Is it correct to state that: $$a^x=\frac{1}{a^{-x}}=\left(\frac{1}{a}\right)^{-x}$$ and: $$a^{-x}=\frac{1}{a^{x}}=\left(\frac{1}{a}\right)^{x}$$
Even though $$\left(\frac{1}{a}\right)^{-x}=\frac{1^{-x}}{a^{-x}}$$ and $$\left(\frac{1}{a}\right)^{x}=\frac{1^x}{a^x}$$ ?
Thanks!
Yes, this is true as for any finite value of $m$, we have $1^m=1$, so the latter reduces to $$\left(\frac1a\right)^{-x}=\frac{1}{a^{-x}}=a^x$$ And by our original identity, $$\left(\frac1a\right)^{-x}=\frac{1}{\left(\frac1a\right)^x}=\frac{1}{\frac{1}{a^x}}=a^x$$ So, there is no contradiction.
Hope this helps. Ask anything if not clear :)
