Let $x>0$ be real. Find all solutions of $x^{0.3} = \ln x $ or more general equation $x^a= \ln x$ , where $ 0 < a < 1 $.

  • $\begingroup$ By taking $\ln$, this equation becomes $0.3 \ln x = \ln(\ln x)$ : it seems that it's impossible to find an analytic solution. $\endgroup$ – paf Aug 9 '16 at 17:48
  • $\begingroup$ I don't know any analytical solutions. If numerical ones are ok, then you are looking for fixed points of the function $f(x) = \sqrt[\alpha]{\ln x}$ (or of a suitable local inverse, if that's more convenient). $\endgroup$ – avs Aug 9 '16 at 17:49
  • $\begingroup$ I don't think it is easy to express in terms of elementary functions. if $a=1$ you could use the W function. $\endgroup$ – mathreadler Aug 9 '16 at 17:51
  • $\begingroup$ WolframAlpha gives two solutions with Productlog function and its analytic continuation. One is $x = e^{\frac{-10}{3}W(-.3)}$ and $ x = e^{-\frac{10}{3}W_{1}(-.3)}$, where $W(z)$ is the product log function and $W_k(z)$ is the analytic continuation of product log function. $\endgroup$ – Mathsira Aug 9 '16 at 17:51

Lets solve it:

$$ x^a = \ln x \\ e^{x^a} = x \\ x = e^{x^a} \\ x^a = e^{a x^a} \\ a x^a = a e^{a x^a} \\ a x^a e^{-a x^a} = a \\ -a x^a e^{-a x^a} = -a $$

Let $s = -a x^a $. We then have the equation: $$ s e^s = -a $$

Thus we can solve for $s$ if we say: $s = W(a)$, where $W$ is the inverse function of $f(s) = s e^s$, that is, $W = f^{-1}$, also called, Lambert-W Function. Be careful, there are two branchs in the real domain.

Therefore, we can now continue: $$ -a x^a = W\left(-a\right) $$

Thus: $$ x^a = -\frac{1}{a} W\left(-a\right) $$


Solutions of $x^a = \ln(x)$ are $(-W(-a)/a)^{1/a}$ where $W$ is a branch of the Lambert W function. In the case $a= 3/10$, there are two real solutions (corresponding to the principal and $-1$ branches of Lambert W), approximately $5.110722365$ and $379.0962301$.

EDIT: There are two real solutions if $0 < a < 1/e$, one if $a < 0$, and none if $a > 1/e$.

  • $\begingroup$ Thank you. Could you please write the solution with more explanations ? $\endgroup$ – Mathsira Aug 9 '16 at 18:01
  • $\begingroup$ Physicist137 explained it well. $\endgroup$ – Robert Israel Aug 9 '16 at 18:24

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