Question 962792
The entire expression as one single term?


{{{E^2=ln(e^(E^2))}}}, makes use of the inverse relationship between exponential and logarithm functions.


{{{E^2-4ln(x)+ln(y)}}}


FIXING MY MISTAKE:
{{{cross(ln(e^(E^2))-ln(x)+ln(y))}}}
{{{ln(e^(E^2))-ln(x^4)+ln(y)}}}


{{{cross(ln(e^(E^2))+ln(y)-ln(x))}}}
{{{ln(e^(E^2))+ln(y)-ln(x^4)}}}


{{{ln(e^(E^2)ln(y/x^4))}}}----Now corrected.


The argument INSIDE the natural log function IS  {{{highlight(e^(E^2)ln(y/x^4))}}}.