I was talking with a friend about how to find a more explicit formula for the relationship between the positive numbers $a$ and $b$ in the equation $$ \log_b a = (a/b)^{1/2} $$ after some rearranging we get that this is equivalent to $$ a^{a^{-1/2}}=b^{b^{-1/2}}. $$ The function $f(x)=x^{x^{-1/2}}$ achieves its global maximum of $e^{2/e}$ at $x=e^2$ and is strictly increasing on $(0,e^2)$ and strictly decreasing on $(e^2,\infty)$. Moreover $\lim\limits_{x\to-\infty}f(x)=-\infty,$ $\lim\limits_{x\to-\infty}f(x)=1$ and $f(1)=1$.

Hence for any $a \in (1,e^2)$ there exists exactly one other number $b=b(a) \in (e^2,\infty)$ so that $f(a)=f(b).$

Here is a plot for the number $a$ with $f(a)=2:$

enter image description here

$a$ is the $x$-value of the first intersection of the two lines, $b$ is the $x$-value of the second intersection.

The function $b:(1,e^2)\to(e^2,\infty)$ is differentiable, strictly decreasing and satisfies $\lim\limits_{a\downarrow 1}b(a)=\infty$ and $\lim\limits_{a\uparrow e^2}b(a)=e^2.$ I think it should be possible to find its derivative given what we have but I'm not sure how to do this.

Question: Is it possible to find an explicit expression for the function $b$?

Also, is there a name for this way of defining a function?

  • $\begingroup$ I'm sure there's no closed form for $b$ in terms of powers, exponential, logs, and arithmetical operations on $a$, though I wouldn't be able to prove it. Finding the derivative of $b$ with respect to $a$ is a simple matter of implicit differentiation (but the answer will involve both $b$ and $a$). $\endgroup$ – Gerry Myerson Jun 15 '18 at 9:12

Welcome to the world of Lambert function !

The solutions are given by

$$b=\frac{4 a }{\log ^2(a)}W\left(\pm\frac{\log (a)}{2 \sqrt{a}}\right)^2$$ $$a=\frac{4 b}{\log ^2(b)} W\left(\pm\frac{\log (b)}{2 \sqrt{b}}\right)^2$$

The plot consists in a straight line for $0 \leq a \leq e^2$ corresponding to $b=a$ and for $a > e^2$ it is a decreasing function "looking like an hyperbola".

  • $\begingroup$ Thank you! Lambert function seems right. When I play with the first expression I get: $$b=\frac{4 a }{\log ^2a}W\left(\frac{\log a}{2 \sqrt{a}}\right)^2$$ $$\implies\sqrt b \frac{\log a}{2 \sqrt{a}}=W\left(\frac{\log a}{2 \sqrt{a}}\right)$$ $$\implies\frac{\log a}{2 \sqrt{a}}=W^{-1}\left(\sqrt b \frac{\log a}{2 \sqrt{a}}\right)=\sqrt b \frac{\log a}{2 \sqrt{a}}e^{\sqrt b \frac{\log a}{2 \sqrt{a}}}$$ $$\implies1=\sqrt b e^{\sqrt b \frac{\log a}{2 \sqrt{a}}}$$ $$\implies0=\frac{1}{2}\log b + \frac{\sqrt b}{2 \sqrt{a}}\log a$$ $$\implies\log_b a =-(a/b)^{1/2}$$ Am I doing something wrong? $\endgroup$ – Epiousios Jun 16 '18 at 8:16
  • $\begingroup$ And for $b=\frac{4 a }{\log ^2a}W\left(-\frac{\log a}{2 \sqrt{a}}\right)^2$ I get the contradiction $$-1=\sqrt b e^{\sqrt b \frac{\log a}{2 \sqrt{a}}}$$ in the fourth line of the above. $\endgroup$ – Epiousios Jun 16 '18 at 8:24

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