Is there an analytic solution for the equation $\log_{2}{x}+\log_{3}{x}+\log_{4}{x}=1$? I am looking for a close form solution for below equation.

$$\log_{2}{x}+\log_{3}{x}+\log_{4}{x}=1.$$ 

I solve it by graphing, but I don't know is there a way to find $x$ analytically  ?
It's $$\ln{x}\left(\frac{1}{\ln2}+\frac{1}{\ln3}+\frac{1}{\ln4}\right)=1$$ or
$$\ln{x}=\frac{1}{\frac{1}{\ln2}+\frac{1}{\ln3}+\frac{1}{\ln4}}$$ or
$$x=e^{\frac{1}{\frac{1}{\ln2}+\frac{1}{\ln3}+\frac{1}{\ln4}}}$$
$$ log_v {u}= \dfrac{log_a u}{log_a v} $$

valid for any arbitrary base $a$, need not necessarily be $10$ or $e$.

So

$$\log_a{x}\left(\dfrac{1}{\log_a2}+\dfrac{1}{\log_a3}+\frac{1}{\log_a4}\right)=1$$

or

$$x=a^\left({\dfrac{1}{\dfrac{1}{\log_a2}+\dfrac{1}{\log_a3}+\dfrac{1}{\log_a4}}}\right)=f(a)$$

EDIT1:

apparently but in fact $ \ne any \,f(a)$

Consequently $x$ is indeterminate... as we are free to choose any base.

Sorry, since $ a^{log_a u} = \,u $ it has a cancelling effect ... hastily over sighted on pure visual/structural inspection of the above. (  I wished only to say "consequently $x$ is independent of base... as we are free to choose any base " ) etc. ...because I placed more value on base independence as relevant for this question.

g[u_] = Log[u] (1/Log[3] + 1/Log[4] + 1/Log[2]);
ParametricPlot[{a, a^(1/g(a))}, {a, 1, 12}, GridLines -&gt; Automatic]


that yields a base free  $ x \approx 1.3844.$