Can we take negative numbers as base in taking log?

Question

Can we take the negative numbers as base in taking log?

For example:$$\log_{(-2)}4=2$$ $$\log_{(-3)}81=4$$ etc.

Answer 1

In the context of complex numbers, yes:

$\log_b(z)$ is a complex number $w$ such that $b^w = z$. Since by definition $b^w = \exp(w \log(b))$, (where $\log$ is any branch of the natural logarithm), $$\log_b(z) = \frac{\log(z) + 2 \pi i n}{\log(b) + 2 \pi i m}$$ for integers $m$ and $n$.

Answer 2

Usually, in a pre-calculus or calculus class (don't ask me about Complex Analysis, I don't know anything there), the following definition is used: $$\log_b(x) = \dfrac{\ln(x)}{\ln(b)}$$ Particularly, since $\ln(b)$ is not defined for negative $b$ values, $\log_{(-2)}4$ would be undefined. As mentioned in the comments, we need not use $\ln$ here - we can use a logarithm of any base - but regardless, logarithms of any base are not defined for an input negative value.

If you defined $\log_b(x) = y$ using the usual $b^{y} = x$ relationship, you would think that this would make sense since $(-2)^{2} = 4$, but by convention, the base is taken to be a positive value. Note:

The logarithm of a positive real number $x$ with respect to base $b$, a positive real number not equal to $1$

Answer 3

I understand the reasoning behind the OP's question: the expression $\log_b(a)=c$ is equivalent to the exponential expression $b^c=a$. Since (for example) $(-3)^4=81$, it seems reasonable to say that $\log_{-3}(81)=4$.

What this reasoning overlooks is that we want logarithms to also be defined for powers of the base that are not integer powers. For example, $\log_2(10)$ is defined even though there is no integer power of $2$ that equals $10$.

This is where negative bases fail us: In general, it is hard to make sense of an expression like $(-2)^r$ if $r$ is not an integer. Notice that even for a "friendly" fraction like $r=\frac{1}{2}$ this expression has no real-valued interpretation. How much harder would it be to assign a meaning to it if $r$ were an irrational number?

Since exponents with negative bases can't be well-defined in real numbers, logarithms with negative bases can't be either.

Answer 4

Absolutely, when working in the field of complex numbers. For real numbers $a$ and $b$, integers $z$, and $i^2=-1$

$$\log_{-a}(b)=\frac{ln(b)+i2z}{ln(a)+i}=\frac{(\ln(b)+i2z)(\ln(a)-i)}{(\ln(a))^2+^2}=\frac{(\ln(a))(\ln(b))+i\ln(b)+i2z\ln(a)+2^2z}{(\ln(a))^2+^2}=\frac{(\ln(a))(\ln(b))+2^2z}{(\ln(a))^2+^2}+\frac{i(2z\ln(a)-\ln(b))}{(\ln(a))^2+^2}$$

Here is something to note. If $a$ is an even power of $b$, with $a$ and $b$ positive real numbers, then $\log_{-a}(b)$ has an even integer value. Take for example, $log_{-2}(\frac14)$

$$log_{-2}(\frac14)=-log_{-2}(4)\frac{-2\ln(2)-i2z}{\ln(2)+i}=\frac{(-2\ln(2)-i2z)(\ln(2)-i)}{(\ln(2))^2+^2}=\frac{-2(\ln(2))^2-i\ln(2)-i2z\ln(2)-2^2z}{(\ln(2))^2+^2}=\frac{-2(\ln(2))^2-2^2z}{(\ln(2))^2+^2}-\frac{i\ln(2)(2z-1)}{(\ln(2))^2+^2}$$

As we can see, when $z=1$, the imaginary part vanishes ($2*1-2=0$, making the nominator of the imaginary part $0$) So we are left with

$$\frac{-2(\ln(2))^2-2^2}{(\ln(2))^2+^2}=-2\frac{(\ln(2))^2+^2}{(\ln(2))^2+^2}=-2$$

We can easily check that $(-2)^{-2}=\frac14$

For $z=0$, the imaginary part does not vanish. We get

$$\frac{-2(\ln(2))^2}{(\ln(2))^2+^2}+\frac{i\ln(2)}{(\ln(2))^2+^2}$$

This is about $-0.09284+i0.2104$