Question #6c76f

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Answer 1 · 2016-11-08T19:17:13

#y^(1/y)=(z/x)^(1/x)#

#z/x=y^(x/y)# applying the #log# function

#log(z/x)=x/y log y# or

#1/xlog(z/x)=1/y log y# and finally

#(z/x)^(1/x)=y^(1/y)# this is the best I can do.

Answer 2 · 2016-11-08T20:37:33

#y=1/(e^(W(1/xln(x/z)))#

where W is the Lambert function (on Wikipedia)

Let #h=1/y#

Then #z/x=1/(h^(xh))\ \ \ \ # so #\ \ \ h^(xh)=x/z\ \ \ # and #\ \ \ h^h=(x/z)^(1/x)#

#hlnh=1/xln(x/z)#

Using the example 5 in Wikipedia

#h=e^(W(1/xln(x/z))#

but #\ \ \ h=1/y\ \ \ # so

#y=1/(e^(W(1/xln(x/z)))#