Why is $\ln 1 = 0 $?

Question

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 $$\ln 1 = 0$$

Answer 1

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.)

Answer 2

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

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

Q.E.D.

Answer 3

$$ e^0=1$$ 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: $$\ln{1}=0$$

Answer 4

$\ln(x)$ 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?

Answer 5

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

So if $x=0$ ...

Answer 6

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 $\log_x(1)=0$.

Answer 7

In any base, $$\log(1)=\log(1\cdot1)=\log(1)+\log(1).$$

---

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