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
$\endgroup$ 42 Answers
$\begingroup$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 $[0,\infty)$.
$\endgroup$ 2 $\begingroup$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 $1$.
