Let $y = \log x$.
$\begin{array}{rcl}
\dfrac{3}{3y – 2} + 7 & = & \dfrac{2}{2y + 1} \\
3(2y + 1) + 7(3y-2)(2y+1) & = & 2(3y-2) \\
6y + 3 + 42y^2 -7y – 14 & = & 6y – 4 \\
42y^2 – 7y – 7 & = & 0 \\
6y^2 – y – 1 & = & 0 \\
(3y + 1)(2y – 1) & = & 0
\end{array}$
Therefore, $y = \dfrac{-1}{3}$ or $y = \dfrac{1}{2}$.
For $y = \dfrac{-1}{3}$, we have
$\begin{array}{rcl}
\log x & = & \dfrac{-1}{3} \\
-\log x & = & \dfrac{1}{3} \\
\log x^{-1} & = & \dfrac{1}{3} \\
\log \dfrac{1}{x} & = & \dfrac{1}{3}
\end{array}$
For $y = \dfrac{1}{2}$, we have
$\begin{array}{rcl}
\log x & = & \dfrac{1}{2} \\
-\log x & = & \dfrac{-1}{2} \\
\log x^{-1} & = & \dfrac{-1}{2} \\
\log \dfrac{1}{x} & = & \dfrac{-1}{2}
\end{array}$
Hence, $\log \dfrac{1}{x} = \dfrac{1}{3}$ or $\log \dfrac{1}{x} = \dfrac{-1}{2}$.
