Logx/y-z=logy/z-x=logz/x-y find x^x.y^y.z^z=?

Question

14461 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-04T19:11:09

#x^x.y^y.z^z=1#

#logx/(y-z)=logy/(z-x)=logz/(x-y) =k#

Therefore,

#logx=(y-z)k# #color(white)(ddddd# #logy=(z-x)k# #color(white)(ddddd# #logz=(x-y)k#

Assume, #A=x^x.y^y.z^z#

Taking log,

#logA=log(x^x.y^y.z^z)#

#=>logx^x+logy^y+logz^z#

#=>xlogx+ylogy+zlogz#

#=>x(y-z)k+y(z-x)k+z(x-y)k#

#=>xky-xkz+ykz-ykx+zkx-zky#

#=>0#

#=> logA=0#

#=>x^x.y^y.z^z=1#