It is given that x^y = y^x

Take the logarithm to the base e, or ln, of both the sides

=> ln [ x^y] = ln [ y^x]

Use the property of log that log a^b = b*log a

=> y*ln x = x*ln y

divide both the sides by xy

=> [y*ln x]/xy = [x*ln y]/xy

=> (ln x)/x = (ln y)/y

** If x^y = y^x, then th****is proves that (ln x)/x = (ln y)/y**

It is given that x^y = y^x.

Take the natural logarithm of both the sides:

ln(x^y) = ln(y^x)

Use the property of logarithm ln a^x = x*ln a

y*ln x = x*ln y

Divide both sides by x*y

(y*ln x)/(x*y) = (x*ln y)/x*y)

ln x/x = ln y/y

This proves that ln x/x = ln y/y if x^y = y^x.

x ^ y = y ^ x ... (Given)

Taking log to the base e on both the sides, we get

y* ln (x) = x * ln (y) [as ln (a ^ b) = b*ln(a)]

Hence, **ln(x) /x = ln(y) / y**