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

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?

• "Where does the constant k come from?" Plug $e^{2.30258509299405}$ into you calculator. I think you will be pleasantly surprised. Jul 22, 2020 at 1:25

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

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

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

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

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

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

1. 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?)

1. 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 $$\ln 10 \approx 2.30258509299405....$$

1. 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.