Show that $8 ^ \sqrt3\in\Bbb R$, and $(-8)^\sqrt3\not\in\Bbb R$.


closed as off-topic by John Doe, Arturo Magidin, anomaly, Saucy O'Path, Paul Frost Apr 27 at 23:39

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  • $\begingroup$ V3 = root of 3 .. $\endgroup$ – GoGame RJ Apr 27 at 22:05
  • $\begingroup$ You could take natural log of each and manipulate them, then show that the exponential of the second has an imaginary part $\endgroup$ – Henry Lee Apr 27 at 22:10
  • $\begingroup$ Any idea? ._. ll $\endgroup$ – GoGame RJ Apr 27 at 22:11
  • 2
    $\begingroup$ What's $(-8)^{\sqrt 3}$? $\endgroup$ – Saucy O'Path Apr 27 at 22:13
  • 2
    $\begingroup$ Welcome to math.stackexchange. You will find your experience much more valuable if rather than just copying the question you have (making it sound as if you are assigning us homework), you provide (i) context: where did you encounter this problem? What background has been covered in the course/textbook/article, or that you have, related to the problem? That will help people answering provide you with an answer at the appropriate level that uses appropriate tools; and (ii) show your effort/work: what have you done, tried, or where are you stuck? That will help answers be relevant. $\endgroup$ – Arturo Magidin Apr 27 at 22:23

x^y = exp(y*ln(x))

(-8)^(sqrt3) = exp((sqrt3)*ln(-8))

ln(-8) is not in real, hence exp((sqrt3)*ln(-8)) is also not in real.

  • $\begingroup$ $\frac{2i\pi}{\sqrt 3}$ is not real, hence $1$ is not real. I don't know about you, but this sounds perfectly fine to me. $\endgroup$ – Saucy O'Path Apr 27 at 23:10
  • $\begingroup$ I think I got it $\endgroup$ – GoGame RJ Apr 28 at 0:12

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