1. Solve the equations:

(i) $\frac{1}{2^{x}}+\frac{1}{3^{x}}=\frac{1}{2^{x}+3^{x}}, x \in \mathbb{R}$.

(ii) $x+2^{x}+\log _{2} x=7, x \in(0, \infty)$.
(i) The equation is equivalent to $\left(2^{x}+3^{x}\right)^{2}=4 \cdot 2^{x} \cdot 3^{x}, n \in \mathbb{N}^{\text {* }}$

From where $2^{x}=3^{x}$ and then $x=0$

(ii) The function $f:(0, \infty) \rightarrow \mathbb{R}$ defined by $f(x)=x+2^{x}+\log _{2} x$ is a strictly increasing function.

$f(2)=7$