Six. If the sum of the absolute values of the coefficients of the first three terms in the expansion of $\left(x^{18 x}-2\right)^{n}$ is 73, and its middle term is -160000, find the value of $x$.


Six. The first three terms of the expansion are:
$$
\begin{array}{l}
x^{n 1 g x},-2 C_{n}^{1} x^{(n-1) 1 \varepsilon x}, \\
4 C_{n}^{2} x^{(n-2) 1 \varepsilon x} .
\end{array}
$$

From the condition, $1+2 \mathrm{n}+\mathrm{C}_{9}^{2}=73$,
simplifying gives $n^{2}=36$,
$$
\therefore \mathrm{n}= \pm 6
$$
(negative value is discarded).
So the term in the expansion is the $\frac{6}{2}+1=4$ term.
From the condition, $\mathrm{T}_{4}=(-1)^{3} \mathrm{C}_{8}^{3} \cdot 2^{3} \cdot \mathrm{X}^{(6-3) 18 \mathrm{x}}$
$$
=-160000
$$

Thus, $-160 x^{318 x}=-160000$.
Simplifying gives $\lg x= \pm 1$.
From $\lg x=1$, we get $x=10$;
From $\lg x=-1$, we get $x=\frac{1}{10}$,
$$
\therefore \quad x_{1}=10, \quad x_{2}=\frac{1}{10} \text {. }
$$