7.221. $4^{\lg x+1}-6^{\lg x}-2 \cdot 3^{\lg x^{2}+2}=0$.

7.221. $4^{\lg x+1}-6^{\lg x}-2 \cdot 3^{\lg x^{2}+2}=0$.
## Solution.

Domain of definition: $x>0$.

From the condition, we have $4 \cdot 2^{2 \lg x}-2^{\lg x} \cdot 3^{\lg x}-18 \cdot 3^{2 \lg x}=0$. Dividing it by $3^{2 \lg x}$, we get $4 \cdot\left(\frac{2}{3}\right)^{2 \lg x}-\left(\frac{2}{3}\right)^{\lg x}-18=0 \Rightarrow\left(\frac{2}{3}\right)^{\lg x}=-2$ (no solutions), or $\left(\frac{2}{3}\right)^{\lg x}=\left(\frac{2}{3}\right)^{-2} \Rightarrow \lg x=-2$. Then $x=10^{-2}=0.01$.

Answer: 0.01.