5. Let $a, b, c$ be positive numbers, and $3^{a}=4^{b}=6^{c}$, then ( ).
(A) $\frac{1}{c}=\frac{1}{a}+\frac{1}{b}$
(B) $\frac{2}{c}=\frac{2}{a}+\frac{1}{b}$
(C) $\frac{1}{c}=\frac{2}{a}+\frac{2}{b}$
(D) $\frac{2}{c}=\frac{1}{a}+\frac{2}{b}$
5. B

Let $3^{a}=4^{b}=6^{c}=k$, then $a=\log _{k} 3, b=\log _{4} k, c=$ $\log _{6} k$, at this time
$$
\begin{array}{l}
\frac{2}{a}+\frac{1}{b}=2 \log _{k} 3+\log _{k} 4=\log _{k} 36, \\
\frac{2}{c}=2 \log _{k} 6=\log _{k} 36, \\
\therefore \frac{2}{c}=\frac{2}{a}+\frac{1}{b} .
\end{array}
$$