2. Solve the equation $\frac{1}{4^{x}+1}+\frac{1}{2^{x} \cdot 3^{x}-1}=\frac{2^{x}}{4^{x} \cdot 3^{x}-2 \cdot 2^{x}-3^{x}}$.
Solution.

The equation is defined on the domain $D=\mathbb{R}^{*} \backslash\left\{x \in \mathbb{R} \mid 4^{x} \cdot 3^{x}-2 \cdot 2^{x}-3^{x}=0\right\}$. (1p)

The equation can be written in the form

$$
2^{x}\left(6^{x}+3^{x}+2^{x}-1\right)\left(6^{x}-3^{x}-2^{x}-1\right)=0, x \in D
$$

It follows that $6^{x}+3^{x}+2^{x}-1=0$ or $6^{x}-3^{x}-2^{x}-1=0$. (2p)

i. We solve in $\mathbb{R}$ the equation $6^{x}+3^{x}+2^{x}-1=0$ or $6^{x}+3^{x}+2^{x}=1$.

Let the function $f: \mathbb{R} \rightarrow(0, \infty), f(x)=6^{x}+3^{x}+2^{x}$. $f$ is strictly increasing on $\mathbb{R}$ and $f(-1)=1$. We deduce that the equation $f(x)=1, x \in \mathbb{R}$, has the unique root $x_{1}=-1$. (2p)

ii. We solve the equation $6^{x}-3^{x}-2^{x}-1=0$, equivalent to $f(-x)=1(x \in \mathbb{R})$. We deduce that the considered equation has the unique root $x_{2}=1$.

Since $-1,1 \in D$, the given equation has the solutions $x_{1}=-1$ and $x_{2}=1$. (2p)