Can $y=10^{-x}$ be converted into an equivalent $y=mathrm{e}^{-kx}$?

Question

I was dealing with the values:
| Digits | Expression | Value                 |
|--------|------------|-----------------------|
| 1      | 10⁻¹       | 0.1                   |
| 2      | 10⁻²       | 0.01                  |
| 3      | 10⁻³       | 0.001                 |
| 4      | 10⁻⁴       | 0.0001                |
| 5      | 10⁻⁵       | 0.00001               |
| 6      | 10⁻⁶       | 0.000001              |
| 7      | 10⁻⁷       | 0.0000001             |
| 8      | 10⁻⁸       | 0.00000001            |
| 9      | 10⁻⁹       | 0.000000001           |
| 10     | 10⁻¹⁰      | 0.0000000001          |
| 11     | 10⁻¹¹      | 0.00000000001         |
| 12     | 10⁻¹²      | 0.000000000001        |
| 13     | 10⁻¹³      | 0.0000000000001       |
| 14     | 10⁻¹⁴      | 0.00000000000001      |
| 15     | 10⁻¹⁵      | 0.000000000000001     |

And then I plotted the results in Excel on a log scale:

Now, I already know the formula for this graph, it's:
$$ y = 10^{-x} $$
But was curious to see how well an "exponential" trendline would fit, and it fits very well:

The $R^2$ is $1$, even for $15$ decimal places.
So it seems that:
$$y = 10^{-x} ↔ y = e^{-2.30258509299405x} $$
The question
So I have to wonder:

is there an algebraic transformation of: $$y = 10^{-x} → y = e^{-kx} $$

Where does the constant $k$ come from?

Does it have an expression?

Or is this all a very interesting coincidence?

fleablood · Answer

To answer your questions one after another
1)is there an algebraic transformation of: $y = 10^{-x} → y = e^{-kx}$
Well, you yourself discovered $y= 10^{-x} = e^{-2.30258509299405x}$

Where does the constant $k$ come from?

That constant $k$ is the unique value $k$ so that $e^k =10$.
If $e^k = 10$ then $10^{-x} = (e^k)^{-x} = e^{-kx}$.
If $b > 0$ and $b ne 1$ and if $M > 0$ there will always be one unique $k$ so that $b^k = M$.  And so there is a unique $k$ so that $e^k =10$.  That $k$ is $approx 2.30258509299405....$. (It's actually an irrational number... that's not surprising, is it?)

Does it have an expression?

Yes.  $k = ln 10$.
If $k$ is the unique number so that $b^k = M$ we refer to $k$ as $log_b M$.  If $b = e$ we call this the "natural logarithm" and write it as $k = ln M = log_b M$.
As it turns out $ln 10 approx 2.30258509299405....$

Or is this all a very interesting coincidence?

Not in the least bit a coincidence.  But absolutely interesting.  And VERY important.  $ln {}$ is one of the most important functions there is.

Nikodem · Answer

$ y = 10^{-x} = e^{log(10^{-x})} = e^{-x log(10)} $
so
$k = log(10) = 2.3025...$

Riemann'sPointyNose · Answer

Yes, it can be. Notice that
$${y=10^{-x}geq 0}$$
Hence ${log(10^{-x})}$ is well defined, and so
$${10^{-x}=e^{log(10^{-x})}=e^{-xlog(10)}=e^{-log(10)x}}$$
And so
$${k=log(10)approx 2.30258509....}$$

runway44 · Answer

Hint. $10=e^{ln 10} phantom{stuff}$

Can $y=10^{-x}$ be converted into an equivalent $y=mathrm{e}^{-kx}$?

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