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The function that I'm trying to find the domain of is $$f(x) = 11.8x^{\frac{\log\left(\frac{39}{11.8}\right)}{\log(11)}}.$$ I know that this function can be evaluated for all non negative real numbers but I'm having trouble determining whether this function can take in negative numbers or not. I was wondering whether someone could explain if negative numbers can be inputted into this function and if they can't why not?

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    $\begingroup$ For non-integer $\alpha$, $x^{\alpha}$ becomes complex-valued for negative $x$. $\endgroup$ – Kenneth Goodenough May 5 '16 at 23:52
  • $\begingroup$ @NickS It does behave differently than $x^{2}$ if you input this function into a graphing calculator you find out it can't be evaluated at negative numbers whereas $x^{2}$ is defined everywhere $\endgroup$ – user262291 May 5 '16 at 23:54
  • $\begingroup$ @KennethGoodenough what about if you have $x^{.4}$ it isn't complex valued for a negative x. You can plug that function into a graphing calculator and see that it has a real value for any real number. $\endgroup$ – user262291 May 5 '16 at 23:56
  • $\begingroup$ oops, apologies, I misevaluated the power. $\endgroup$ – Nick S May 5 '16 at 23:56
  • $\begingroup$ $x^r$ with $x<0$ and $r \in \mathbb R$, will be complex when x<0 and r is irrational. And will be complex for certain rational $r$ as well. $\endgroup$ – Doug M May 5 '16 at 23:57
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A function where the exponent of $x$ is not a integer value is complex-valued for negative values of $x$.

Evaluating your exponent gives a value of approximately $0.4985$ so it will have non-real values for $x<0$.

Edit: Some non-integers may give real values but these cases are all rationals where as your value is irrational.

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