7.230. $\log _{x+1}(x-0.5)=\log _{x-0.5}(x+1)$.
Solution.

Domain of definition: $\left\{\begin{array}{l}0<x+1 \neq 1, \\ 0<x-0.5 \neq 1\end{array}\right.$ or $0.5<x \neq 1.5$.

Multiplying both sides of the equation by $\log _{x+1}(x-0.5) \neq 0$, we get

$$
\begin{aligned}
& \log _{x+1}^{2}(x-0.5)=1 \Rightarrow \log _{x+1}(x-0.5)=-1 \Rightarrow \\
& \Rightarrow x-0.5=\frac{1}{x+1}, 2 x^{2}+x-3=0, x_{1}=-\frac{3}{2} \text { (does not satisfy the domain of definition), }
\end{aligned}
$$

$x_{2}=1 ;$ or $\log _{x+1}(x-0.5)=1, x-0.5=x+1$, no solutions.

Answer: 1 .