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

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

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    Could you explain your reasoning on the examples that you provided? It doesn't seem like they make sense.... The direct answer to your question is going to be "no" by the usual definition of log. Maybe if you explain more what you want, someone can give information about a generalization that will be appropriate. – 2017-01-06
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    $\log_2 4= 2$ NOT $\log_{-2} 4=2$ – 2017-01-06
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    @Brick. We know that $\log_ab=c$ means $a^c=b$. Hence $\log_{(-2)}4=2$ means $(-2)^2=4$. – 2017-01-06
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    @JustinBieber By convention, $a^c = b$ with $a$ positive. See my answer. – 2017-01-06
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    Thanks for clarifying your thoughts. The convention mentioned by @Clarinetist ensure other properties like $\log_a b = \log b / \log a$. Your version would create both ambiguity and break other properties expected of log. Answers along these lines are now up. – 2017-01-06
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    this is because function definition in real number. if you can get negative or zero number as base then log is not be a real function – 2017-01-06

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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 $n$.

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    The first answer which deals with the question in appropriate generality. – 2017-01-06
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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.

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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 $1$

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    On the right side of your equation above, *any* log would work. It doesn't have to be natural log. That any log works is a theorem though, not a definition. – 2017-01-06
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    @Brick Of course. – 2017-01-06