$7.28 \frac{2-\lg 4+\lg 0.12}{\lg (\sqrt{3 x+1}+4)-\lg 2 x}=1$.

$7.29 0.5\left(\lg \left(x^{2}-55 x+90\right)-\lg (x-36)\right)=\lg \sqrt{2}$.
7.28 Given the domains of the logarithmic function and the square root, we have the system of inequalities $3 x+1 \geq 0$ (under this condition $\sqrt{3 x+1}+4>0$), $x>0$, the solution of which is $\boldsymbol{x}>0$. This equation is equivalent to the system

$$
\left\{\begin{array}{l}
x>0 \\
\lg (\sqrt{3 x+1}+4)-\lg 2 x \neq 0 \\
2-\lg 4+\lg 0.12=\lg (\sqrt{3 x+1}+4)-\lg 2 x
\end{array}\right.
$$

Using the hint $5^{0}$, we transition to the equation

$$
\lg \frac{100 \cdot 0.12}{4}=\lg \frac{\sqrt{3 x+1}+4}{2 x}, \text{ or } 3=\frac{\sqrt{3 x+1}+4}{2 x}
$$

After multiplying both sides of the equation by $2 x$ (no extraneous roots will be introduced since $x>0$), we obtain the irrational equation

$$
\sqrt{3 x+1}=6 x-4
$$

Since $\sqrt{3 x+1} \geq 0$, then $6 x-4 \geq 0$, from which $x \geq \frac{2}{3}$.

Square both sides of equation (2):

$3 x+1=36 x^{2}-48 x+16$, or $36 x^{2}-51 x+15=0$.

From here, we find $x_{1}=1, x_{2}=\frac{5}{12}$ (does not satisfy the condition $x \geq \frac{2}{3}$). It remains to verify that for $x=1$, the second condition of system (1) is satisfied: $\lg (\sqrt{3 x+1}+4)-\lg 2 x=\lg 6-\lg 2=\lg 3 \neq 0$. Thus, $x=1$.

Answer: $x=1$.