16. Given that $a$ and $b$ are the two positive roots of the equation $\log _{3 x} 3+\log _{77} 3 x=-\frac{4}{3}$. Then the value of $a+b$ is ().
(A) $\frac{4}{27}$
(B) $\frac{10}{27}$
(C) $\frac{4}{81}$.
(D) $\frac{10}{81}$
(E) $\frac{28}{81}$
16.D.

Let $t=1+\log _{3} x$. Transform the original equation to contain only logarithms with base 3, we get
$$
\begin{array}{l}
\frac{\log _{3} 3}{\log _{3} 3+\log _{3} x}+\frac{\log _{3} 3+\log _{3} x}{\log _{3} 27}=-\frac{4}{3} \\
\Leftrightarrow \frac{1}{t}+\frac{t}{3}=-\frac{4}{3} \Leftrightarrow t^{2}+4 t+3=0 \\
\Leftrightarrow(t+1)(t+3)=0 .
\end{array}
$$

Therefore, $2+\log _{3} x=t+1=0$ or $4+\log _{3} x=t+3=0$.
So, $\log _{3} x=-2$ or $-4, x=\frac{1}{9}$ or $\frac{1}{81}$.
Thus, $\frac{1}{9}-\frac{1}{81}=\frac{10}{81}$.