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

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

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
• Perhaps this article might prove itself useful. – Lucian Jan 23 '14 at 13:20

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 $\log_2(e) = 1/\ln(2)$

• $\log_2$ or $\operatorname{lb}$ please... I have never seen people writing $\ln_2$... – peterwhy Jan 23 '14 at 13:17
• 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. – Namaste 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
• i'm serious !!! – rope Jan 23 '14 at 18:34

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 $e^1=e$.

• 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

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..