Question

What is the logarithm of a negative number?

Answer 1

Explanation:

Logarithms of negative numbers are not defined in the real numbers, in the same way that square roots of negative numbers aren't defined in the real numbers.

If you are expected to find the log of a negative number, an answer of "undefined" is sufficient in most cases.

It is possible to evaluate one, however, the answer will be a complex number. (a number of the form $a+bi$, where $i=\sqrt{-1}$)

Starting with a general case-

${\log }_b\left(-x\right)=?$

We will use the change-of-base rule and convert to natural logarithms

$$\begin{gathered} {\log }_b\left(-x\right)=\frac{\ln \left(-x\right)}{\ln \left(b\right)} \\ \Rightarrow {\log }_b\left(-x\right)=\frac{\ln \left(-1.x\right)}{\ln \left(b\right)} \\ \Rightarrow {\log }_b\left(-x\right)=\frac{\ln \left(x\right)+\ln \left(-1\right)}{\ln \left(b\right)} \end{gathered}$$

We know, Euler's identity states that

$e^{i\mathrm{\pi }}=-1$

For now, let us simply take the natural log of both sides of Euler's Identity:

$$\begin{gathered} \ln \left(e^{i\mathrm{\pi }}\right)=\ln \left(-1\right) \\ \Rightarrow i\mathrm{\pi }=\ln \left(-1\right) \end{gathered}$$

So we can insert this in our equation.

${\log }_b\left(-x\right)=\frac{\ln \left(x\right)+i\mathrm{\pi }}{\ln \left(b\right)}$

Hence, we have a formula for finding logs of negative numbers.