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Is is possible to solve the equation $\ln x=a+bx^{-1}$ using the Lambert W function? I understand that the lambert W function is the solution for equations like $\ln x=bx^{-1}$, which does not apply here.

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  • $\begingroup$ According to CAS the solution is: ${{\rm e}^{{\it LambertW} \left( {\frac {b}{{{\rm e}^{a}}}} \right) +a} }$. $\endgroup$
    – joro
    Commented Oct 10, 2014 at 7:16

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Yes it seems so, $\ln x=a+bx^{-1}$ being equivalent to $$\ln (x/e^a) = (b/e^a)/(x/e^a). $$

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