I'm looking for an example of a transcendental equation with a unique "nice" solution that can be identified as correct by inspection. My first thought was $$3^x = x^3$$ but this equation has two solutions, one of which is trivial and the other of which requires $W$-functions.

However, it seems to me that if I introduce a constant factor on one side of the equation, I ought to be able to get the two function $f(x)=3^x$ and $g(x)=kx^3$ to be tangent to each other at a single point of intersection. More broadly, it seems like if $a$ and $b$ are any positive numbers (and $b$ is odd) there should be some constant $k$ such that $a^x = kx^b$ has a unique positive solution.

I doubt there's a simple, closed form solution for this general problem that doesn't involve $W$ functions, but I don't really care about the general case; I just want one clear example. So the question:

Can you provide a specific example of an equation of the form $x^a = k\cdot b^x$ for which $a, b, k$ are reasonably "nice", and the equation has a unique solution that's also reasonably "nice"? Alternatively, can you confirm that no such example exists?

  • $\begingroup$ desmos.com/calculator/oseyzn7ffs $\endgroup$ – Mason Nov 8 '18 at 3:53
  • $\begingroup$ @Mason This is what I get for not specifying what I mean by "nice". Sure, I know how to find a numerical approximation for $k$. I'd like an example where $k$ is simple enough that you can verify the solution by hand. $\endgroup$ – mweiss Nov 8 '18 at 3:58
  • $\begingroup$ Is $x=a=b=e,k=1$ nice enough? $\endgroup$ – Ian Nov 8 '18 at 4:01
  • $\begingroup$ I thought maybe someone would enjoy/benefit from a graph illustrating the general idea of your question. I think I well-enough understand your question. $\endgroup$ – Mason Nov 8 '18 at 4:02

Telling you what you already know: $a^x$ "starts" at a value of $1$ (when $x=0$), whereas $k x^b$ starts at $0$. Since exponentials out-pace powers, if $k x^b$ were to cross $a^x$, from below to above, then it would have to cross back again later. Therefore, we must keep the functions from crossing at all: where they meet, they must be tangent. (Also, to keep the power below the exponential for $x<0$, we should require $b$ to be a odd integer, and $k$ to be some positive value.)

Okay, then ... There must be a place, say, $x=c > 0$, where the curves' values match and their slopes match, so we consider the relation and its derivative:

$$a^c = k c^b \qquad\text{and}\qquad a^c\log a = b k c^{b-1} \tag{1}$$ These imply $$k c^b \log a = b k c^{b-1} \quad\to\quad c\log a = b \tag{2}$$ where $c$ and $a$ are such that $b$ is that odd integer we need. Therefore, $$a^c = k c^{c\log a}\quad\to\quad k = \left(\frac{a}{c^{\log a}}\right)^c \tag{3}$$ giving the desired relation this form:

$$a^x = \left(\frac{a}{c^{\log a}}\right)^c x^{c\log a} \tag{4}$$

The solution, $x=c$, is as nice as you want it to be, although $a$ could be a little ugly to get that integer in $(2)$. Optimal niceness seems to be achieved by taking, $a=e$, so that $\log a = 1$, and we have

$$e^x = \left(\frac{e}{c}\right)^c\cdot x^c \tag{$\star$}$$

where we restrict $c$ itself to a nice odd integer. (Note that $c=1$ gives the nicest of all possible worlds.)

Here's a plot with $c=3$:

enter image description here

  • $\begingroup$ Got it. Perhaps worth mentioning that we need $c$ to be odd; otherwise there would be a second, negative solution. $\endgroup$ – mweiss Nov 8 '18 at 4:21
  • $\begingroup$ Ah, good point. $\endgroup$ – Blue Nov 8 '18 at 4:23

Following your suggestion of tangency, we need $$x_0^a=kb^{x_0}\ ,\qquad ax_0^{a-1}=kb^{x_0}\ln b$$ which implies $$a=x_0\ln b\ .$$ Because of the logarithm, it seems unlikely that there are any "nice" possibilities other than $b=e$. This leads to various possibilities such as $$x^2=4e^{x-2}$$ which has a unique positive solution but also has a negative solution, and $$x^3=27e^{x-3}$$ which perhaps will do what you want.

  • $\begingroup$ I like the solution, but I am puzzled by the fact that the first example ($x^2 = 4e^{x-2}$) has two solutions. If the two functions are tangent to one another, shouldn't there be a unique solution? $\endgroup$ – mweiss Nov 8 '18 at 4:11
  • $\begingroup$ @mweiss No, look at $x^3$ and $x^2$: they intersect and are tangent at $x=0$ but they have another intersection at $x=1$. Tangency is necessary for uniqueness in this situation however. $\endgroup$ – Ian Nov 8 '18 at 4:14
  • $\begingroup$ Oh, of course! Silly of me, thanks. $\endgroup$ – mweiss Nov 8 '18 at 4:15

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