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Question Number 54013 by zambolly19 last updated on 27/Jan/19

$$2^{\mathbf{x}}=-4.\mathbf{solve}\mathbf{for}\mathbf{x}$$

Answered by Smail last updated on 27/Jan/19

$$xln2=ln(-4)$$$$x=\frac{ln\left(4e^{i\pi }\right)}{ln2}=\frac{2ln2+i\pi }{ln2}=2+i\frac{\pi }{ln2}$$$$x=2+\frac{i\pi }{ln2}$$

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