1) a,b must be real or they won't concatenate

2) if a<0, a must be an integer, then b must be a non-negative integer, therefore -a, which is >0 in this case, is also a solution, so we can limit our search to look for solutions in the non-negative real numbers for a.

3) a cannot be zero, becasue 0^0 is undefined

4) 0<a<1 is not a solution, becasue b must be a non-negative integer to concatenate with a decimal, and in this case a^b decreases while a&b increases.  b=0 never works

5) a>=1  implies  b must be >=0 or a&b won't exist, a=1 is not a solution

5a) a= integer > 1.  b= integer=0 or 1 are never solutions for any such a

5b) a=integer >1, b= integer>=2.  Easy to see no solutions by inspection

5c) a=integer>1, b=non integer>0  ????

5d) a=rational non-integer >1.  b must be an integer>1 ????

5e) a= irrational number >1,b must be integer >1 ????

SO

 - a is a non integer >1, b must be an integer>1 - SUSPECT NO SOLUTION?

OR

a=integer>1, b=positive non-integer ???