$7.57 \quad 4^{\lg x+1}-6^{\lg x}-2 \cdot 3^{\lg x^{2}+2}=0$.
7.57 Given that $\lg x^{2}=2 \lg x$ (since $x>0$), we will transform the given equation as follows:

$$
\begin{aligned}
& 2^{2 \lg x+2}=2^{\lg x} \cdot 3^{\lg x}-2 \cdot 3^{2 \lg x+2}=0, \text { or } \\
& 4 \cdot 2^{2 \lg x}-2^{\lg x} \cdot 3^{\lg x}-18 \cdot 3^{2 \lg x}=0 .
\end{aligned}
$$

Divide both sides of the last equation by $3^{2 \lg x} \neq 0$:

$4\left(\frac{2}{3}\right)^{2 \lg x}-\left(\frac{2}{3}\right)^{\lg x}-18=0$.

Let $\left(\frac{2}{3}\right)^{\lg x}=y>0$, we get the quadratic equation $4 y^{2}-y-18=0$, the roots of which are $y_{1}=\frac{9}{4}$ and $y_{2}=-2$ (not suitable, since $y>0$). Therefore, $\left(\frac{2}{3}\right)^{\lg x}=\frac{9}{4}=\left(\frac{2}{3}\right)^{-2}$, from which $\lg x=-2$, i.e., $x=0.01$.

Answer: $x=0.01$.