7.223. $\log _{5}\left(2^{1.5 x-2.5}+2^{1.5 x-0.5}-0.01 \cdot 5^{3 x+1}\right)=3 x-1$.
## Solution.

Domain of definition: $2^{1.5 x-2.5}+2^{1.5 x-0.5}-0.01 \cdot 5^{3 x+1}>0$. By the definition of the logarithm, we get

$$
\begin{aligned}
& 2^{1.5 x-2.5}+2^{1.5 x-0.5}-0.01 \cdot 5^{3 x+1}=5^{3 x-1} \Leftrightarrow \\
& \Leftrightarrow \frac{2^{1.5 x}}{2^{2.5}}+\frac{2^{1.5 x}}{2^{0.5}}-0.05 \cdot 5^{3 x}=\frac{5^{3 x}}{5} \Leftrightarrow \frac{2^{1.5 x}}{2^{2.5}}+\frac{2^{1.5 x}}{2^{0.5}}=\frac{5^{3 x}}{5}+\frac{5^{3 x}}{20} \Leftrightarrow \\
& \Leftrightarrow \frac{2^{1.5 x}}{2^{2.5}}=\frac{5^{3 x}}{2^{2} \cdot 5} \Leftrightarrow 2^{1.5 x-0.5}=5^{3 x-1} \Rightarrow\left\{\begin{array}{l}
1.5 x-0.5=0, \\
3 x-1=0,
\end{array}\right.
\end{aligned}
$$

from which $x=\frac{1}{3}$.

Answer: $\frac{1}{3}$.