Solve the following:

1) 2^x+y = 6^y

2) 3^x = 3(2^y+1)

1)   2^x+y  = 6^y

taking log on both sides 

(x+y)log2 = ylog6 

log6 = log3*2 = log3+log2

(x+y)log2 = y(log2+log3)

 xlog2 = ylog3

  2^x = 3^y

solution is only one , (x,y) = (0,0)