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I’m a little unsure how to simplify the following expression: $$ x\left(\frac{y^{3}}{x^{4}}\right)^{1/4} $$

According to the answer, this should get you $\;\; x y^{3/4} x^{-1} = y^{3/4} $.

My intuition tells me that when we bring up $ \,x^{4} \,$ from the denominator, we get $ \,x^{-4}\, $ (with a negative power). In general, I’m unsure how $ x $ on the outside should multiply with what is inside the brackets. (Following BEDMAS, brackets should go first, right?)

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  • $\begingroup$ Please use LaTeX formatting next time. $\endgroup$ – nbubis Feb 26 '13 at 19:16
  • $\begingroup$ I would also suggest using \frac instead of /. $\endgroup$ – k1next Feb 26 '13 at 19:16
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You have $$ x\Bigl(\frac{y^3}{x^4}\Bigr)^{\frac{1}{4}}$$ You can leave out the braces, if you apply the exponent $\frac14$ to each, nominator and denominator. Thus $$ x\Bigl(\frac{y^3}{x^4}\Bigr)^{\frac{1}{4}}=x\frac{y^\frac{3}{4}}{x^1}$$ Now as $\frac{1}{x} = x^{-1}$ you get $$ x\Bigl(\frac{y^3}{x^4}\Bigr)^{\frac{1}{4}}=x{y^\frac{3}{4}}{x^{-1}}=x^{1-1} y^\frac34= y^\frac34$$

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  • $\begingroup$ x^1/4*X^4 should be x(^1)? if I'm reading that right ? if so this solution is more elegant. $\endgroup$ – peter_gent Feb 26 '13 at 19:37
  • $\begingroup$ No $x^\frac14 \cdot x^4 = x^{\frac14+4}=x^\frac{17}{4}$. But $\bigl(x^\frac14 \bigr)^{4} = x^{\frac14\cdot 4} = x^1 = x$. $\endgroup$ – k1next Feb 26 '13 at 19:48
  • $\begingroup$ good solution anyway. thanks, this is much more straightforward. $\endgroup$ – peter_gent Feb 26 '13 at 19:58
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    $\begingroup$ I wouldn't say, that my solution is better than the one of @amWhy. Note, that he is more detailed and really takes his time with his comments. $\endgroup$ – k1next Feb 26 '13 at 20:01
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$$ \begin{align} x\left(\frac{y^3}{x^4}\right)^{1/4} \\ \\ & = x(y^3x^{-4})^{1/4} \\ \\ & = xy^{3(1/4)}x^{-4(1/4)} \\ \\ & = xy^{3/4}x^{-1} \\ \\ & = xx^{-1}y^{3/4} \\ \\ & = \frac xx (y^{3/4}) \\ \\ & = y^{3/4} \\ \\ \end{align} $$

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  • $\begingroup$ so the y^3 multiplies with ^-4 to give you 3/4? What about the remaining X on the outside or the ^1\4 power ? $\endgroup$ – peter_gent Feb 26 '13 at 19:21
  • $\begingroup$ No, the $y^3$ becomes $y^{3(1/4)} = y^{3/4}$ and $x^{-4}$ becomes $x^{-4(1/4)} = x^{-1}$ I edited to make this clearer to you. $\endgroup$ – Namaste Feb 26 '13 at 19:24
  • $\begingroup$ Does that make sense now? You distribute the exponent $1/4$ over $y^3x^{-4}$ $\endgroup$ – Namaste Feb 26 '13 at 19:27
  • $\begingroup$ can anyone explain why you can gather powers like in step 2? Is it because of commutivity ? (sorry laggy internet on this one) $\endgroup$ – peter_gent Feb 26 '13 at 19:31
  • $\begingroup$ Do you mean placing x next to $x^{−1}$? Yes, multiplication is commutative, so you can rearrange/commute factors. ...Perhaps you mean why $(y^3x^{-4})^{1/4} = y^{3/4}x^{-4/4} = y^{3/4}x^{-1}\,$? Because one of the laws of exponents is $(ab)^c = a^cb^c$. So $(a^db^e)^c = a^{c \cdot d}b^{c\cdot e}$ $\endgroup$ – Namaste Feb 26 '13 at 19:38

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