Hello let : $ X=2^x$ and $ Y=3^x$ and $ (E): 9^x+4^x+1=6^x+3^x+2^x$ $ (E) \Leftrightarrow X^2+Y^2 +1 = XY+X+Y$ on the other hand : $ X^2+Y^2 +1\geq 2XY+1$ Hence , $ (X-1)(1-Y) \geq 0$ therefore : $ X-1 \leq 0$ and $ Y-1 \geq 0$ Or $ X-1 \geq 0$and $ Y-1 \leq 0$ From That inequalities : $ x=0$ By introducing expononciel function ( $ 2^x= e^{xln(2)}$ ) *Excuse me for me poor english

Another way is to write it as $ \frac{(3^x-2^x)^2 + (3^x-1)^2 + (2^x-1)^2}{2} = 0,$ after which $ x = 0$ is obviously the only solution.