Solve a
^{b} = concatenate(a,b) or prove that there is no solution.
For example if 2^{10} were equal to 210, this would be a solution.
No limits are placed on a and b. But (let a&b = a concatonate b)
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 nonnegative 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 nonnegative 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 nonnegative 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 noninteger >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 noninteger ???

Posted by Kenny M
on 20201114 19:48:20 