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Yes I know, and believe, and have used it for all the time I have done mathematics as fun as well as a subject. But why is it that $$\ln 1 = 0$$

KaRJ XEN
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7 Answers7

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One can define $\ln(x)$ as the unique number $y$ satisfying that $$ e^y = x. $$

So, $\ln(1) = 0$ because $e^0 = 1$.

You can generalize this to a logarithm with base $b>0$. So $\log_b(x) = y$ means that $b^y = x$.

Another definition of the natural logarithm is that $$ \ln(x) = \int_1^x \frac{1}{t}\; dt. $$ And again, $$ \ln(1) = \int_1^1 \frac{1}{t}\; dt = 0. $$

(These definitions are of course equivalent.)

Thomas
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  • Thanks, and I get your argument but my question still remains unanswered; Why in the first place is $$e^0 = 1$$ (Why is it that any number raised to $0$ is 1)? – KaRJ XEN Oct 29 '15 at 18:59
  • @KaRJXEN: No, your question was, why is $\ln(1) = 0$? They are technically two different questions. Now the answer to the question why $e^0 = 1$ is that this is a definition. You define $a^n$ ($a>0$) as the product of $n$ with itself $n$ times when $n$ is a natural number. Then you define $a^0 = 1$. One reason for this definition is that then the rule $a^na^m = a^{n+m}$ always holds. – Thomas Oct 29 '15 at 19:01
  • Thanks and I appreciate. But I thought asking $\ln 1 = 0$ would be equivalent to asking $e^0 = 1$. Is there anywhere I could get a proof or an insight on why anything raised to a zero is 1? – KaRJ XEN Oct 29 '15 at 19:05
  • @KaRJXEN: Sure, just search the site. Then you will find, for example, http://math.stackexchange.com/questions/238300/how-to-understand-why-x0-1-where-x-is-any-real-number – Thomas Oct 29 '15 at 19:08
  • @KaRJXEN: Also, feel free to upvote answers that you appreciate :) Also, you can accept an answer if it answers your question. Both of these things give reputation to the people that took the time to answer your question. – Thomas Oct 29 '15 at 19:10
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In any base, $$\log(1)=\log(1\cdot1)=\log(1)+\log(1).$$

Alternatively,$$\text{antilog}(0)=1.$$

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$$ e^0=1$$ Since the $ \ln$ function is defined as the inverse of the exponential function, i.e., when we write $\ln{x}$ we ask to what power must $e$ be raised to produce $x$. So: $$\ln{1}=0$$

  • Thanks, and I get your argument but my question still remains unanswered; Why in the first place is $$e^0 = 1$$ (Why is it that any number raised to $0$ is 1)? – KaRJ XEN Oct 29 '15 at 18:50
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$\ln(x)$ is defined as the value $y$ such that $x=e^y$.

In this case, $\ln(1) = y$ such that $1=e^y$.

Can you see why $y=0$ is the solution now?

wythagoras
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JacobCheverie
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$\ln(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dots+(-1)^{k-1}\dfrac{x^k}{k}+\dots$ for $|x|\lt1$.

So if $x=0$ ...

JMP
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Let's cover how $\ln$ can be defined.

  1. $e^{\ln(1)}=1=e^0$, since $e^x$ is an increasing function, it follows that $\ln(1)=0$.

  2. $\ln(a)=\ln(1\times a)=\ln(1)+\ln(a)\Rightarrow\ln(1)=0$

  3. $\ln(x)=\int_1^x\frac{1}{t}dt\Rightarrow\ln(1)=\int_1^1\frac{1}{t}dt=0$, without integrating as the lower and upper limits of the integral are the same.

  4. $\ln(x^r)=r\ln(x)$, in particular, $\ln(1)=\ln(1^2)=2\ln(1)$, hence $\ln(1)=0$.

  5. $\log_a(b)=\frac{1}{\log_b(a)}$, since $\log_1(x)$ cannot be defined, it makes sense for it to be impossible to divide by $\log_x(1)$. Since it is impossible to divide by $0$, it makes sense to let $\log_x(1)=0$.

Element118
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$$\ln{1}=0$$ $$\log_e{1}=0$$ $$e^0=1.$$

Lesson here: anything to the $0$th-power is $1$.

Q.E.D.

Julian Rachman
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