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

Domain of definition: $\left\{\begin{array}{l}0<4 x+1 \neq 1, \\ 0<9 x \neq 1\end{array} \Leftrightarrow 0<x \neq \frac{1}{9}\right.$.

Switch to base 7. We have

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

$\Leftrightarrow 9 x=\frac{1}{4 x+1} \Leftrightarrow 36 x^{2}+9 x-1=0$,

from which $x_{1}=\frac{1}{12}, x_{2}=-\frac{1}{3} ; x_{2}=-\frac{1}{3}$ does not satisfy the domain of definition.

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