. any help would be greatly appreciated.

Question

This is my first time using this web site. Could anyone help me with the solution to [3x^2/3y^1/4]^3   i dont know if i am just over complicating the simple or what. any help would be greatly appreciated.

anonymous · Answer

thank you for your prompt reply. that said i guess i wrote the problem wrong.

anonymous · Answer

the x is to the two thirds power(2/3) and the y is just y to the one fourth power(1/4) ...please forgive  me if i wrote it wrong.

anonymous · Answer

oh yeah also it is not x over y it is x times y .

anonymous · Answer

how did  you draw?

unklerhaukus · Answer

$$\left[3x^{2/3} y^{1/4}\right]^3 $$?

unklerhaukus · Answer

$$\left[ab^nc^d\right]^f=\left[a^fb^{n\times f}c^{d\times f}\right]$$

anonymous · Answer

Yes sir your first response is my problem. I am having Internet issues today.

precal · Answer

see the blue middle key that states "Draw", press that and use your mouse to draw

precal · Answer

seems I did try to help you after all

precal · Answer

@precal
will send me a notification
were you able to figure this one out?
Let me know and I will look at it again.

anonymous · Answer

I was not able to figure it out. @precal

precal · Answer

what page number and problem, easier for me to look it up

anonymous · Answer

It's  page 2 problem fifteen of the packet.

precal · Answer

ok let me look at it

precal · Answer

|dw:1346599631577:dw|
ok apply your exponent 3 to all of your bases