# Does this rule I found really work?

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 $$x,y,n \in \mathbb Z$$.

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

Let $$n=-1$$, $$x=2$$ and $$y=1$$. Then you have

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

but

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

• @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. – B. Goddard Jul 16 '19 at 21:19
• 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. – B. Goddard Jul 16 '19 at 21:37
• I think the restrictions you want are $n>0$, $x\neq0$. – David K Jul 17 '19 at 17:14
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 $$7^8$$.