Why must the base of a logarithm be a positive real number not equal to 1? Why must the base of a logarithm be a positive real number not equal to 1? and why must $x$ be positive?
Thanks.
 A: I am not a mathematician at all, but during a quick reflexion, I just found myself a simple explanation : 
I have always learned that $\log a(x) = \ln(x)/\ln(a)$
As everything inside a ln function, a must be strictly positive.
And as $\ln(a)$ is in the denominator, $\ln (a) \ne 0$, so $a \ne 1$
That's why the base must be positive and different from 1.
I'm pretty sure it is not rigorous at all, but maybe it will be easier to remind?
Thank you for your comments
A: By definition, $\log_bx$ is the number for which, if you take $b$ to that power, you get $x$. Symbolically:
$$b^{\log_b x} = x$$
For example, what power do we need to raise $2$ to in order to get $4$? Well, it's $\log_24 = 2$. What power do we need to raise $81$ to in order to get $9$? Well, it's $\log_{81}9 = 0.5$.
Ask yourself what $\log_1x$ means. It's the power, say $p$, for which $1^p=x$.
Unless $x=1$, there is no solution, and when $x=1$ any power will do, so $\log_11$ is any number.
For the same reason $\log_0$ doesn't make sense because we can't solve $0^y=x$ unless $x=0$, and when $x=0$, any power will do, so $\log_00$ could be any number.
Why can logarithms only be applied to positive arguments? Well, $\log_2(-1)$ would be the power, say $p$, for which $2^p = -1$. Hopefully, you can see that $2^p > 0$ for all real numbers $p$.
A: This explanation is using my own logic, so please don't think it is textbook by any means!
Consider a hypothetical negative base of $-4$, so the undefined (non-existent) function $y$ $=$ $log$$_{-4}$$(x)$. This logarithm would be the inverse of the function $y$ $=$ $(-4)$$^x$, which can only be evaluated for exponents that can be written as a fraction where the denominator is odd. Remember a rational exponent, such as $(-4)$$^{a/b}$, represents a radical, namely $\sqrt[b]{(-4)^a}$, and a negative number can only be evaluated for an odd root (using real numbers). For example, $(-4)$$^{1/2}$ means $\sqrt{-4}$ which is a non-real answer.
Thus, an exponential function with a negative base, such as $y$ $=$ $(-4)$$^x$ isn't much of a function at all (it is not continuous), since it can only be evaluated at very specific x-values. So, a logarithm with a negative base, like $y$ $=$ $log$$_{-4}$$(x)$ would also only work for very specific arguments (due to its connection to the non-continuous $y$ $=$ $(-4)$$^x$) and such a logarithmic function would also not be continuous.
It is for such reasons that we only consider logarithms with positive bases, as negative bases are not continuous and generally not useful. Hope this insight makes sense and is somewhat helpful!
A: Because the logarithm is the inverse function of the exponential operation, i.e.: if $a^b=c$, then $b=\log_a(c)$.
As you can see, if $a=1$, $1^b=1, \forall b\in\mathbb{R}$, and it would not make sense to study this case.
As regards its sign: if $a<0$, then $a=(-1)\cdot (-a)$, thus:
$$
a^b=(-1)^b\cdot (-a)^b,
$$
that will lead to an alternation in sign, and it would be more difficult to study.
If $a>0$ so, also $c>0$.
A: I think you can take a log of a negative number depending on the base.  for example, solve the solution of〖-2〗^x=-8   which can be rewritten as 〖log〗_(-2) (-8)=3
