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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
$$x_1=frac{W_0left(frac{1}{log (a)}right)}{log (a)}qquad text{and}qquad x_2=frac{W_{-1}left(frac{1}{log (a)}right)}{log (a)}$$ 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 $lim_{xto 0} , f(x)=infty$. 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
$$left(
begin{array}{cc}
a & text{ root} \
0.10 & 0.399013 \
0.15 & 0.436709 \
0.20 & 0.469622 \
0.25 & 0.500000 \
0.30 & 0.528956 \
0.35 & 0.557154 \
0.40 & 0.585043 \
0.45 & 0.612961 \
0.50 & 0.641186 \
0.55 & 0.669965 \
0.60 & 0.699535 \
0.65 & 0.730133 \
0.70 & 0.762013 \
0.75 & 0.795457 \
0.80 & 0.830785 \
0.85 & 0.868378 \
0.90 & 0.908699 \
0.95 & 0.952326
end{array}
right)$$

For convenience's sake set $a=1/b$ so that $bin(1,infty)$. 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 $sim15.16$.