What gives the natural logarithm its name? What's "natural" about it?

Question

I know that the natural logarithm is defined as $\ln(x)$ or $\log_{e}(x)$, where $e$ is the Euler number.

But what is so "natural" about it? Is there an explanation on why that name was chosen?

Answer 1

Base $e$ for exponentials and logarithms is distinguished by the fact that it is the unique base $c$ such that the derivative of $f(x)=c^x$ at $x=0$ is equal to one.

Answer 2

Another way of looking at it is that, to find the instantaneous rate of change of a function, f(x) we take the limit of (f(x+h)- f(x))/h as h goes to 0. In particular, for $f(x)=a^x$, that is $\frac{a^{x+h}- a^x}{h}= \frac{a^xa^h- a^x}{h}= a^x\left(\frac{a^h- 1}{h}\right)$. Notice that we have $a^x$, with no h, times $\left(\frac{a^h- 1}{h}\right)$ which has no x.

The limit of that is $a^x\left(\lim_{h\to 0}\frac{a^h- 1}{h}\right)$.

Whatever a or x is that limit is a number so we can say:

The derivative of $a^x$ is a number, that depends on a, times $a^x$.

Now what is that number? Look at some examples. If a= 1 then a^h= 1 for all h so $\frac{1- 1}{h}= 0$.

If a= 2 and suppose h= 0.001, then $a^h= 2^{0.001}= 1.000693387$ (hallelujah for calculators!) so $2^h- 1= 0.000693387$ and $\frac{2^h- 1}{0.001}= 0.6934. The limit as h goes to 0 is close to 0.69.

If a= 3, h= 0.001, then $a^h= 3^{0.001}= 1.001099$, $3^{0.001}-1= 0.001099$ and $\frac{3^{0.001}- 1}{0.001}= 1.099$. Again the limit as h goes to 0 is close to 1.099.

Notice that for x= 2, that value is less than 1 and for x= 3, that limit is greater than 1. Don't you suspect that there is an a, between 2 and 3, such that it is exactly 1?

For one more check, try a= 2.7. $a^h= 2.7^{0.001}= 1.00099374521$, $2.7^{0.001}- 1= 0.00099374521$, and $\frac{2.7^{0.001}- 1}{0.001}= 0.99374521$.

Now that's awfully close to 1 and still just a little less than 1. We can conjecture that there is a number, slightly larger than 2.7, which I am going to call, for no particular reason, $e$, for which $\lim_{h\to 0}\frac{e^h- 1}{h}= 1$. So the derivative of $e^x$ is just $e^x$ again! The function $f(x)= e^x$ is its own derivative!

Of course, any time we have a function, f(x), would like to have it "inverse" function, $f^{-1}(x)$, so that we can solve equations- to solve f(x)= a, take $f^{-1}(f(x))= x= f^{-1}(a)$.

$ln(x)$, the "natural logarithm" is defined as that inverse function to $e^x$. It is "natural" because it is the inverse to the "natural" exponential $e^x$.

One caveat, e, a positive number, to any power, is positive so while its domain is "all real numbers" its range is "all positive numbers. That means that the domain of ln is positive numbers, ln(-1) is undefined (until you move to the complex numbers).

Answer 3

An expontential function is one where it's rate of growth is proportional to its current value, that is if it's current value at time $t$ is $g(t)$ then it's rate of growth is $c\cdot g(t)$ where $c$ is a constant value (and the same constant value for any time $t$).

Such functions are of the form $b^t$ and their rates of growth will be $c_b$, some constant based on the value of $b$.

The most "natural" such exponent would be the one where $c_b = 1$ and then rate of of $b^t$ would be the exact value of $b^t$ itself.

And that natural base and that natural exponent (the logarithmic inverse function that derives from it) all do exist and happens to be the base $e$.

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Oh, guess an even more basic answer is there is, on the surface of things, no "natural" way of defining $b^x$ if $x$ is irrational. You certainly cant multiply $b$ times itself some irrational number of times.

But if we work backwards and define a function $LN(x) = \int_1^x \frac 1t dt$ so that $LN'(x) = \frac 1x$ (thus solving the question: If $(x^k)' = kx^{k-1}$ then the anti-deriviative of $x^k$ would be $\frac {x^{k+1}}{k+1}$ so the anti derivative of $x^{-1}$ should be .... $\frac {x^0}{0}???????$ ....?

Then $LN(x)$ is invertable and the inverse function we can call $EXP(x)$ so that $LN(EXP(x))= x$ and $EXP(LN(x)) = x$.

The we can define $b^x$ as $EXP(LN(b)\cdot x)$

That is..... $b^x= $ the value $M$ must be so that $\int_1^M \frac 1t dt= x\int_1^b \frac 1t dt$ would be

and (perhaps surprisingly the first time we ever see this) that satisfies all the properties we expect $b^x$ to have. Namely: $b^{x+y} = b^xb^y$ and if $x \in \mathbb N$ then $b^x= b$ times itself $x$ times.

If we define $e$ as $EXP(1)$ then $e^x = EXP(x)$ and $\log_e y = LN(y)$ and $e^x$ and $LN(y)$ are the most "natural" of all exponents and logs as they were the one all definitions are derived from.