Why must the base of a logarithm be a positive real number not equal to 1? and why must $x$ be positive?


  • 13
    $\begingroup$ Can you solve $1^x=2$ for $x$? Can you solve $x^2=-1$ for $x$ a real number? $\endgroup$ – Pedro Tamaroff Feb 25 '14 at 16:34
  • $\begingroup$ I can solve (-2)^3=-8. So why isn’t acceptable to write log_(-2)(-8)=3? $\endgroup$ – Jose Ramirez Jun 18 at 3:34
  • $\begingroup$ @JoseRamirez You mean just writing $\log_{-2}{-8}$ by itself while not allowing $y=\log_{-2}{x}$ as a function? I wonder. Maybe you have a point. $\endgroup$ – linearAlg Aug 3 at 12:36

By definition, $\log_bx$ is the number for which if you take $b$ to that power, you get $x$. In symbols: $$b^{\log_b x} = x$$ 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 $\log_01$ 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 $p$.

  • $\begingroup$ I assume that when $b$ is negative there are multiple solutions, for example $\log_{-1} 1$ will equal $0$ and $2$, is that true? $\endgroup$ – Mahmoud Ahmed Feb 25 '14 at 17:18
  • 1
    $\begingroup$ @MahmoudAhmed That is not true. We leave the logarithm undefined in that case, as you said in your original question: the logarithm is only defined for a positive base. We could say that $\log_{-1} 1 $ is 0 and 2, and 4, and 6, and -18, but there is no value in doing that. $\endgroup$ – MJD Feb 25 '14 at 17:26
  • 27
    $\begingroup$ This doesn't answer the question about why the base of the logarithm cannot be negative. $\endgroup$ – bryn Oct 20 '14 at 2:31
  • 7
    $\begingroup$ What happens if we say ${log_{(-2)}({-8})}=3 ?$ $\endgroup$ – user118494 May 7 '16 at 14:11
  • $\begingroup$ @MJD : Why cannot we just strike off $-1$ from negative bases just like we strike off $1$ from positive bases ? $\endgroup$ – user118494 May 7 '16 at 14:15

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!

  • $\begingroup$ Thanks, this is an excellent explanation. $\endgroup$ – undefined Jan 20 '18 at 2:29
  • $\begingroup$ This is an old question but couldn't $y=(-4)^{a/b}$ also be evaluated for $a$ even and $b$ either even or odd? The reason being that $(-4)^a$ where $a$ is even would be a positive number which can thus appear under even or odd root. $\endgroup$ – pseudomarvin Mar 16 '18 at 9:14
  • 1
    $\begingroup$ @pseudomarvin, I believe Robin means that $(-4)^{a/b}$ has a solution in general only when $b$ is odd. If $a$ is even, then $b$ need not to be odd, but that is not true in general. $\endgroup$ – gwg Aug 25 '18 at 20:53

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

  • 4
    $\begingroup$ This question already has a well acepted answer . Your commentary does not provide any new insights $\endgroup$ – Shailesh Nov 19 '15 at 1:43

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 $c>0$.

  • $\begingroup$ What is a*? The absolute value of a? (I've never seen this notation, hence the question). $\endgroup$ – PGupta May 21 '18 at 3:42
  • $\begingroup$ @Teacher123 I've updated the answer, there was indeed no reason of that notation $\endgroup$ – 7raiden7 May 21 '18 at 5:45

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


protected by J. M. is a poor mathematician Mar 9 at 12:26

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.