How to solve 1. 4^x+6^x=9^x ?

2. x-2/x-1 = 1-2/x-1 ?

Hi Giridhar,

2^2x+2^x3^x=3^2x

divide both sides by 2^x3^x

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

1/y+1=y

y²-y-1=0

y=(1+√5)/2 _ve is not valid.

(3/2)^x=(1+√5)/2

x=[log(1+√5)/2]/[log 3/2]