There is an answer when $c_1=0$ at (Solving $\frac{\log(x)}{x}=c$, where $c < e^{-1}$). How could we solve the following?
$$\frac{\log(x)+c_1}{x}=c_2, \text{where } x > 1,~c_1>0, ~\text{and}~ c_2 < e^{-1}$$
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You can have a solution in terms of Lambert W function. – Mhenni Benghorbal Sep 05 '14 at 01:01
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@MhenniBenghorbal It makes sense to have such a solution, but could you explain it more? – Elrond Gimli Sep 05 '14 at 01:07
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Have you give it a try? – Mhenni Benghorbal Sep 05 '14 at 01:36
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See the answer. – Mhenni Benghorbal Sep 05 '14 at 01:39
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@MhenniBenghorbal Thanks! – Elrond Gimli Sep 05 '14 at 01:52
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Gimili : You are welcome. – Mhenni Benghorbal Sep 05 '14 at 01:52
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Related problems. Here is how. Make the change of variables $\ln(x)=u$
$$ \frac{\log(x)+c_1}{x}=c_2 \implies (u+c_1){e^{-u}}=c_2 . $$
Another change of variables $ u+c_1 = z $ gives
$$ ze^{-z}= c_2e^{-c_1} $$
Replace $z=-y$ gives
$$ ye^y=-c_2e^{-c_1} \implies y = W(-c_2e^{-c_1}), $$
where $W(x)$ is the Lambert W function. Now you need to go backward with your substitutions. The final answer should be
$$ x =e^{-c_1}\, e^{-W(-c_2e^{-c_1})} $$
Mhenni Benghorbal
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There is a typo, which is missing the negative sign for $c_1$ before the final answer $$ ye^y=-c_2e^{-c_1} \implies y = W(-c_2e^{-c_1}) $$ – Elrond Gimli Sep 05 '14 at 02:24
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