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How do you do this simple algebraic question?
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\[\frac{ a^{2x}b ^{3x}(\sqrt{a^{3}})^x }{ b^{x-y}a}\]
Start with\[(\sqrt{a^3})^x\]\[(a^{\frac{3}{2}})^x\]\[a^{\frac{3}{2}x}\]
Gather the a's\[a^{2x}a^{\frac{3}{2}x}a^{-1}\]\[a^{\frac{4}{2}x}a^{\frac{3}{2}x}a^{-1}\]\[a^{\frac{7}{2}x-1}\]
\[a ^{\frac{ 7x-2 }{ 2 }}b^{2x-y}\]
Is that correct?
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and the b's\[b^{3x}b^{-(x-y)}\]\[b^{3x-x+y}\]\[b^{2x+y}\]
haha darn. so close. don't know why I added the -1 for the a's.
I don't know? You asked how I would do it.
We're both right. I think.
Thanks man! You helped me. I forgot how to deal with the square roots.
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\[a^{\frac{7}{2}x-1}b^{2x+y}\]
\[\sqrt{x}=x^{\frac12}\]
Thanks again. :)
I never use \(\sqrt\cdot\) I think they're ugly.
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