The Taylor Series of $e^{ix}$, from the Series' of Sine and Cosine
$$ \log_2(-16) = \log_2(16) + \log_2(-1) = 4 + \frac{\ln(-1)}{ln(2)}= 4 + \frac{i\pi}{ln(2)} $$The last step above is solved by using Euler's fomula for the imaginary exponent, $e^{ix} = \cos(x) + i\sin(x)$, which is derived using the Taylor series, in a subsequent article.
$$ e^{ix} = \cos(x) + i\sin(x) = -1 \rightarrow x=\pi + 2k\pi, k \in \mathbb{Z} \rightarrow \ln(-1)=i\pi $$$$ y=2^x \rightarrow \ln(y) = x\ln(2) \rightarrow e^{\ln(y)} = e^{x\ln(2)} \rightarrow 2^x = e^{x\ln(2)} $$
$$ 2^{4+\frac{i\pi}{ln(2)}} = 16\left(2^{\frac{i\pi}{ln(2)}}\right) = 16e^{i\pi} = -16 $$
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