7.058. $\frac{\log _{5}(\sqrt{2 x-7}+1)}{\log _{5}(\sqrt{2 x-7}+7)}=0.5$.
## Solution.

Domain of definition: $2 x-7 \geq 0, x \geq \frac{7}{2}$.

From the condition

$\log _{5}(\sqrt{2 x-7}+1)=\frac{1}{2} \log _{5}(\sqrt{2 x-7}+7) \quad \log _{5}(\sqrt{2 x-7}+1)=\log _{5} \sqrt{\sqrt{2 x-7}+7}$, hence

$\sqrt{2 x-7}+1=\sqrt{\sqrt{2 x-7}+7} \Rightarrow(\sqrt{2 x-7})^{2}+2 \sqrt{2 x-7}+1=\sqrt{2 x-7}+7$, $(\sqrt{2 x-7})^{2}+\sqrt{2 x-7}-6=0$.

Solving this equation as a quadratic equation in terms of $\sqrt{2 x-7}$, we find $\sqrt{2 x-7}=-3, \varnothing$; or $\sqrt{2 x-7}=2$, from which $x=5.5$.

Answer: 5.5.