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I am trying to understand the different rules for multiplying exponents by fractional exponents and raising whole numbers by the power of fractional exponents. I have an idea but I'm trying to assure myself. Take the following example:

$(64x^4)^\frac{1}{3}$

Do I first find the cube root of 64 and bring it to the power of 1, which is 4? And then for the exponent of $x$, multiply it by 1 and divide it by 3, which is $\frac{4}{3}$ - resulting in:

$4x^\frac{4}{3}$

I am just trying to make sure I've got this down because I'm having a test next week. Thanks!

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Why are you dividing the exponent by 2? –  Drew Christianson Sep 25 '11 at 15:20
    
Sorry, the 3 looked like a two. I changed it back –  David Sep 25 '11 at 15:42
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You are interested in $(64x^4)^{\frac{1}{3}}$, which can be decomposed as $64^{\frac{1}{3}}(x^4)^{\frac{1}{3}}$. Then you have dealt with $64^{\frac{1}{3}}=4$ correctly. For $(x^4)^{\frac{1}{3}}$, the law of exponents $(a^b)^c=a^{bc}$ gives $(x^4)^{\frac{1}{3}}=x^{\frac{4}{3}}$, giving a final result $(64x^4)^{\frac{1}{3}}=4x^{\frac{4}{3}}$ It looks like you misread the $3$ as a $2$ along the way.

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You were right, I did mistake 3 for 2. Other than that, I got the answer right. Thanks! –  David Sep 25 '11 at 15:54
    
Since you're talking about haveing a test next week it might be significant (and should possibly be mentioned here) that the order of computation of exponents can only be interchanged freely if x is assumed to be a positive real. If x is negative (or even complex) the order is significant.(But perhaps it is usual in the course to assume x being a positive real number so far) –  Gottfried Helms Sep 26 '11 at 7:13
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