Constant e: The Unique Intersection of Exponential and Linear Functions

[Jhenrique]

Find a number "k" such that exist only one intersection between the curve exponential k^x e and the straight x·k. This number is the constant e!

Have you noticed this? This relationship must have many important implications and indirect that we already know...

 

[PSarkar]

The $e$ after $x^k$ is unclear. Is it $x^{ke}$ or $x^k \cdot e$?

 

[Jhenrique]

Correction:

"between the curve exponential k^x and the straight x·k"

I press "e" not wanting... sorry!

 

[PSarkar]

Interesting. Do you know the proof? If you don't then I can post it if you are interested.

 

[Jhenrique]

I don't know, I just observed... Yes, I'm interesting!

 

[PSarkar]

Actually, there is also only one intersection for $0 < k \leq 1$. Anyway, here's the proof:

Let $k > 0$. For $0 < k \leq 1$, we have $k^x$ constant or decreasing while $kx$ is increasing. Hence it is easy to show that there is only one intersection, in fact at $x = 1$.

Now assume $k > 1$ and let $f(x) = k^x - kx$ so that $f(x) = 0$ whenever $k^x = kx$. Now we can notice a few things.

Firstly, as $x \to -\infty$, we have $k^x \to 0$ and $kx \to -\infty$, and so $f(x) \to \infty$.

Secondly, as $x \to \infty$, we have $\frac{k^x}{kx} \to \infty$ since $k > 1$ and so $f(x) \to \infty$.

Lastly, we can see that $f(1) = k^1 - 1 \cdot k = 0$ for any $k$.

Considering the limits shown above, if $f(x) < 0$ for some $x$, then there must be at least two zeros of $f$ by the intermediate value theorem. Thus, if there is only one intersection, we must have $f(x) \geq 0$ for all $x$. So $f$ is a minimum at its only zero (which must be $1$ by our last observation above). Since $f$ differentiable, this means we must have $f'(1) = 0$. Since $f'(x) = \log(k)k^x - k$ so the criteria $f'(1) = 0$ implies $\log(k)k - k = 0$. But $k > 0$ and so $\log(k) = 1 \implies k = e$ which you observed.