I have a hard time understanding why $\ln e=1$

Question

Can someone explain to me why the natural logarithm of e is exactly equal to the first nonzero but positive integer?

Because $e^1 = e$? You always have $\log_a a = 1$, since $a^1 = a$. — Daniel Fischer, Jan 23 '14 at 13:00

score 6 · Answer 1 · answered Jan 23 '14 at 13:05

6

It depends on how you define the natural logarithm; but, let's do it this way:

By definition, $\ln(x)$ is the unique number $y$ such that $e^y=x$. In other words, the natural logarithm $g(x)=\ln(x)$ is the inverse function for the exponential function $f(x)=e^x$.

So, $\ln(e)=1$ because $e^1=e$.

answered Jan 23 '14 at 13:05

Nick Peterson

32,430

Why $e^1=e$ ? ? – rope Jan 23 '14 at 14:45
Because $x^1=x$ for any $x$! Remember that for any positive integer $n$, you can think of $x^n$ as the result of multiplying $n$ copies of $x$; if $n=1$, then you have only one copy. – Nick Peterson Jan 23 '14 at 14:46

score 0 · Answer 2 · answered Sep 23 '21 at 01:51

$In()$ means the logarithm with base $e$ if I say $\log_{e}(e)$ I am asking how many times I should multiply $e$ to 1 to get $e$. Or in other words we are trying to figure out the value of x in :-

$$e^x=e$$

We know anything raised to the first power or 1 is itself. You would probably know it.

So I can say:-

$$e^x=e \\e^1=e \\\therefore x=1 $$

And this means:-

$$In(e)=1$$

You can think about it in exponential form if it is not clear in logarithm as I showed you.

You can say $x^1=x$ as it says multiply x to 1 1 times and that is $1×x=x$ I know that people would say that we do not multiply by 1 in exponential but the number itself well what I think is adding and additional 1 causes no change and it is useful for negative exponents so according to me $x^2=1×x×x$

score 0 · Accepted Answer · edited Jan 23 '14 at 13:26

0

that is the definition of $\ln$ (logarithm in base e):

if you take $\log_2$ (logarithm in base 2) then $\log_2(2)=1$ and $\log_2(e) = 1/\ln(2)$

edited Jan 23 '14 at 13:26

amWhy

209,954

answered Jan 23 '14 at 13:03

Christophe

137

3

$\log_2$ or $\operatorname{lb}$ please... I have never seen people writing $\ln_2$... – peterwhy Jan 23 '14 at 13:17
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I've edited your answer for the sake of posterity: please be advised by peterwhy's comment regarding using appropriate notation. The "gist" of your answer is spot on; but it's in everyone's best interest to use notation that is going to clarify and not mystify. – amWhy Jan 23 '14 at 13:27
the best answer on mathematics.stackexchange so far – rope Jan 23 '14 at 14:42
you're absolutely a genius – rope Jan 23 '14 at 15:20
don't make fun of me, rope ! it is my first answer ever (-; – Christophe Jan 23 '14 at 17:01
1

i'm serious !!! – rope Jan 23 '14 at 18:34

score -2 · Answer 4 · answered Jan 23 '14 at 13:20

-2

A logarithm (being a function) along with a base $b$, when given an input $x$, could be interpreted as "What is $y$ such that $x=b^y$".
So if $b=e$, when evaluating $\log_e(e)=\ln(e)$, we actually ask "What is $y$ such that $e=e^y$". Therefore $\ln(e)=y=1$ must hold..

answered Jan 23 '14 at 13:20

user76568

4,542

You can try adding more details – Mohd Saad Sep 23 '21 at 04:53

I have a hard time understanding why $\ln e=1$

4 Answers4