$7.69 \log _{4 x+1} 7+\log _{9 x} 7=0$.

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$7.69 \log _{4 x+1} 7+\log _{9 x} 7=0$.
7.69 Due to the restrictions imposed on the base of the logarithm, we have the system of inequalities $4 x+1>0$, $4 x+1 \neq 1$, $9 x>0$, $9 x \neq 1$, from which

$x>0, x \neq \frac{1}{9}$.

We will transform the given equation to the form

$\frac{1}{\log _{7}(4 x+1)}+\frac{1}{\log _{7}(9 x)}=0$.

Multiplying both sides of the equation by $\log _{7}(4 x+1) \cdot \log _{7}(9 x) \neq 0$, under the conditions (1), we get

$\log _{7}(9 x)+\log _{7}(4 x+1)=0$, or $\log _{7}(9 x(4 x+1))=\log _{7} 1$, or $9 x(4 x+1)=1$.

From this, we find $x_{1}=-\frac{1}{3}$ (does not satisfy conditions (1)), $x_{2}=\frac{1}{12}$. Therefore, $x=\frac{1}{12}$.

Answer: $x=\frac{1}{12}$.