7.184. $\log _{a} x+\log _{a^{2}} x+\log _{a^{3}} x=11$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}x>0, \\ 0<a \neq 1 .\end{array}\right.$

Switch to base $a$. We have $\log _{a} x+\frac{1}{2} \log _{a} x+\frac{1}{3} \log _{a} x=11 \Leftrightarrow$ $\Leftrightarrow \log _{a} x=6$, from which $x=a^{6}$.

Answer: $a^{6}$, where $0<a \neq 1$.