How to find the number of solutions of $(0.01)^x=\log_{0.01}x$?
I drew the graph of $a^x$ and $\log_ax$, with $0<a<1$, and thought they intersect just once.
But the answer given is $3$.
Wolfram and Desmos have confirmed it.
All three solutions are between 0 to 1.
I tried taking $x$ as $0.01, 0.1, 0.5$ to see if I could make sense of the graph but not able to do so.
On
If you consider that you look for the zero's of function $$f(x)=a^x-\frac{\log (x)}{\log (a)}$$ you have $$f'(x)=a^x \log (a)-\frac{1}{x \log (a)}$$ which cancels potentially twice at $$x_1=\frac{W\left(\frac{1}{\log (a)}\right)}{\log (a)}\qquad \text{and} \qquad x_2=\frac{W_{-1}\left(\frac{1}{\log (a)}\right)}{\log (a)}$$ $W(.)$ being Lambert function.
These roots exist is the real domain if $$\frac{1}{\log (a)} \geq -\frac 1e\qquad \implies \qquad a \leq e^{-e}$$ which is your case with $a=0.01$.
By the second derivative test, $x_1$ correspond to a maximum and $x_2$ to a minimum. So, three real roots located at $0.0131$, $0.278$ and $0.941$.
If $a=e^{-e}$, two roots (one of them being double and if $a>e^{-e}$ a single root.
On
(Some graphs to complement Claude's answer.)
The scale on the Wolfram graph is bad. Plotting the two sides in Desmos shows the intersections in a much clearer way:
As mentioned by Claude, the choice of the constant $0.01$ is important in determining how many solutions there are, with a transition at $e^{-e}$:
(in the above graph, $c=e^{-e}$ is achieved at $d=-e\approx -2.71$.)
It might be a bit less maddening to work with these functions as $$ f(x) \ = \ 0.01^x \ = \ (10^{-2})^x \ = \ 10^{-2x} \ \ \ \text{and} \ \ \ g(x) \ = \ \log_{0.01} x \ = \ -\log_{100} x \ = \ -\frac12 · \log_{10} x \ \ . $$ This makes it easier to see that $ \ x \ = \ 10^{-2y} \ \Rightarrow \ \log_{10} x \ = \ -2y \ \Rightarrow \ y \ = \ -\frac12·\log_{10} x \ \ , \ $ indicating that $ \ f(x) \ $ and $ \ g(x) \ $ are inverses of one another. The domain and range of $ \ f(x) \ $ are $ \ \mathbb{R} \ $ and $ \ y \ > \ 0 \ \ , \ $ and the domain and range of $ \ g(x) \ $ are $ \ x \ > \ 0 \ $ and $ \ \mathbb{R} \ \ , \ $ respectively. Both functions are continuous on their domains and have values everywhere in the first quadrant, so it is reasonable that one of the roots of the equation is located on the line $ \ y \ = \ x \ \ . \ $ [This is the intersection point given numerically as $ \ ( \ \approx 0.27799 \ , \ \approx 0.27799 \ ) \ \ ] \ . $
The other two solutions are suggested by the Intermediate Value Theorem. We have $ \ f(0) \ = \ 1 \ \ $ and $ \ \lim_{x \ \rightarrow \ 0+} \ g(x) \ = \ +\infty \ \ , \ $ while $ \ f(0.1) \ = \ 10^{-0.2} \ = \ \frac{1}{10^{1/5}} \ \approx \ \frac{1}{1.6} \ \approx \ \frac58 $ $ > \ g(0.1) \ = \ -\frac12·\log_{10}(0.1) \ = \ \frac12 \ \ . \ $ So there is another intersection point in the interval $ \ (0 \ , \ 0.1) \ \ . \ $ Owing to the inverse relation between the functions, there will be a third intersection in the interval where $ \ 0 \ < \ y \ < \ 0.1 \ \ . \ $ Beyond this point, $ \ f(x) \ $ approaches its asymptote, the $ x-$axis, while $ \ g(x) \ $ crosses that line at the intercept $ \ (1 \ , \ 0) \ $ and $ \ \lim_{x \ \rightarrow \ +\infty} \ g(x) \ = \ -\infty \ \ . \ $ These are then the only three intersection points, the two additional ones being $ \ ( \ \approx 0.01309 \ , \ \approx 0.94149 \ ) \ $ and $ \ ( \ \approx 0.94149 \ , \ \approx 0.01309 \ ) \ \ . $
[ADDENDUM -- A little more should perhaps be said to show that there is in fact an intersection on $ \ y \ = \ x \ \ , \ $ in view of Calvin Khor's animation. We can estimate $$ \ f(0.25) \ = \ 10^{-2 \ · \ 0.25} \ = \ \frac{1}{10^{1/2}} \ \approx \ \frac{1}{3.2} \ = \ \frac{1}{16/5} \ = \ \frac{5}{16} \ = \ 0.3125 $$ $$ > \ g(0.25) \ = \ -\frac12·\log_{10}(\frac14) \ = \ \frac12·\log_{10}(2^2) \ = \ \log_{10}(2) \ \approx \ 0.301 \ \ . \ $$ As the functions are mutual inverses, there are numbers $ \ c \ $ in the vicinity of $ \ 0.3 \ $ for which $ \ g(c) \ > \ f(c) \ \ , \ $ so the IVT guarantees the existence of an intersection point of the function curves somewhere in the interval $ \ (0.25 \ , \ 0.3) \ \ . \ ] $