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

1 Answer
Apr 4, 2018

xx.yy.zz=1

Explanation:

logxyz=logyzx=logzxy=k

Therefore,

logx=(yz)k ddddd logy=(zx)k ddddd logz=(xy)k

Assume, A=xx.yy.zz

Taking log,

logA=log(xx.yy.zz)

logxx+logyy+logzz

xlogx+ylogy+zlogz

x(yz)k+y(zx)k+z(xy)k

xkyxkz+ykzykx+zkxzky

0

logA=0

xx.yy.zz=1