Question #6c76f

2 Answers
Nov 8, 2016

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

Explanation:

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.

Nov 8, 2016

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

where W is the Lambert function (on Wikipedia)

Explanation:

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)))