I am having problems understanding how to answer questions containing fractional exponents to a given power ie $(2x^{1/2})^6$, i do not understand how to go about answering the question. I know this is an easy topic, but i would really appreciate the help
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$\begingroup$ (2x^1/2)^6=2x^3. remember (x^y)^z=x^yz $\endgroup$– user148697May 13, 2014 at 0:49
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$\begingroup$ @jonnytan999: Wrong; Try $(x,y,z) = (-1,6,1/2)$. $\endgroup$– user21820May 13, 2014 at 11:09
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2 Answers
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The general rule for exponents is $x^{a/b} = \sqrt[b]{x^a} \forall x \in \mathbb{N}$. For example, $4^{2/3} = \sqrt[3]{4^2} = \sqrt[3]{8}$. You can apply this to your problem to see that $(2x^{1/2})^6 = 64x^3.$
answered May 13, 2014 at 0:52
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1$\begingroup$ This is wrong. $(-1)^{6/2} = (-1)^3 = -1 \ne 1 = \sqrt{1} = \sqrt[2]{(-1)^6}$. $\endgroup$ May 13, 2014 at 11:07
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how to answer questions containing fractional exponents
In the exact same manner in which you answer questions containing non-fractional exponents, since the base is obviously a positive number, given the fact that both $2$ and $\sqrt x$ are always $\ge0$. $(ab)^c=a^cb^c$ for all exponents, fractional or not, it doesn't matter. Likewise, $(a^b)^c=a^{bc}$, again, for all exponents. By combining the two, we have $(ab^c)^d=a^db^{cd}$. In this case, $a=2,b=x,c=$ $=\dfrac12$ and $d=6$. And $\dfrac12\cdot6=3$.
