Solved by googlebot in post #4 
              
              
                logsumexp exists to tackle this case using identity: 
log(exp(a)+exp(b)) = c + log(exp(a-c) + exp(b-c)) 
c=max(a,b) 
You can adapt this for scaling and mean with: K*exp(a) = exp(log(K))*exp(a) = exp(a+log(K)) 
Or just use .clamp() on problematic tensors.