How I could show that :$\log1=0$?

$\begingroup$

I would be like somone to show me or give me a prove for this :

Why $\ln 1=0$ ?

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!

$\endgroup$ 1

5 Answers

$\begingroup$

$$ \begin{align} \log 1 = \log(1\cdot1) & = \log 1+\log 1 \\[10pt] \text{So } \log 1 & = \phantom{{}-}\log 1 + \log 1 \\[2pt] -\log 1 & \phantom{=}-\log 1 \\[6pt] 0 & = \log 1 \end{align} $$

$\endgroup$ 9 $\begingroup$

I will denote the natural logarithm of $x$ by $\ln x$.

There are several definitions of the natural logarithm, so I will look at two of them.

1) $\ln(-): (0,\infty)\rightarrow \mathbb{R}$ is the inverse of the exponential function $e^{-}:\mathbb{R}\rightarrow (0,\infty)$, where $(0,\infty)$ denotes the set of positive real numbers. Then because $a^0=1$ for every real number $a$, particularly $e$, it follows that $\ln (1)=0$.

2) $\ln x=\int_1^x{ \frac{dt}{t}}$. Then Evaluating at $1$ gives $\ln 1=\int_1^1{\frac{dx}{x}}=0$.

$\endgroup$ 2 $\begingroup$

Using the fact that $$e^0=1,$$ we may take the natural logarithm of both sides to give $$\ln(e^0)=\ln(1) \iff 0\ln(e)=\ln(1) \iff0=\ln(1).$$

$\endgroup$ 6 $\begingroup$

The way people usually learn about logarithms, before they learn any of their other properties, is that $\log_b(a)$ means "the power to which $b$ must be raised in order to get $a$".

In particular, $\ln(1)$ or $\log_e(1)$ means "the power to which $e$ must be raised in order to get $1$."

So what power do you raise $e$ to, if you want the result to be $1$? The answer is $0$, because $e^0=1$ (and no other power of $e$ will work). Therefore, $\log_e(1)=0$.

$\endgroup$ $\begingroup$

$\ln x = a \Leftrightarrow x = e^a$

$\ln 1 = 0 \Leftrightarrow 1 = e^0$

And since $e^0 = 1$ is a true statement, $\ln 1 = 0$.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy