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:

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

2. Where does the constant $k$ come from?

3. Does it have an expression?

4. Or is this all a very interesting coincidence?

Answer 1

Yes, it can be. Notice that

$${y=10^{-x}\geq 0}$$

Hence ${\log(10^{-x})}$ is well defined, and so

$${10^{-x}=e^{\log(10^{-x})}=e^{-x\log(10)}=e^{-\log(10)x}}$$

And so

$${k=\log(10)\approx 2.30258509....}$$

Answer 2

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

Answer 3

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

Answer 4

To answer your questions one after another

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

Well, you yourself discovered $y= 10^{-x} = e^{-2.30258509299405x}$

2. 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?)

3. 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 $\ln 10 \approx 2.30258509299405....$

4. 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.