1B. Solve the equation

$$
4^{x}+9^{x}=2 \cdot 6^{x}
$$
Solution. We have $4^{x}+9^{x}=2 \cdot 6^{x} \Leftrightarrow 2^{2 x}+3^{2 x}=2 \cdot 2^{x} \cdot 3^{x}$. We will divide the last equation by $2^{x} \cdot 3^{x}$ and get $\left(\frac{2}{3}\right)^{x}+\left(\frac{3}{2}\right)^{x}=2$. Now we introduce the substitution $\left(\frac{2}{3}\right)^{x}=t$ and get $t+\frac{1}{t}=2$, where $t \neq 0$. Further, we get $t^{2}-2 t+1=0$, or $(t-1)^{2}=0$. Thus, $t=1$, from which for $x$ we get $\left(\frac{2}{3}\right)^{x}=1=\left(\frac{2}{3}\right)^{0}$, or $x=0$.