Show for $x>0$ and $ p, q \in \mathbb N$ that $$(x^{kp})^{1/(kq)} = (x^p)^{1/q}$$ given $(x^p)^{1/q} = (x^{1/q})^p$.
Perhaps use injectivity of $f(x)=x^{kp}$?
Any help would be appreciated.
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$x$ is what? $p$ is what? $q$ is what? $k$ is what? Also, $1/kq$ means $1/(kq)$ and not $(1/k)q$, right? – GEdgar Feb 05 '15 at 01:34
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x is a real greater than zero, p,q are natural numbers, and 1/kq=1/(kq) – Sidrow Feb 05 '15 at 01:37
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4 out or 5 explanations provided. How about putting them in the question itself? – GEdgar Feb 05 '15 at 01:38
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Usually exponents of arbitrary base are defined as$$x^k=e^{k\ln x}.$$ Therefore $$(x^{kp})^{1/kq}=(e^{kp\ln x})^{1/kp}=e^{\frac{1}{kq}\ln (e^{kp\ln x})}.$$ We know that $$e^{\ln x}=x.$$ I think now you can continue form here. Also I found a similar question.
