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Simplify the expression and eliminate any negative exponents, assuming that all letters denote positive numbers. (2x^4 * y^(-4/5))^3(8y^2)^(2/3) I try to post it in equation form too.

Mathematics
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Could you draw it out?
\[(2x^4y^\frac{ -4 }{ 5 })^3(8y^2)^\frac{ 2 }{ 3 }\]
if it was \[(a ^{4})^{3}\] What have you learned to do with the exponents?

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Other answers:

Multiply 4 and 3, right?
Yes, what if it was \[(4a ^{4})^{3}\] ?
4 to the third times a to the 12
Yes, what happens with \[(8y ^{2})^{2/3}\] ?
\[8^\frac{ 2 }{ 3 }y ^\frac{ 4 }{ 3 }\]
Ok, so which part do you have trouble with?
Well, the answer in the back of the book is 14, and no matter what I do, i can't get 14! I'll take a picture of my work real quick and maybe you can see what I did then?
Sure
\[ \large (2x^4y^{-4/5})^3(8y^2)^{2/3}=2^3x^{4\times3}y^{-4/5\times3} 8^{2/3}y^{2\times2/3} \]
\[ \large 8x^{12}y^{-12/5}8^{2/3}y^{4/3} \]
\[ \large =8^{1+2/3}x^{12}y^{4/3-12/5}=8^{5/3}x^{12}y^{-16/15} \]
|dw:1347151652056:dw|
\[2x^{12}*\frac{ 8 }{ y}\]
Here's what I have... I've just tried a few times and I'm really having difficulty. This is my most recent bit.
1 Attachment
\[ \large =2^5x^{12}y^{-16/15} \]
According to the back of the book, the answer is \[(32x^\frac{ 12 }{ 1 })/(y^ \frac{ 16 }{ 15 })\]
I just don't have an idea of how to get it.
\[ \large \frac{32x^{12}}{y^{16/15}} \] 2^5=32
|dw:1347152276782:dw|
You broke down your y's by saying \[y ^{4/3}= y \sqrt[3]{y}\] and such...
\[y ^{-1}=\frac{ 1 }{ y }\]
|dw:1347152573807:dw|
ok, thanks everyone! i have a better understanding now.

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