$7.26 \frac{\log _{5}(\sqrt{2 x-7}+1)}{\log _{5}(\sqrt{2 x-7}+7)}=0.5$.
7.26 Given the domain of the square root, we conclude that $2 x-7 \geq 0$. For these values of $x$, we have $\sqrt{2 x-7} \geq 0 ; \quad \sqrt{2 x-7}+1>0, \quad \sqrt{2 x-7}+7>0 \quad$ and $\sqrt{2 x-7}+7 \geq 7$, i.e., $\log _{5}(\sqrt{2 x-7}+7) \neq 0$. Multiplying both sides of the equation by $\log _{5}(\sqrt{2 x-7}+7)$, we get

$\log _{5}(\sqrt{2 x-7}+1)=0.5 \log _{5}(\sqrt{2 x-7}+7)$, or

$\log _{5}(\sqrt{2 x-7}+1)=\log _{5} \sqrt{\sqrt{2 x-7}+7}$.

According to hint $4^{0}$, we transition to the system

$$
\left\{\begin{array}{l}
2 x-7 \geq 0 \\
\sqrt{2 x-7}+1=\sqrt{\sqrt{2 x-7}+7}
\end{array}\right.
$$

Introducing a new variable $\sqrt{2 x-7}+1=y>0$, we arrive at the equation $y=\sqrt{y+6}$ or $y^{2}-y-6=0$, which has roots $y_{1}=3$ and $y_{2}=-2$ (the latter does not satisfy the condition $y>0$). Thus, we are left to solve the equation $\sqrt{2 x-7}+1=3$, from which $x=5.5$, which satisfies the inequality $2 x-7 \geq 0$.

Answer: $x=5.5$.