What is the value of $a$ for which $\frac{1}{\text{log}_2a} + \frac{1}{\text{log}_3a} + \frac{1}{\text{log}_4a} = 1$?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 36$
We use the change of base formula to show that
\[\log_a b = \dfrac{\log_b b}{\log_b a} = \dfrac{1}{\log_b a}.\]
Thus, our equation becomes
\[\log_a 2 + \log_a 3 + \log_a 4 = 1,\]
which becomes after combining:
\[\log_a 24 = 1.\]
Hence $a = 24$, and the answer is $\textbf{(D)}.$