# How many solutions are there for the equation $a^x = \log_a x$, where $0 < a < 1$?

How many solutions are there for the equation $$a^x = \log_a x$$, where $$0 < a < 1$$?

When I first saw this quiz for japanese high school students, I wondered there was only 1 solution for the equation for any $$0 < a < 1$$.

But I was wrong:

Then, for what values of $$a$$ such that $$0 < a < 1$$ are there 3 solutions for the equation?

Too complex for a quiz.

Consider that you look for the zero's of function $$f(x)=a^x-\frac{\log (x)}{\log (a)}$$ Its derivative is given by $$f'(x)=a^x \log (a)-\frac{1}{x \log (a)}$$ this cancels at two points given by $$x_1=\frac{W_0\left(\frac{1}{\log (a)}\right)}{\log (a)}\qquad \text{and}\qquad x_2=\frac{W_{-1}\left(\frac{1}{\log (a)}\right)}{\log (a)}$$ where appears Lambert function. In the real domain, we need $$a \lt e^{-e}$$. When this is the case, $$f(x_1)>0$$ and $$f(x_2) < 0$$ and in this range $$\lim_{x\to 0} \, f(x)=\infty$$. So, for $$0 < a < e^{-e}$$, there are three roots (the first one between $$0$$ and $$x_1$$; the second one between $$x_1$$ and $$x_2$$; the third one above $$x_2$$); for $$a=e^{-e}$$, there is a triple root and for $$a>e^{-e}$$, there is a single root.

Edit

Since this is an interesting numerical problem, I give you below the three roots for a faw values of $$a$$ $$\left( \begin{array}{cccc} a & \text{first root} & \text{second root} & \text{third root} \\ 0.00500 & 0.005883 & 0.256675 & 0.969312 \\ 0.01000 & 0.013093 & 0.277987 & 0.941488 \\ 0.01500 & 0.021585 & 0.292615 & 0.913335 \\ 0.02000 & 0.031462 & 0.304205 & 0.884194 \\ 0.02500 & 0.042894 & 0.314008 & 0.853652 \\ 0.03000 & 0.056133 & 0.322619 & 0.821327 \\ 0.03500 & 0.071532 & 0.330371 & 0.786783 \\ 0.04000 & 0.089601 & 0.337471 & 0.749451 \\ 0.04500 & 0.111117 & 0.344056 & 0.708514 \\ 0.05000 & 0.137359 & 0.350225 & 0.662661 \\ 0.05500 & 0.170721 & 0.356048 & 0.609472 \\ 0.06000 & 0.216898 & 0.361580 & 0.543230 \\ 0.06500 & 0.303124 & 0.366862 & 0.436682 \\ 0.06510 & 0.306379 & 0.366965 & 0.433018 \\ 0.06520 & 0.309837 & 0.367069 & 0.429151 \\ 0.06530 & 0.313538 & 0.367172 & 0.425041 \\ 0.06540 & 0.317536 & 0.367275 & 0.420633 \\ 0.06550 & 0.321911 & 0.367378 & 0.415848 \\ 0.06560 & 0.326787 & 0.367481 & 0.410562 \\ 0.06570 & 0.332376 & 0.367584 & 0.404564 \\ 0.06580 & 0.339098 & 0.367686 & 0.397432 \\ 0.06590 & 0.348099 & 0.367789 & 0.388021 \\ 0.06591 & 0.349246 & 0.367799 & 0.386833 \\ 0.06592 & 0.350471 & 0.367810 & 0.385567 \\ 0.06593 & 0.351791 & 0.367820 & 0.384206 \\ 0.06594 & 0.353233 & 0.367830 & 0.382723 \\ 0.06595 & 0.354836 & 0.367840 & 0.381079 \\ 0.06596 & 0.356672 & 0.367851 & 0.379202 \\ 0.06597 & 0.358881 & 0.367861 & 0.376952 \\ 0.06598 & 0.361865 & 0.367871 & 0.373927 \end{array} \right)$$

For $$a=e^{-e}$$, the triple root is $$0.367882$$.

For the case of a single root $$\left( \begin{array}{cc} a & \text{ root} \\ 0.10 & 0.399013 \\ 0.15 & 0.436709 \\ 0.20 & 0.469622 \\ 0.25 & 0.500000 \\ 0.30 & 0.528956 \\ 0.35 & 0.557154 \\ 0.40 & 0.585043 \\ 0.45 & 0.612961 \\ 0.50 & 0.641186 \\ 0.55 & 0.669965 \\ 0.60 & 0.699535 \\ 0.65 & 0.730133 \\ 0.70 & 0.762013 \\ 0.75 & 0.795457 \\ 0.80 & 0.830785 \\ 0.85 & 0.868378 \\ 0.90 & 0.908699 \\ 0.95 & 0.952326 \end{array} \right)$$

• Thank you very much, Claude Leibovici. My imperfect solution agrees with your solution. I guess your solution is perfect, but unfortunately, I don't know Lambert function at all. – tchappy ha Dec 1 '18 at 9:23
• @tchappyha. It is a fantastic function with a lot of applications. Search on this site (MSE) and have a look on Wikipedia. – Claude Leibovici Dec 1 '18 at 9:26
• Thank you very much for the information, Claude Leibovici. – tchappy ha Dec 1 '18 at 9:27
• IMO this post should be the answer, nice shot. Could you detail in what sense root is triple when $x=e^{-e}$ (I guess it is a kind of multiplicity such as for polynomials)? Thank you. – jlandercy Dec 1 '18 at 19:21
• Thank you very much again for computation, Claude Leibovici. – tchappy ha Dec 2 '18 at 0:07

For convenience's sake set $$a=1/b$$ so that $$b\in(1,\infty)$$. The equation becomes solving

$$b^{-x}=-\log_b(x).$$

Let us restrict attention to $$x>0$$, because that's where all the roots lie (if any). In addition the LHS is always positive, hence the RHS is as well, so any root is in $$(0,1)$$. Now rewrite the equation in its equivalent form,

$$x=\frac1{b^{b^{-x}}}.$$

Proceed to study the expression on the right, and consider its gradient at its point of intersection with $$y=x$$ (the one which is "always there'). Hence when the gradient is greater than $$1$$, then the graph had to "cross over" the line $$y=x$$, and then "cross back"; there are three solutions. Otherwise, there is one unique solution. For a visualisation, check this Desmos plot. Unfortunately, there's no nice expression for the value beyond which $$b$$ has three solutions, but the numerical value is around $$\sim15.16$$.

• Could you try for a=0.05 – Atharva Kathale Dec 1 '18 at 2:52
• Thank you very much, YiFan. – tchappy ha Dec 1 '18 at 9:20