Rafael,
Normally for simple expressions you can use Replace[] (also typed /.):
For example:
In[13]:= eqn = exp1 == (r P^(-1))/Log[S/N0]
Out[13]= exp1 == r/(P Log[S/N0])
In[14]:= eqn /. S/N0 -> xx
Out[14]= exp1 == r/(P Log[xx])
HOWEVER, this relies on the pattern matching features of Mathematica and can fail on expressions that simplify in a variety of ways (such as your case with Log[S/N0]/r -> T)
This fails because
In[15]:= FullForm[eqn]
Out[15]Equal[exp1,Times[Power[P,-1],r,Power[Log[Times[Power[N0,-1],S]],-1]]]
EDITED:
Note that your expression has a different form than the original so you can't do a pattern match without getting a bit more clever and account for the simplified terms in the multiplication. Another approach is to use Reduce[]. Reduce is a bit more complicated to use but it does respect the mathematical relationships.
In your case:
In[18]:= Reduce[{exp1 == (r P^(-1))/Log[S/N0], T == Log[S/N0]/r}, {T}]
Out[18]= P Log[S/N0] != 0 && exp1 == r/(P Log[S/N0]) && exp1 != 0 &&
T == 1/(exp1 P)
Which gives you the answer with additional information that the denominator can't go to zero. The downside of this is that you then need to parse the result to pick out what you want.
Regards,
Neil
