Logarithms with negative bases for real numbers For real numbers, a logarithm finds the exponent that when put on the base gives the input, in this case $a$. $$\log_a(a^b)=b.$$
As far as I know, logarithms cannot be found for negative numbers: $$\log(a) \,, \quad a>0.$$
$a>0$ is a general requirement, as far as I am aware.
My question is: does this requirement apply universally or only to some numbers? For instance, I fully see that for a number like $2^n$, no value of the exponent $n$ will ever make this power negative. It is therefor not allowed (not possible) to write, say:
$$\log_2(-8),$$
since no exponent $n$ in $2^n$ will ever give this negative number. This applies to all positive bases.
But for some negative bases it is possible. Was the base not $2$ but instead $-2$,
$$\log_{-2}(-8),$$
then I would argue that $n=3$ is a fine solution, since obviously: $(-2)^3=-8$.
But even with negative bases, certain negative values seem to be impossible, such as: $$\log_{-2}(-4).$$
No exponent $n$ in $-2^n$ will result in $-4$, as far as I can see.
So, it seems to be in summary that for all positive bases, the logarithm can't be found for negative numbers. But for some negative bases it can. Am I right in this? Is the requirement of a positive number within the logarithm restricted to all positive and some negative numbers? Or do we deal with negative bases in a different manner that I have not been introduced to yet?
 A: The definition that you mentioned about logarithms is true.
$$a^{b}=x \implies \log_{a}(x)=b$$.
And to solve your problem, you must need to understand new kind of number system called as complex numbers.
"A complex number is a number of form $a+ib$ where $a,b \in \mathbb{R}$ and $i=\sqrt{-1}$."
Generally in high school it's taught that $\sqrt{-1}$ is undefined. But the thing is they do work and their applications are widely used in various braches of math like analytic number theory,...
You may have already seen the formula $e^{i\pi}=-1$, which as many mathematicians quote to be one of beautiful equations in math. It's because is involved three constants which themselves are irrational and are of great importance.
There is a beautiful formula concerning complex numbers
$$e^{ix}=cos(x)+isin(x)$$
It's about complex numbers. Now coming to your problem of negative bases. Complex numbers can be used to solve them.
For example,
$$\ln(-4)=\ln(-1)+\ln(4)=i\pi+ln(4)=ln(4)+i\pi $$.
You can extend this idea to negative bases as well.
In case you are interested to learn much more about the complex numbers I suggest the following videos by khan academy.
https://youtube.com/playlist?list=PLXSlB4yMaoJtM2gG5Mas5mMjwX_B51vsB
$$\mathbf{continuation:-}$$
Apologise for incomplete answer. The thing is that, in the question you said even giving an example that $\log_{2}(-8)$ doesn't exist. When you are talking about real numbers then you are right. But it isn't true that it doesn't have solutions. The example you have has a complex solution at $3+\frac{iπ}{ln2}$. Similarly $\log_{-2}(-4)$ also has a complex solution but not real. So no matter if base is positive or negative and no matter if input is positive or negative. Solution do exist.
