7.061. $\sqrt{2} \cdot 0.5^{\frac{5}{5^{x}+10}}-16^{\left.\frac{1}{2(\sqrt{x}+1}\right)}=0$.
## Solution.

Domain of definition: $x \geq 0$.

From the condition

$$
2^{\frac{1}{2}} \cdot 2^{-\frac{5}{4 \sqrt{x}+10}}=2^{\frac{2}{\sqrt{x}+1}}, 2^{\frac{1}{2}-\frac{5}{4 \sqrt{x}+10}}=2^{\frac{2}{\sqrt{x}+1}}
$$

it follows that

$$
\frac{1}{2}-\frac{5}{4 \sqrt{x}+10}=\frac{2}{\sqrt{x}+1} \Rightarrow(\sqrt{x})^{2}-3 \sqrt{x}-10=0
$$

Solving this equation as a quadratic equation in $\sqrt{x}$, we find $\sqrt{x}=-2, \varnothing$; or $\sqrt{x}=5$, from which $x=25$.

Answer: 25.