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March 25th, 2016, 09:32 PM  #1 
Member Joined: Mar 2015 From: Los Angeles Posts: 73 Thanks: 7  what bases and exponents are defined on real numbers
I'm interested in knowing what bases and exponents have a meaning/definition in real numbers. Tell me if any of this is right. I think that $0^0$ is undefined. $ 0^a = 0 $ but only for $ a > 0 $. $b^0 = 1$ for all real $b$ except $b=0$. Is this right so far? Then we get into negative bases and rational exponents. Suppose we have $$ b \in \mathbb{R}, b <0, m \in \mathbb{N}, n \in \mathbb{N} $$ Then $ b^{m \over n} $ has a real solution only if $n$ is odd. Correct? It has imaginary or complex solutions if $n$ is even. Also, what about this? For $ b \in \mathbb{R}, b < 0, a \in \mathbb{R}$, then $b^a$ generally does not have a solution unless the above conditions are met. But does it have complex solutions? Like, does $ (1)^\pi $ have any complex solutions? 
March 26th, 2016, 05:33 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,511 Thanks: 2514 Math Focus: Mainly analysis and algebra 
$(1)^\pi=\left(\mathrm e^{\mathrm i \pi}\right)^\pi=\mathrm e^{\mathrm i \pi^2}$ which is a complex number of unit modulus and argument $\pi^2$. Once you get into negative bases, the construction may exist in the real numbers or not as you point out, but we don't usually deal with functions having negative bases because they are not connected (smoothly) connected. 
March 26th, 2016, 07:23 AM  #3  
Member Joined: Mar 2015 From: Los Angeles Posts: 73 Thanks: 7  Quote:
$(1) = e^{ 3 i \pi}$ and $ (1)^\pi = e^ {3 i \pi^2 }$ which is a different number as the angle $3 i \pi^2$ is not coterminal with $ i \pi^2$. Or does this have no practical use?  

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