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Can I say that the answer is $3$?

When I exponentiated both sides I got:

$$ (-2)^3 = -8$$

But I am being told that negative logarithms cannot occur. Why is this?

an4s
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Bob Smith
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    The base of logarithm must always be strictly positive and different from 1. Otherwise you can calculate ""logarithm"" only in some very specifical cases like the one you posted. If I asked you the ""logarithm"" in base -2 of -3 ?(if we stay in real numbers) – Kandinskij Dec 28 '19 at 21:40
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    Look in your book for the "definition" of logarithm. Maybe $(-2)^3$ makes sense, but $(-2)^{3.1}$ doesn't (in the real numbers). For most $y$ there is no (real) solution $x$ to the equation $(-2)^x = y$. So the convention is that $\log_a$ is defined only for $a>1$. Maybe (in strange circumstances) also for $0 < a < 1$. – GEdgar Dec 28 '19 at 21:41
  • Does this answer your question? Why aren't logarithms defined for negative $x$? – an4s Dec 28 '19 at 21:42

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