# Number of real solutions for this logarithmic equation [duplicate]

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.

• did you make a plot of the functions ? Commented Aug 5, 2019 at 15:40
• @GCab I did, but I'm interested in a more analytical solution. Commented Aug 5, 2019 at 15:49
• It's not duplicate, of course! In the linked question the topic starter solved this problem and looks for possibility to find a third root by some formula. In the starting topic looked for solution of the problem at all. Commented Aug 6, 2019 at 2:27
• Commented Aug 6, 2019 at 7:48

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 $$0, we have, that there exist at least one solution for any $$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 $$x=.25, x=.36424..., x=.5$$

• Thanks. Quick question: is there a way to analytically solve f(x), or otherwise intuitively "know" that it has three x-intersects without plotting it? Commented Aug 5, 2019 at 17:44
• @Jane: Assuming I didn't make a mistake (I did this quickly, and need to get back to something else), the equation can be put into the form $e^{ax+b} = \ln x$ for real number constants $a$ and $b,$ and I strongly suspect this equation can't be solved in general, although of course it still might be solvable (in terms of elementary functions evaluated at "elementary numbers") for the specific values of $a$ and $b$ that correspond to your equation. For example, it's solvable for $a = 0,$ although that doesn't help in your case. Commented Aug 5, 2019 at 17:58
• @DaveL.Renfro I see. That's frustrating. I was given this question under circumstances where I didn't have access to a calculator/plotting device, so I wasn't sure how I was supposed to reason out that there were three intercepts without plotting it. It looks like analytically solving it might not be a clear option, either. Commented Aug 5, 2019 at 18:42
• with the assumption of $x=2^k$ we can easily get the two solutions $x=1/2$ and $x=1/4$ then considering the end behavior of the function $y=16^x \ln x$ we realize that we need the third solution as well. Commented Aug 5, 2019 at 19:56
• @Jane: My comment had to do with whether any or all of the solutions can be expressed explicitly in some form (ratio of integers, quadratic surds, real radicals, radical form, are algebraic numbers, Chow's closed form numbers, etc.), which is what you first comment just above seems to be asking, and I wasn't dealing with simply determining the number of real solutions. I think that can be taken care of rather easily by hand-sketches based on rigorous real analysis properties, such as intermediate value property, behavior of exponentials and logs, etc. Commented Aug 6, 2019 at 7:37