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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?

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4 Answers

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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 $e^1=e$.

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

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$In()$ 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 $x^2=1×x×x$

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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..

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