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  ?
 A: Let $y = log_{10}x$
Then 
$y(\frac{1}{log2}+\frac{1}{log3}+\frac{1}{log4}) = 1$
Find y
and then $x = 10^y$
A: Hint: $$\log[b](x) = \dfrac{\ln(x)}{\ln(b)}$$
A: $$\tag1 
\log_{2}{x}+\log_{3}{x}+\log_{4}{x}=1$$ 
$$\tag2
\frac{\log{x}}{\log{2}}+\frac{\log{x}}{\log{3}}+\frac{\log{x}}{\log{4}} = 1$$
$$\tag3
\left(\frac{1}{\log{2}}+\frac{1}{\log{3}}+\frac{1}{\log{4}}\right)\log{x} = 1$$
$$\tag4
\left(\frac{1}{\log{2}}+\frac{1}{\log{3}}+\frac{1}{\log{4}}\right) = \frac{1}{\log{x}}$$
$$\tag5
\left(\frac{1}{\log{3}} + \frac{3}{\log{4}}\right) = (\log{x})^{-1}$$
$$\tag6
\left(\frac{1}{\log{3}} + \frac{3}{\log{4}}\right)^{-1} = \log{x}$$
$$\tag7
10^{\left(\frac{1}{\log{3}} + \frac{3}{\log{4}}\right)^{-1}}= x$$

$$10^{\frac{1}{\frac{1}{\log{3}} + \frac{3}{\log{4}}}}= x$$

A: 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
$$x=e^{\frac{1}{\frac{1}{\ln2}+\frac{1}{\ln3}+\frac{1}{\ln4}}}$$
A: Just to give a slightly different approach, let $x=e^u$.  Then
$$1=\log_2x+\log_3x+\log_4x=u(\log_2e+\log_3e+\log_4e)$$
so
$$x=e^u=e^{1/(\log_2e+\log_3e+\log_4e)}$$
Note: $e$ here can be any positive number, not just $2.718281828\ldots$.  For example, we might write
$$x=4^{1/(\log_24+\log_34+\log_44)}=4^{1/(3+\log_34)}$$
Since $3\lt4\lt9$, we have $1\lt\log_34\lt2$, with the value of $\log_34$ closer to $1$ than it is to $2$.  If we use the crude approximation $\log_34\approx1$, we get
$$x\approx4^{1/4}=\sqrt2\approx1.414$$
Since $\log_34$ is actually bigger than $1$, the exact value of $x$ is somewhat less than $\sqrt2$. Closer calculation gives
$$x\approx4^{1/4.2618595}\approx4^{0.234639}\approx1.384416$$
A: $$ log_v {u}= \dfrac{log_a u}{log_a v} $$
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 -> Automatic]

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