Is $0 \times \ln(0) =\ln(1) $ true?

Question

Can we affirm that:

$0 \times \ln(0) = \ln(0^0) = \ln(1) = 0$?

The problem is $\ln(0)$ is supposed to be undefined but it works

No, $0\cdot \ln(0)$only has meaning if $0$, $\cdot$ and $\ln(0)$ have meaning. — Git Gud, Dec 22 '15 at 23:17

score 6 · Accepted Answer · answered Dec 22 '15 at 23:19

6

The interpretation that $0\times \ln(0)=0$ comes from one possible definition. We know

$$\lim_{x\rightarrow 0^+}x\ln(x)=0,$$

which can easily be shown using l'hopital's rule on $\ln(x)/(1/x)$. So formally we can define:

$$f(x)=x\ln x$$

for $x>0$, and

$$f(0)=0.$$

The advantage of the above definition is that the resulting function $f(x)$ is continuous on $[0,\infty)$.

answered Dec 22 '15 at 23:19

Alex R.

32,771

2

Consider also $g(x)=x\ln(e^{-1/x})$. Since $e^{-1/x}\to 0+$ as $x\to 0+$, this is another interpretation of $0\times \ln(0)$, and yet $g(x)\equiv -1$ for all $x>0$. – vadim123 Dec 22 '15 at 23:50
It really depends on the conventions and situation. $ln(0)$ can be defined as $\infty$ to make $log$ continuous one the one-point compactification of $\mathbb{R}$, and then $0 \cdot \infty$ can be defined as $0$ (as is done for the purposes of measure theory, for instance). The point is: the definition should be useful for some context, and the context in this question is unclear. – Aloizio Macedo Dec 23 '15 at 00:58

score 0 · Answer 2 · answered Dec 23 '15 at 01:44

Can we affirm that $0\times \ln(0)=\ln(0^0)=\ln(1)$?

In the context of basic properties of exponents and logarithms, the answer is no because $\ln(0)$ has no common meaning (as you observed), the rule $\ln(x^y)=y\ln(x)$ assumes $x>0$ and $0^0$ has no common meaning too.

So, to validate the calculation $$0\times \ln(0)=\ln(0^0)=\ln(1),$$ you have to

define $\ln(0)$;
prove that the rule $\ln(x^y)=y\ln(x)$ is valid for $x=y=0$ (using your definition of $\ln(0)$);
define $0^0$ to be $1$.

Is $0 \times \ln(0) =\ln(1) $ true?

2 Answers2