7.201. $3 \lg 2+\lg \left(2^{\sqrt{x-1}-1}-1\right)=\lg \left(0.4 \sqrt{2^{\sqrt{x-1}}}+4\right)+1$.
Solution.

Domain of definition: $\left\{\begin{array}{l}2^{\sqrt{x-1}-1}-1>0, \\ x-1 \geq 0\end{array} \Leftrightarrow x>2\right.$.

Rewrite the equation in the form

$$
\begin{aligned}
& \lg 8+\lg \left(2^{\sqrt{x-1}-1}-1\right)=\lg \left(0.4 \sqrt{2^{\sqrt{x-1}}}+4\right)+\lg 10 \Leftrightarrow \\
& \Leftrightarrow \lg \left(8 \cdot\left(2^{\sqrt{x-1}-1}-1\right)\right)=\lg \left(4 \sqrt{2^{\sqrt{x-1}}}+40\right) \Leftrightarrow \\
& \Leftrightarrow 8\left(2^{\sqrt{x-1}-1}-1\right)=4\left(\sqrt{2^{\sqrt{x-1}}}+10\right) \Leftrightarrow\left(2^{\frac{\sqrt{x-1}}{2}}\right)^{2}-2^{\frac{\sqrt{x-1}}{2}}-12=0
\end{aligned}
$$

Solving this equation as a quadratic equation in terms of $2^{\frac{\sqrt{x-1}}{2}}$, we get $2^{\frac{\sqrt{x-1}}{2}}=-3$ (no solutions), or $2^{\frac{\sqrt{x-1}}{2}}=2^{2}$, from which $\frac{\sqrt{x-1}}{2}=2, \sqrt{x-1}=4$, $x-1=16, x=17$.

Answer: 17.