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Question Number 191203 by otchereabdullai last updated on 20/Apr/23

$$If{3}^x=-1findthevalueofx$$

Answered by Frix last updated on 20/Apr/23

$$3^x=-1$$$${\mathrm{e}}^{x\ln 3}={\mathrm{e}}^{\mathrm{i}(2n+1)\pi }$$$$x\ln 3=\mathrm{i}(2n+1)\pi$$$$x=\mathrm{i}\frac{(2n+1)\pi }{\ln 3}$$

Answered by Skabetix last updated on 20/Apr/23

$$3^x=i^2$$$$xln\left(3\right)=ln\left(i^2\right)$$$$x=\frac{ln\left(i^2\right)}{ln\left(3\right)}$$

Commented by Skabetix last updated on 21/Apr/23

$$◼ln\left(i\right)=ln\left({(-1)}^{\frac{1}{2}}\right)=\frac{1}{2}ln(-1)=\frac{1}{2}\times \pi i$$$$soln\left(i^2\right)=2\times \frac{1}{2}\times \pi i=\pi i$$

Commented by Frix last updated on 21/Apr/23

$$\mathrm{The}\mathrm{value}\mathrm{of}\ln \left({\mathrm{i}}^2\right)=?$$

Answered by anr0h3 last updated on 21/Apr/23

$$$$$$-1=e^{i\cdot \pi }$$$$3^x=e^{i\cdot \pi }$$$$ln\left(3^x\right)=ln\left(e^{i\cdot \pi }\right)$$$$x\cdot ln\left(3\right)=i\cdot \pi$$$$x=\frac{\left(\sqrt{-1}\right)\cdot \left(\pi \right)}{ln\left(3\right)}$$$$$$$$$$$$$$

Commented by Frix last updated on 21/Apr/23

$$-1={\mathrm{e}}^{(2n+1)\pi \mathrm{i}}$$$$\ln (-1)=(2n+1)\pi \mathrm{i}$$$$\mathrm{Complex}\mathrm{logaritms}\mathrm{are}\mathrm{not}\mathrm{unique}$$$$[\mathrm{why}\mathrm{do}\mathrm{you}\mathrm{use}\sqrt{-1}\mathrm{instead}\mathrm{of}\mathrm{i}\mathrm{at}\mathrm{the}\mathrm{end}?]$$

Commented by anr0h3 last updated on 21/Apr/23

that's correct, ln(-1) = ln(i^2) and ln(i^2)=(2n+1)πi

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