7.242. $\left(16 \cdot 5^{2 x-1}-2 \cdot 5^{x-1}-0.048\right) \lg \left(x^{3}+2 x+1\right)=0$.
Solution.

Domain: $x^{3}+2 x+1>0$.

From the condition $16 \cdot 5^{2 x-1}-2^{x-1}-0.048=0$ or $\lg \left(x^{3}+2 x+1\right)=0$. Rewrite the first equation as

$\frac{16}{5} \cdot 5^{2 x}-\frac{2}{5} \cdot 5^{x}-0.048=0 \Leftrightarrow 16 \cdot 5^{2 x}-2 \cdot 5^{x}-0.24=0$.

Solving this equation as a quadratic in terms of $5^{x}$, we get $5^{x}=-\frac{3}{40}$ (no solutions), or $5^{x}=5^{-1} \Leftrightarrow x_{1}=-1$ (does not satisfy the domain). From the second equation, we have

$x^{3}+2 x+1=1 \Leftrightarrow x^{3}+2 x=0 \Leftrightarrow x\left(x^{2}+2\right)=0, x_{3}=0, x^{2}+2 \neq 0$.

Answer: 0.