Question Number 17713 by tawa tawa last updated on 09/Jul/17
$$\mathrm{Evaluate}:\:\:\:\:\left(−\sqrt{\mathrm{3}}\right)^{\left(−\sqrt{\mathrm{2}}\right)} \\ $$
Commented by b.e.h.i.8.3.417@gmail.com last updated on 09/Jul/17
![a=(−(√3))^((−(√2))) lna=−(√2)ln(−(√3))=−(√2)ln(i^2 (√3))= =−(√2)[2lni+(1/2)ln3]=−(√2)[2i(π/2)+(1/2)ln3]= =−((√2)/2)(2iπ+ln3) ⇒a=e^(−((√2)/2)(2iπ+ln3)) =(e^(2iπ+ln3) )^((−(√2))/2) = =[(e^(iπ) )^2 .e^(ln3) ]^(−((√2)/2)) =[(−1)^2 .3]^(−((√2)/2)) =3^((−((√2)/2))) note: 1)i=cos90+isin90=e^(i(π/2)) ⇒lni=i.(π/2) 2)e^(iπ) +1=0 (Euler′s formula)](https://www.tinkutara.com/question/Q17730.png)
$${a}=\left(−\sqrt{\mathrm{3}}\right)^{\left(−\sqrt{\mathrm{2}}\right)} \\ $$$${lna}=−\sqrt{\mathrm{2}}{ln}\left(−\sqrt{\mathrm{3}}\right)=−\sqrt{\mathrm{2}}{ln}\left({i}^{\mathrm{2}} \sqrt{\mathrm{3}}\right)= \\ $$$$=−\sqrt{\mathrm{2}}\left[\mathrm{2}{lni}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}\right]=−\sqrt{\mathrm{2}}\left[\mathrm{2}{i}\frac{\pi}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{3}\right]= \\ $$$$=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{2}{i}\pi+{ln}\mathrm{3}\right) \\ $$$$\Rightarrow{a}={e}^{−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{2}{i}\pi+{ln}\mathrm{3}\right)} =\left({e}^{\mathrm{2}{i}\pi+{ln}\mathrm{3}} \right)^{\frac{−\sqrt{\mathrm{2}}}{\mathrm{2}}} = \\ $$$$=\left[\left({e}^{{i}\pi} \right)^{\mathrm{2}} .{e}^{{ln}\mathrm{3}} \right]^{−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} =\left[\left(−\mathrm{1}\right)^{\mathrm{2}} .\mathrm{3}\right]^{−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} =\mathrm{3}^{\left(−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)} \\ $$$${note}: \\ $$$$\left.\mathrm{1}\right){i}={cos}\mathrm{90}+{isin}\mathrm{90}={e}^{{i}\frac{\pi}{\mathrm{2}}} \Rightarrow{lni}={i}.\frac{\pi}{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){e}^{{i}\pi} +\mathrm{1}=\mathrm{0}\:\:\:\:\:\left({Euler}'{s}\:{formula}\right) \\ $$
Commented by tawa tawa last updated on 09/Jul/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$