1) Taking logs of both equations gives

(A+B)*lnA = 12*ln(B)

3*ln(A) = (A+B)*lnB

2) Dividing LHS by LHS and RHS by RHS is legitimate unless ln(A) - ln(b) = 0, and checking we see that (1,1) is in fact a solution. Otherwise,

(A+B)/3 = 12/(A+B)

3) Rearranging and solving gives A+B = +/- 6

Clearly, A+B cannot equal -6, so A+B = 6

4) Substituting (A+B) = 6 in either equation and taking roots gives

A = +/- B^2

5) If A and B are both positive, the only solution is {4,2)

If B is negative, the only solution is (9, -3)

And there can be no solution where A is negative, because that makes abs(A) less than B, so A cannot = -B^2

6) (0,0) is not a solution because 0^0 is undefined.

(-1,-1) satisfies the first equation but not the second.

Only solutions are (1,1) and (4,2) and (9,-3)

*Edited on ***March 29, 2015, 10:11 am**