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Here's a proof: $$\left(b^{x}\right)^{y}=\exp\left[y\left(\ln b^{x}\right)\right]=\exp\left[xy\left(\ln b\right)\right]=b^{xy}$$

But if for example $b=-1$, then $(b^2)^{1/2}=1$, while $b^{2\times1/2}=b^1=-1$, so that $(b^x)^y\neq b^{xy}$.

So where does the above proof fail for $b<0$?

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    $\ln(b)$ is not defined – Kenta S Mar 02 '20 at 03:12
  • @KentaS: I thought we can define the complex natural logarithm function like this? –  Mar 02 '20 at 03:14
  • Notice that the complex logarithm can take infinitely-many values, so $(b^x)^y$doesn't necessarily equal $b^{xy}$ for complex numbers; it depends on how you choose which log-value to take – Kenta S Mar 02 '20 at 03:19

2 Answers2

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In general, we have $$\ln(b^x) = x\ln(b) + 2\pi i k$$ for some $k \in \mathbb{Z}$ (which may depend on both $x$ and $b$), for any fixed branch of the complex logarithm.

In your case, this gives $$(b^x)^y = \exp[y(\ln b^x)] = \exp[y(x\ln(b)+2\pi ik)] = b^{xy}\cdot\exp[2\pi iyk]$$

If $yk \in \mathbb{Z},$ then we can show $(b^x)^y = b^{xy}$ for that branch, but usually this isn't reasonable to assume unless $y \in \mathbb{Z}$.

Brian Moehring
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The equation $\ln(b^y) = y \ln(b)$ is not true for negative $b$.

Robert Israel
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