"4^x+6^x=9^x
两边除9^x
(4/9)^x+(6/9)^x=1
[(2/3)^x]^2+(2/3)^x-1=0
令a=(2/3)^x
a^2+a-1=0
a=(-1-√5)/2,a=(-1+√5)/2
a=(2/3)^x>0
所以a=(-1+√5)/2
(2/3)^x=(-1+√5)/2
所以x=log(2/3) [(-1+√5)/2]