Is there proof show that $\log x$ is undefined and make no sense at $ x=0$?

Question

I was asked by someone: why $\log x$ is undefined at $x=0 $?

Is there proof show that $\log x$ is undefined at $x=0$?

Note(01):: log is the inverse function of the exponential function.

note(02): I edited my question as I meant why it's not make a sens at $x=0$ ?

Thank you for your help .

Answer 1

We could define $\log0$ in whatever way we like, but a sensible definition should preserve the main property of the logarithm, that is, $$ \log(xy)=\log x+\log y $$ Suppose we set $\log0=a$; then, taking $y=0$ in the formula above, we have $$ \log(x0)=\log x+\log0 $$ that is, $$ a=\log x+a $$ and we conclude that $\log x=0$. But $x$ can be any positive number! So defining the logarithm at $0$ to be some real number, forces $\log x=0$ for any other $x>0$.

Not really a useful function, I believe you can agree, and certainly not the inverse to the exponential function.

Answer 2

"Undefined" means not defined. We have simply not defined what what $\log(0)$ means. There is no proof for this.

You CAN show why it would not make sense to define $\log(0)$, but this is NOT a proof that $\log(0)$ is not defined.

As for a "visual proof", just look at the graph of $\log(x)$ and note that we want to $\log(x)$ to be continuous.

Answer 3

Visual demonstration: As avid19 stated, look at the graph of $\log x$. It is negative for $0 < x < 1$, and as $x$ gets closer and closer to zero, it clearly goes below any value that you might define $\log 0$ to be.

Algebraic demonstration: $y = \log x$ means $x = e^y$. (Assuming $\log x$ is natural log, but it really doesn't matter that much.) Whatever value of $y$ you might choose for $L \equiv \log 0$, notice that you can find an $x < e^y$ such that $\log x \ll L$. Thus it doesn't make sense to define $\log 0$.

This is just the examination of the graph of $\log x$, put into words.

Answer 4

Since the logarithm is defined as the inverse function of the exponential function, the domain of $\log x$ is exactly the range of $e^x$, i.e. $\mathbb{R}^+$.

Answer 5

The natural logarithm is defined as

$$ \log x = \int_{1}^{x} \frac{1}{t} \, dt $$

Naively plugging in $0$, we get

$$ \log 0 = \int_{1}^{0} \frac{1}{t} \, dt = - \int_{0}^{1} \frac{1}{t} \, dt, $$

However, the area under $y = 1 / t$ from $0$ to $1$ diverges to $\infty$, which would imply

$$ \log 0 = - \infty $$

Intuitively, that's why $\log 0$ is undefined, which is inline with your original question.

Answer 6

For the exponential function we all agree that $exp(- \infty) = 0$. Now the logarithm is the inverse of the exponential function. Applying it to both sides we obtain: $log(0) = - \infty$. This is perfectly sound, but mathematicians prefer to exclude the point $x = 0$ from the domain. That is all, really.