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