How many solutions $x\in\Bbb R$ does the equation $$(1/16)^x=\log_{1/16}x$$ have?
I've tried several methods to manipulate this equation, but to little avail.
Edit: I'm looking for a solution that does not require plotting the curve. I've heard about the Lambert W function, but if there's a way to reason this out that doesn't require "happening" to know this piece of information, I'd be delighted.
Thanks so much for your help.
You have an equation of type $a^x = \log_a(x)$. Clearly, due to this log, we can consider only $x>0$. In our specific case, $a=\frac{1}{16}$, so we can consider only $a\in(0,1)$. For such $a$, $\forall_{x \ge 1} \log_a(x) \le 0$. So again, we can consider only $x \in (0,1)$.
For such $a,x$ equation $a^x = \log_a(x)$ is equivalent to equation $a^{a^x} = x$.
Let $id_{(0,1)}:(0,1) \to (0,1)$, $id_{(0,1)}(x) = x$. We know the behavior of this function and its simple derivative on $(0,1)$. For any $a\in(0,1)$ define $f_a : (0,1) \to (0,1)$, $f_a(x) = a^{a^x}$. Then $f_a'(x) = a^{a^x}a^x \ln^2(a) $ Which clearly is positive for any $a,x \in (0,1)$. So $f_a$ is increasing for any $a \in (0,1)$
Now, due to $id_{(0,1)}(0+) = 0, id_{(0,1)}(1-) = 1, f_a(0+) = a, f_a(1-) = a^a$ ( by for example $0+$ I mean the limit as $ x \to 0^+$), and $0<a<a^a<1$, we have, that there exist at least one solution for any $a\in(0,1)$.
Notice, that if $y$ is a solution, then $\theta_a(y) = a^y$ is also a solution, because $a^{a^{\theta_a(y)}} = \theta_a(y)$ gives us $a^{a^{a^y}} = a^y$, and using $a^{a^y} = y$, we get $a^y = a^y$, which is true. However, we cannot simply "create" infinite number of solutions, because $\theta_a(\theta_a(y)) = y$ (due to $y$ being the solution to $a^{a^y}$. What I want to say, is that, unless one solution $y_0$ which is also a solution to the equation $a^y = y$ ( because then $\theta_a(y) = y$ ) (and this equation has a solution, due to decreasing behavior of $a^y$ and its limits at $0^+,1^-$) we have just for any other solution $y \neq y_{0}$ another solution $\theta_a(y)$. So the number of solutions to our equation must be odd (since we already know there is at least one, and every other goes in pairs).
Well, looking at $(y,\theta_a(y))$ we see that ( I'm still and will be considering case $y \neq y_0$ of course) exactly one of $y,\theta_a(y)$ is less than $y_0$ and one is bigger than $y_0$ (if for example $y < y_0$ then $a^y > a^{y_0}$ ( due to $a \in (0,1)$) but $a^{y_0} = y_0$ as it is special solution). So it is enought (and even reasonable to avoid double counting) to look (for any $a \in (0,1)$) at interval $(0,y_0)$ (here $y_0$ of course depend on $a$).
Let's look at the second derivative of $f_a$. $f_a''(x) = \frac{d}{dx}(a^{a^x+x}\ln^2(a)) = a^{a^x+x}\ln^3(a)(a^x\ln(a) + 1)$ Sign of it depends only on the term $a^x\ln(a) + 1$ which is $<0$ for $x < \frac{-1}{\ln(a)} = c_a$. Due to sign of $f_a''$, we have that function $f_a$ is convex on $(0,c_a)$ and concave on $(c_a,1)$. Last thing, look at $d_a(x) = a^{a^x} - x$ and it's derivative $d_a'(x) = a^{a^x}a^x\ln^2(a) - 1$. We need to determine number of solutions to $a^{a^x+x}\ln^2(a)=1$. Term involving $x$ is $a^x + x$, which is decreasing on $(0,c_a)$ and increasing on $(c_a,1)$ (we've already examined behavior of $a^x\ln(a) + 1$ which is a case there). So $d_a'$ has at most 2 solutions, and that means $d_a$ can have at most $3$ solutions.
Now when we know, that possible number of solutions is either $1$ or $3$, all we need to examine (due to limits at $x=0$ and $x=1$ ) is the special $a_0$ for which $f_{a_0}(y_0) = id_{(0,1)}(y_0)$ and $f'_{a_0}(y_0) = 1$ (because there are only $2$ cases, either $f_a$ "cuts" curve $(x,id_{(0,1)}(x))$ at point $y_0$ from above (and then there is only one solution) or cuts that curve from below and there are $3$ solutions (due to the fact, that it must have cut that curve before to be below it).
So (1) : $a_{0}^{a_0^{y_0}} = y_0$ and (2): $a_{0}^{a_0^{y_0}+y_0}\ln^2(a_0) = 1$.
Plugging (1) into (2) we get $y_0 \cdot a_0^{y_0} \cdot \ln^2(a_0) = 1$. But $y_0$ is also a special solution, so the solution to equation(3): $a^y = y$, and that means:
$(y_0 \ln(a_0))^2 = 1$, so $y_0 \ln(a_0) = -1$ => $ y_0 = \frac{-1}{\ln(a_0)}$ and plugging it into our (3) we get : $\frac{-1}{\ln(a_0)} = (a_0)^{-\frac{1}{\ln(a_0)}}$
By swapping and times numerator/denominator, we get: $-\ln(a_0) = a_0^{\frac{1}{\ln(a_0)}}$ and letting $a_0 = e^{z}$, we see that $-z = (e^{z})^{\frac{1}{z}} = e$, so $a_0 = e^{-e}$.
To conclude, for $a \in (0,e^{-e})$ we have $3$ solutions, and for $a \in [e^{-e},1)$ we have $1$ solution. In our case $\frac{1}{16} < e^{-e}$, because $e^e < 16$, we do have $3$ solutions.
You have three solutions.
Note that $$(1/16)^x=\log_{1/16}x\iff 16^x\ln x = \ln \frac {1}{16}$$
The function $$ f(x) = 16^x\ln x$$ is an $S$-shaped curve intersecting $y=\ln \frac {1}{16}$ at three points.
The solutions are $$x=.25, x=.36424..., x=.5$$
