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I was playing a bit whit exponents. I maybe found a working formula for calculating $n^y$ if you know $n^x$. The formula may already be discovered, but the formula I found is: $$ (n^x)^\frac{y}{x} = n^y $$ Ok, so the formula should work if $n,x,y \in \mathbb N$

I am not sure if it does work whit negative and decimal numbers tho.

Ok, my questions are:
1. Can the formula be used whit decimal, and negative numbers (x,y,n)
2. Can you prove that the formula works if $x,y,n \in \mathbb N$, if it is possible also for $x,y,n \in \mathbb Q$, $x,y,n \in \mathbb R$ and $x,y,n \in \mathbb Z$.

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  • $\begingroup$ @DavidG.Stork, thank you so much. I am new to exponents and all that. I dont really know all the rules yet. $\endgroup$ – CppPythonDude Jul 16 '19 at 20:47
  • $\begingroup$ Of course, it's a good question, and it takes more work than one might imagine to prove this for arbitrary $x,y,n\in\mathbb{R}$ $\endgroup$ – Thomas Winckelman Jul 16 '19 at 20:54
  • $\begingroup$ @DavidG.Stork The rule fails if $n$ is negative and $x$ is even. $\endgroup$ – B. Goddard Jul 16 '19 at 21:01
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Let $n=-1$, $x=2$ and $y=1$. Then you have

$$((-1)^2)^{1/2} = 1^{1/2} = 1$$

but

$$(-1)^1 = -1.$$

So your rule fails.

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  • $\begingroup$ Ok, that is true. But what if we |x| (absolute value) the result? $\endgroup$ – CppPythonDude Jul 16 '19 at 21:11
  • $\begingroup$ @CppPythonDude Then you have a rule that works on half of the fields you mentioned. What about complex numbers? There are 4 fourth roots of $-1$, so I could cook up an example where the absolute value won't fix the problem. $\endgroup$ – B. Goddard Jul 16 '19 at 21:19
  • $\begingroup$ Can i somehow fix the rule then? $\endgroup$ – CppPythonDude Jul 16 '19 at 21:32
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    $\begingroup$ If you add the condition that $x>0$ the rule will work. In general, when you see fractional exponents and negative bases, trouble is lurking at every turn. $\endgroup$ – B. Goddard Jul 16 '19 at 21:37
  • $\begingroup$ I think the restrictions you want are $n>0$, $x\neq0$. $\endgroup$ – David K Jul 17 '19 at 17:14
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That is one of the exponent rules.

The rule is: $(x^a)^b = x^{ab}$, But generally, you can't get $n^y$ if you have $n^x$, per example if you have $7^{12}$, you can't get $7^{20} = 7^{12} \cdot 7^8$, without calculating $7^8$.

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