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In high school we learn that a $a\log[(x)] = \log (x^a)$

From this I would assume $2\log(-1) = \log [(-1)^2]$

However, the first is not real and the second is, according to my calculator and textbook. Why is this?

Xoque55
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Jonathan Milford
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  • The rule $a\log x=\log x^a$ holds for $x,a$ real. You'd have to think first what you mean by $\log z$ when $z$ is any complex number. – Pedro Nov 05 '13 at 01:05
  • What do you mean by $log(-1)$? – WillO Nov 05 '13 at 01:06
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    You also learn $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$, but it turns out this is only true for $a,b\geq 0$ since $$\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\cdot \sqrt{-1} = i^2 = -1$$ cannot be true. – abnry Nov 05 '13 at 01:09
  • If I overlook the rigour for a minute, then $\text{exp}(\pi\cdot \imath) = -1$ and so clearly the result is complex. But, in reality, $\text{log}$ is not defined $\forall x \leq 0$. – Christopher K Nov 05 '13 at 01:11
  • Are you sure the second is real? $\log(1)$ might be $2 \pi \mathbf{i}$! –  Nov 05 '13 at 01:13

3 Answers3

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The formula $a\log x=\log x^a$ requires that $\log x$ exists. Here you want $\log(-1)$; this would a number such that $e^{\log(-1)}=-1$. But it turns out that the exponential of real numbers is always positive, so to write $\log(-1)$ you somehow need to extend the log function to the complex plane.

Notice that if you keep going from your desired equality, you get $$ 2\log(-1)=\log(-1)^2=\log 1 = 0, $$ which would imply that $\log(-1)=0$. What this shows is that the property $a\log x=\log x^a$ does not hold for arbitrary complex numbers when you extend the log function to the complex plane.

Regarding your first question, $\log(-1)$ cannot be real because the exponential of a real number is always positive.

Martin Argerami
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Logs are tricky when you start to discuss negative numbers. To do this properly, you must concern yourself with branches of the log in the complex plane. Note that $e^{i\pi} = -1$ and that the exponential function has period $2\pi i$ in the complex plance.

ncmathsadist
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If the square were already inside the parenthesis like $\log((-1)^2)$ then you would indeed have $0$ as the result.

Beginning with the exponent outside, it is already the case that $\log(-1)$ must be evaluated outside the real numbers, in which case we would use the complex form $\log|a|+i\arg(a)$ which evaluates (at $-1$) to $0+i(2n+1)\pi$. Then the final expression is $2i(2n+1)\pi$.

abiessu
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