Question Number 127363 by mohammad17 last updated on 29/Dec/20
$${prove}\:{that}\:{log}\left(−\mathrm{1}\right)^{\mathrm{3}} =\mathrm{3}{log}\left(−\mathrm{1}\right)\:? \\ $$
Commented by mohammad17 last updated on 29/Dec/20
$$????? \\ $$
Commented by mr W last updated on 29/Dec/20
$${in}\:\mathbb{C}\:{we}\:{have}\:{always} \\ $$$$\mathrm{log}\:{a}^{{b}} ={b}\:\mathrm{log}\:{a} \\ $$
Commented by mohammad17 last updated on 29/Dec/20
$${Yes},{my}\:{techer},{but}\:{our}\:{doctor}\:{wants}\:{to} \\ $$$${prove}\:{it}\:{in}\:{steps} \\ $$
Answered by ebi last updated on 29/Dec/20
$${let}\:{say}: \\ $$$${z}={log}\left(−\mathrm{1}\right) \\ $$$$\left(−\mathrm{1}\right)={e}^{{z}} −−−{take}\:{exponent}\:{on}\:{both}\:{side} \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{3}} =\left({e}^{{z}} \right)^{\mathrm{3}} −−−{take}\:{power}\:\mathrm{3}\:{on}\:{both}\:{side} \\ $$$$\left(−\mathrm{1}\right)^{\mathrm{3}} ={e}^{\mathrm{3}{z}} \\ $$$${log}\left(−\mathrm{1}\right)^{\mathrm{3}} =\mathrm{3}{z}−−−{take}\:{log}\:{on}\:{both}\:{side} \\ $$$${log}\left(−\mathrm{1}\right)^{\mathrm{3}} =\mathrm{3}\:{log}\:\left(−\mathrm{1}\right)−−−{substitute}\:{z}={log}\left(−\mathrm{1}\right) \\ $$$$ \\ $$$$−−−{Generalize}−−− \\ $$$${m}={log}_{{a}} \left({x}\right) \\ $$$${x}={a}^{{m}} −−−{convert}\:{to}\:{exponential}\:{form} \\ $$$${x}^{{n}} =\left({a}^{{m}} \right)^{{n}} −−−{take}\:{power}\:{n}\:{on}\:{both}\:{side} \\ $$$${x}^{{n}} ={a}^{{mn}} \\ $$$${log}_{{a}} \left({x}\right)^{{n}} ={mn}\:−−−\:{convert}\:{to}\:{logarithm}\:{form} \\ $$$${log}_{{a}} \left({x}\right)^{{n}} ={n}\:{log}_{{a}} \left({x}\right)\:−−−\:{substitute}\:{m}={log}_{{a}} \left({x}\right) \\ $$
Commented by mr W last updated on 29/Dec/20
$${z}={log}\left(−\mathrm{1}\right)\neq{log}\left({i}\right) \\ $$
Commented by mohammad17 last updated on 29/Dec/20
$${can}\:{you}\:{give}\:{me}\:{stebs} \\ $$