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I am looking for a close form solution for below equation.


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 $$x=e^{\frac{1}{\frac{1}{\ln2}+\frac{1}{\ln3}+\frac{1}{\ln4}}}$$

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    $\begingroup$ For incognito, which up-votes my answers. If you indeed so like my solutions then do it please not so often, otherwise the system removes your work. Thank you! $\endgroup$ – Michael Rozenberg Oct 10 '17 at 5:13

Let $y = log_{10}x$


$y(\frac{1}{log2}+\frac{1}{log3}+\frac{1}{log4}) = 1$

Find y

and then $x = 10^y$


Hint: $$\log[b](x) = \dfrac{\ln(x)}{\ln(b)}$$


$$\tag1 \log_{2}{x}+\log_{3}{x}+\log_{4}{x}=1$$

$$\tag2 \frac{\log{x}}{\log{2}}+\frac{\log{x}}{\log{3}}+\frac{\log{x}}{\log{4}} = 1$$

$$\tag3 \left(\frac{1}{\log{2}}+\frac{1}{\log{3}}+\frac{1}{\log{4}}\right)\log{x} = 1$$

$$\tag4 \left(\frac{1}{\log{2}}+\frac{1}{\log{3}}+\frac{1}{\log{4}}\right) = \frac{1}{\log{x}}$$

$$\tag5 \left(\frac{1}{\log{3}} + \frac{3}{\log{4}}\right) = (\log{x})^{-1}$$

$$\tag6 \left(\frac{1}{\log{3}} + \frac{3}{\log{4}}\right)^{-1} = \log{x}$$

$$\tag7 10^{\left(\frac{1}{\log{3}} + \frac{3}{\log{4}}\right)^{-1}}= x$$

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

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    $\begingroup$ The step from line 3 to line 4 is interesting. Congrats to the upvoters. $\endgroup$ – Yves Daoust Oct 4 '17 at 8:03
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    $\begingroup$ Lines $1$ and $2$ are fair enough, the remaining needs some revision. $\endgroup$ – fundamentalform Oct 4 '17 at 8:15
  • $\begingroup$ Your are right. I have updated my answer. Thanks! $\endgroup$ – Darío A. Gutiérrez Oct 4 '17 at 9:23

Just to give a slightly different approach, let $x=e^u$. Then




Note: $e$ here can be any positive number, not just $2.718281828\ldots$. For example, we might write


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


Since $\log_34$ is actually bigger than $1$, the exact value of $x$ is somewhat less than $\sqrt2$. Closer calculation gives



$$ log_v {u}= \dfrac{log_a u}{log_a v} $$

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






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.$

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    $\begingroup$ Are you serious ? (By the way, $x=1.3844171570591...$.) $\endgroup$ – Yves Daoust Oct 4 '17 at 8:04
  • $\begingroup$ Edited answer as above. $\endgroup$ – Narasimham Oct 5 '17 at 16:40

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