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Question Number 193170 by aba last updated on 05/Jun/23

$$\mathrm{solve}:7^{\mathrm{x}}=-2$$

Answered by Frix last updated on 06/Jun/23

$$-2={2\mathrm{e}}^{\mathrm{i}\pi (2n+1)}\mathrm{\forall }n\in \mathbb{Z}$$$$7^x={2\mathrm{e}}^{\mathrm{i}\pi (2n+1)}$$$$x\ln 7=\ln 2+\mathrm{i}\pi (2n+1)$$$$x=\frac{\ln 2}{\ln 7}+\mathrm{i}\frac{\pi (2n+1)}{\ln 7}$$

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