# How I could show that :log1=0? I would be like somone to show me or give me a prove for this : Why ln1=0 ? Note that ln is logarithme népérien, the natural logarithm of a number is its logarithm to the base e. Thanks for any replies or any comments!

How I could show that :$\mathrm{log}1=0$?
I would be like somone to show me or give me a prove for this :
Why $\mathrm{ln}1=0$?
Note that $\mathrm{ln}$ is logarithme népérien, the natural logarithm of a number is its logarithm to the base $e$.
Thanks for any replies or any comments!
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Markus Petty
I will denote the natural logarithm of $x$ by $\mathrm{ln}x$
There are several definitions of the natural logarithm, so I will look at two of them.
1) $\mathrm{ln}\left(-\right):\left(0,\mathrm{\infty }\right)\to \mathbb{R}$ is the inverse of the exponential function ${e}^{-}:\mathbb{R}\to \left(0,\mathrm{\infty }\right)$, where $\left(0,\mathrm{\infty }\right)$ denotes the set of positive real numbers. Then because ${a}^{0}=1$ for every real number $a$, particularly $e$, it follows that $\mathrm{ln}\left(1\right)=0$
2) $\mathrm{ln}x={\int }_{1}^{x}\frac{dt}{t}$. Then Evaluating at $1$ gives $\mathrm{ln}1={\int }_{1}^{1}\frac{dx}{x}=0$