Question Number 91936 by MWSuSon last updated on 03/May/20
$${how}\:{to}\:{evaluate}\:{ln}\left({i}\right),\:{i}=\sqrt{−\mathrm{1}}. \\ $$
Commented by MWSuSon last updated on 03/May/20
thank you sir, but what if I'm looking for log(i) in base 2 or 10?
Commented by mr W last updated on 03/May/20
$$\mathrm{log}_{\mathrm{10}} \:\left({i}\right)=\frac{\mathrm{ln}\:\left({i}\right)}{\mathrm{ln}\:\mathrm{10}}=\frac{\pi}{\mathrm{2}\:\mathrm{ln}\:\mathrm{10}}\:{i} \\ $$$$\mathrm{log}_{\mathrm{2}} \:\left({i}\right)=\frac{\mathrm{ln}\:\left({i}\right)}{\mathrm{ln}\:\mathrm{2}}=\frac{\pi}{\mathrm{2}\:\mathrm{ln}\:\mathrm{2}}\:{i} \\ $$$$... \\ $$
Commented by MWSuSon last updated on 03/May/20
oh I see, change of base. Thank you sir.
Commented by mr W last updated on 03/May/20
$${i}={e}^{\frac{\pi{i}}{\mathrm{2}}} \\ $$$$\mathrm{ln}\:\left({i}\right)=\mathrm{ln}\:\left({e}^{\frac{\pi{i}}{\mathrm{2}}} \right)=\frac{\pi}{\mathrm{2}}{i} \\ $$
Commented by MWSuSon last updated on 03/May/20
$${sir}\:{so}\:{the}\:{solution}\:{of}\:{this}\:{equation} \\ $$$$\mathrm{2}^{\left({x}+\mathrm{1}\right)} ={i} \\ $$$${would}\:{proceed}\:{as}\:{follows}? \\ $$$${x}+\mathrm{1}={log}_{\mathrm{2}} {i} \\ $$$${x}={log}_{\mathrm{2}} {i}−\mathrm{1} \\ $$$${x}=\frac{{lni}}{{ln}\mathrm{2}}−\mathrm{1} \\ $$$${x}=\frac{\pi}{\mathrm{2}{ln}\mathrm{2}}{i}−\mathrm{1}? \\ $$$$ \\ $$
Commented by mr W last updated on 04/May/20
$${yes} \\ $$