shelovespiano
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.
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J-Monstur
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Could you draw it out?
shelovespiano
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\[(2x^4y^\frac{ -4 }{ 5 })^3(8y^2)^\frac{ 2 }{ 3 }\]
Denebel
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if it was \[(a ^{4})^{3}\]
What have you learned to do with the exponents?
shelovespiano
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Multiply 4 and 3, right?
Denebel
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Yes, what if it was \[(4a ^{4})^{3}\] ?
shelovespiano
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4 to the third times a to the 12
Denebel
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Yes, what happens with \[(8y ^{2})^{2/3}\] ?
shelovespiano
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\[8^\frac{ 2 }{ 3 }y ^\frac{ 4 }{ 3 }\]
Denebel
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Ok, so which part do you have trouble with?
shelovespiano
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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?
Denebel
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Sure
helder_edwin
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\[ \large (2x^4y^{-4/5})^3(8y^2)^{2/3}=2^3x^{4\times3}y^{-4/5\times3}
8^{2/3}y^{2\times2/3} \]
helder_edwin
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\[ \large 8x^{12}y^{-12/5}8^{2/3}y^{4/3} \]
helder_edwin
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\[ \large =8^{1+2/3}x^{12}y^{4/3-12/5}=8^{5/3}x^{12}y^{-16/15} \]
lopus
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|dw:1347151652056:dw|
lopus
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\[2x^{12}*\frac{ 8 }{ y}\]
shelovespiano
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Here's what I have... I've just tried a few times and I'm really having difficulty. This is my most recent bit.
helder_edwin
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\[ \large =2^5x^{12}y^{-16/15} \]
shelovespiano
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According to the back of the book, the answer is \[(32x^\frac{ 12 }{ 1 })/(y^ \frac{ 16 }{ 15 })\]
shelovespiano
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I just don't have an idea of how to get it.
helder_edwin
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\[ \large \frac{32x^{12}}{y^{16/15}} \]
2^5=32
lopus
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|dw:1347152276782:dw|
Denebel
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You broke down your y's by saying
\[y ^{4/3}= y \sqrt[3]{y}\]
and such...
lopus
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\[y ^{-1}=\frac{ 1 }{ y }\]
Denebel
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|dw:1347152573807:dw|
shelovespiano
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ok, thanks everyone! i have a better understanding now.