Solve the system
5^(y-x)*(y+x) = 1
and
(x+y)^(x-y) = 5
(In reply to
Puzzle Solution: Method 1 by K Sengupta)
Taking logarithm in base 5, we obtain:
log(5)(x+y) = x-y .......(i)
(x-y)*log(5)(x+y) = 1 ....(ii)
Dividing (i) by (ii), we obtain:
1/(x-y) = x-y, so that:
(x-y)^2 = 1, giving:
x-y = +/-1
If x-y = 1, then log(5)(x+y) = 1, so that: x+y = 5
Solving, we have: (x,y) = (3, 2)
If x-y = -1, then log(5)(x+y) = 11, so that: x+y = 1/5
Solving, we have: (x,y) = (-2/5. 3/5).
Consequently, the required solution is given by (x, y) = (3, 2), or (-2/5, 3/5).
Puzzle Solution: Method 2
