There are multiple cases. For example, if #p=ln2#, say, then
#Q=Ae^ln2=2A#
with a possible solution #x=2# and #y=1#.
This question is somewhat vague as if #p# is a constant or a variable. For example, if #p=1#, then the answer is no. There are no rational numbers #x# and #y# such that #e=x^y#, because #e# is irrational.
Case 1: p is a constant
If we allow #p# to be a constant, then
#p=ylnx#
Proves sufficient to replace the term #e^p# with #x^y#, #x,y in QQ#.
Case 2: p is a variable
We can't say that #p=ylnx# anymore, because #y# and #x# would also have to be variables.
Let #p# be a function of some other variable, #alpha#.
#p=f(alpha)#
As #x#, #y# and #e# are all constants, it means that #p# can only be written in terms of them and not #alpha#. As such, if #p# is a variable, the answer is no.
