$Log_{x}{0}$ ?

The function $Log_x n$ is supposed to give an answer $a$ for which the equation $x^a = n$ holds true

This means that when $n = 0$

$x^a = 0$ which doesn't make sense since any number $b$ when put to any power $c$ never equals $0$

$b^c \neq 0$

The closest we can get is an incredibly small number

Let's say that $n = 0.000\ldots 0001$ . This gives us

$x^a = 0.000\ldots 0001$

But since $x > 1$ to get anywhere near that fraction we need to go into negative powers.

$x^a < 1$ when $a < 0$

This means that the smaller the fraction the larger the negative power, meaning as $n \rightarrow 0, a \rightarrow -\infty$

It never reaches $0$ but it gets incredibly close to it

Let $y = \displaystyle \lim_{n \to 0^+ } \bigg ( \log_x n \bigg )$ , then $x^y = \displaystyle \lim_{n \to 0^+ } n = 0$ . Knowing the graph of an exponential function for $f(x) = a^x$ with $a>1$ we see that $y \rightarrow \boxed{-\infty }$ .

Another way to think of this expression is: log n / log x by applying the change of base formula.

Disregarding the log x, by substituting smaller and smaller numbers for n and taking the log of n on a calculator, the answer becomes more and more negative.

If the numerator is becoming increasingly negative, the expression has to become more and more negative unless the denominator is also negative.

However, logarithms do not become negative unless you are taking the log of a number less than one. Since one of the constraints limits it to a positive number, the denominator will always be positive.

Now that it is known that the numerator is become increasingly negative with a denominator never being negative, the fraction can the be deduced to be approaching negative infinity, giving you the answer.