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I was asked the following question and asked to evaluate it: $\log(-8)$. Do I answer that the answer will be a complex number and therefore no real answer exists? Is that true to say?

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It depends what range of answers are acceptable. You are correct that there is no real $x$ such that $e^{x}=-8$. The logarithm can be defined in the complex numbers and there are then solutions: $\ln 8 + (2k+1) \pi i$ for integral $k$.