# Negative logs?

1. Oct 4, 2013

### Acnhduy

Why is it not possible to have the log of a negative number?
Examples would be greatly appreciated.

Furthermore, is it the base that cannot be negative?

Like log-bx

or is it that when you have a number and you log it, the number cannot be negative. log(-x)

Last edited: Oct 4, 2013
2. Oct 4, 2013

### Mentallic

Both really.

If we have a negative number in the base, such as

$$y=\log_{-2}{x}$$

then this is equivalent to $x=(-2)^y$. What's wrong with this though?

Well, let's try and see what happens when we graph $y=(-1)^x$. For all even integers x, we get a result of y=1, and for all odd integers, we get -1. What about all the rationals and irrationals in between? It turns out that we can't do it (they're actually complex numbers, but you can ignore that). For the same reason we can't take the square root of -1, we can't plot the graph $y=(-1)^x$ and hence we can't plot $y=(-a)^x=a^x(-1)^x$ for any values other than the integers because ax is simply some positive number. Because this function is so discontinuous, the logarithm with a negative base will have the problem of being defined in some cases such as

$$\log_{-2}{-8}=3$$
$$\log_{-2}{16}=4$$

but it will fail in most other cases.

That's for negative bases. Taking the log with a positive base of a negative number is pretty easy to figure out.

What does the following equal?
$$\log_{2}{-1}$$

In other words, what value of x solves the equation $2^x=-1$ ? We already know that 2anything is a positive number, so how do we ever end up getting a negative number? The result again is complex, but you can think of it as not having an answer.

3. Oct 4, 2013

### HallsofIvy

Staff Emeritus
$log_a(x)= y$ is equivalent to $a^y= x$. As long as a is positive, $x= a^y$ cannot be negative (if y> 0, x> 1, if y< 0, 0< x< 1) so we cannot have "$log_a(x)$" with x negative. Further, if we allow the base, a, to be negtive, we will run into problems with fractional powers.

If, say, a= -2 then saying $log_{-2}(x)= 1/2$ is equivalent to saying $(-2)^{1/2}= \sqrt{-2}= x$. But the square root of -2 is not a real number.

Of course, if we extend the real number system and allow both x and y to be any complex numbers, those restrictions to do not apply. In the complex number system, we can have log(x) with x any complex number, including the negative real numbers (although we will need to make a "cut" somewhere and, by convention, that is done along the negative real numbers so the "standard" logarithm" will still not be defined for the negative real numbers), and can have $f(x)= a^x$ for a negative.

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