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Yesterday a question popped in my mind before going to sleep and I can't solve it (probably because my math knowledge is lacking). Assuming there is a function

$$f(x) = a^x$$

Its inverse function would look like this:

$$f^{-1}(x) = \log_a x$$

Question:

For which value of $a$ would the graphs of these two functions just touch at only one common point (of tangency)?

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    $\begingroup$ I would prefer that you wrote $f(x)=a^x$ and $f^{-1}(y)=\log_a y$. Also notice this forces $a>0, y>0$. $\endgroup$
    – user284001
    Jul 24, 2018 at 15:18

5 Answers 5

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Basically, the inverse function is obtained by finding the mirror image of $y=a^x$, in $y=x $ .Now note for both the function and its inverse to intersect at just one point, $y=x $ must be the common tangent to both of them. So slope of tangent of $y=a^x $ ,$y=\log_a x$ and $y=x $ must be equal. So,$ 1=a^x \ln a=\ln x{\log_a e } $ now you have 3 equations and 2 unknowns one being 'a' and other being the value of $x $ at which the functions have only one point of intersection . solve finally to get $a^e=e $ which gives the value of $a $ approximately as 1.44.

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  • $\begingroup$ Excellent, +1. I love MSE because it's constantly reminding me how dumb I know to be from time to time. $\endgroup$
    – Saša
    Jul 24, 2018 at 18:21
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As @Jasmine point out, because that $a^x$ and $\log_ax$ are inverses we can add $x$ to get $x=a^x=\log_ax$.

We can say more from this: $a^x=\log_a x\implies a^{a^x}=x$ but also $x=a^x$ hence $a^{a^x}=\log_a x\implies a^{a^{a^x}}$ and so on.

Hence we are searching for $a,x$ such that the sequence $x,a^x,a^{a^x},\cdots$ is a constant, $a\in\Bbb R^+\setminus\{1\},x\in \Bbb R^+$.

As it is shown here we have that $e^{-e} \leq a \leq e^{\frac{1}{e}}$ for $a^{a^{.^{.^{.}}}}$ to converge, hence, by taking the sequence and taking the $1/x$ power of it we get $x^{\frac1x}\in[e^{-e},e^{\frac1e}]$, by guessing $e$ we can find that $x=e,a=e^{\frac1e}$ works. But $x=e$ is not the only $x$ that works. Take any $x$ such that $x^{\frac1x}\in[e^{-e},e^{\frac1e}]$ it can be shown that $x,a=x^{\frac1x}$ are tuple that answer the condition of the post.


Edit, I realized that I made a mistake.

[$(a^{a^{...^x}})^{1/x}=(a^{a^{...^x}/x})\ne (a^{a^{...}})$, although it is interesting to see that if you continue with the logic of the mistake and find the $x$s where the 2 function "kiss" you get correct answer]

But given $x=a^x$ we have $x^{1/x}=a$(In analogy to the mistake, the maximum value of $x^{1/x}$ is $e^{1/e}$ at $x=e$).

As I said in the post $a=x^{\frac1x}$ gives us $a^y=\log_ay=y$, at $y=x$, but we also want them to be tangent so $(a^y)'\mid_{y=x}=\ln(a)a^x=\frac1{\ln(a)x}=(\log_a(y))'\mid_{y=x}$, we know that $a^x=x$ so $\ln(a)a^x=\ln(a)x$, let set $\ln(a)x=z$ and we have $z=\frac1z\implies z=\pm1$, and so $\ln(a)x=\pm1\implies\ln(x^{1/x})x=\pm1\implies\ln(x)=\pm1\implies x=\begin{cases}e\\1/e\end{cases}\\\implies a=\begin{cases}e^{1/e}\\(1/e)^{e}\end{cases}$

(I think that OEIS made a mistake about uniqueness)

See here the graph of the four functions

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Such $a$ likely exists and is somewhere in the following range which can be narrowed down further: $$1.44\lt a \lt 1.45$$

enter image description here

EDIT: $$a\approx1.444667861, \qquad x\approx 2.71827$$

Note that $x$ value is very close to $e$. I don't know how to explain it but it makes this problem really, really interesting!

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  • $\begingroup$ Thank you very much for your answer! But is there any possible way to solve this without using a graphing program? $\endgroup$
    – Ozzy
    Jul 24, 2018 at 16:00
  • $\begingroup$ I doubt it, really. $\endgroup$
    – Saša
    Jul 24, 2018 at 16:20
  • $\begingroup$ Oh okay.. Thank you anyways $\endgroup$
    – Ozzy
    Jul 24, 2018 at 16:33
  • $\begingroup$ I think the key question is: prove or disprove that touching point has $x=e$ . That would be something :) I feel like I'm missing something obvious. $\endgroup$
    – Saša
    Jul 24, 2018 at 16:36
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According to a OEIS comment, the only base $a$ such that $\exists \xi \in A \subset\mathbb R: f(\xi)=f^{-1}(\xi)$ is $a=e^{1/e}$ with $\xi=e$. Numerically this seems to be true, as Oldboy pointed out.

Whether that's unique (for any $a$) needs to be proven.

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I found this to have a solution algebraically: $$a=e^{\frac{1}{e}}$$ For $f(x)$ to be tangent to $f^{-1}(x)$ and to $y=x$, it must have a tangent line with slope $1$ that has a $y$-intercept of $0$. $$f(x)=\log_{a}(x)=\frac{\ln(x)}{\ln(a)}$$ $$f'(x)=\frac{1}{x\cdot\ln(a)}$$ To find $k$ such that $f'(k)=1$ (in terms of a): $$f'(k)=\frac{1}{k\cdot\ln(a)}=1$$ $$k=\frac{1}{\ln(a)}$$ The equation of a tangent line at $(k,f(k))$ is $y-f(k)=f'(k)(x-k)$ $$y-\frac{\ln\left(\frac{1}{\ln(a)}\right)}{\ln(a)}=1\cdot\left(x-\frac{1}{\ln(a)}\right)$$ $$y=x-\frac{1}{\ln(a)}+\frac{\ln\left(\frac{1}{\ln(a)}\right)}{\ln(a)}=x-\frac{1}{\ln(a)}-\frac{\ln(\ln(a))}{\ln(a)}$$ Since the $y$-intercept must be $0$: $$\frac{1}{\ln(a)}+\frac{\ln(\ln(a))}{\ln(a)}=0$$ $$1+\ln(\ln(a))=0$$ $$\ln(\ln(a))=-1$$ $$\ln(a)=\frac{1}{e}$$ $$a=e^{\frac{1}{e}}$$

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