If $log(x)$ is not defined, then how is $x$ defined?

Question

To take an example, suppose we have $$x=\left({1\over -2}\right)^{-2}$$ Taking the logarithm, $$log(x) = -2log\left({1\over -2}\right)$$ Reversing the logarithm, $$x = 10^{-2log\left({1\over -2}\right)}$$ Which is undefined. But clearly, $x=4$. How is this possible?

Answer 1

You raise an interesting question here, and the primary answer is essentially by definition, but also one should note the special role of integers.

In general, the symbol $a^b$ can only be defined (without a myriad of caveats from complex analysis) in the following situations:

• $a>0$ and $b\in\Bbb{C}$, in which case we define $a^b:=e^{b\log a}=\exp(b\log a)$.
• $a\in\Bbb{C}\setminus\{0\}$ and $b\in\Bbb{Z}$ by first defining $a^0=1$, then for positive integers $n$, defining $a^n:=a\cdot a^{n-1}$, and then for a negative integer $n$, defining $a^n:=\frac{1}{a^{|n|}}$. In other words, this is just the middle school definition of “multiply repeatedly”.

Of course these bullet points aren’t disjoint, because if $a>0$ and $b\in\Bbb{Z}$, we now have two possible interpretations for the symbol $a^b$. The first is via epxonentials and logarithms as in the first bullet point, and the second is via recursion in the second bullet point. Fortunately, both approaches give the same value.

Pursuing this line of thought, one might ponder whether this consistency in the definitions can be further extended. And indeed, it can, in the following sense. If $a\in\Bbb{C}\setminus\{0\}$ and $b\in\Bbb{Z}$ (so we’re in the situation of the second bullet point) we can ask whether it is possible to interpret the result $a^b$ using exponentials and logarithms. Now, lets say we choose one specific branch for the logarithm, call it $\lambda_1$. Then, we have one possible output $e^{b\lambda_1(a)}$. If we had chosen a different branch for the logarithm, say $\lambda_2$, then we would get the output $e^{b\lambda_2(a)}$. It is a fact of complex analysis that any two branches of logarithm (on a connected open subset of $\Bbb{C}\setminus\{0\}$) differs by an integer multiple of $2\pi i$. In formulae, this means there is an integer $k$ such that $\lambda_1(\cdot)=\lambda_2(\cdot)+2k\pi i$, so \begin{align} e^{b\lambda_1(a)}=e^{b\lambda_2(a)+b\cdot 2k\pi i}=e^{b\lambda_2(a)}\cdot e^{2kb\pi i}=e^{b\lambda_2(a)}, \end{align} where the final equal sign is because $bk$ is an integer (since both $b$ and $k$ are). So in this case, regardless of the choice of branch for the logarithm, we get the same answer. In other words, although $\log(a)$ by itself does not have a unique value (prior to specifying a branch) $e^{b\log a}$ does have a unique (i.e branch-independent) value when $b$ is an integer. This unique value is exactly the one we expect from the ‘repeated multiplication’ approach given to us by the second bullet point.

Applying this discussion in your case, we see that regardless of how you want to look at it (i.e middle school vs complex analysis) you get the same answer of $4$. Finally, if you haven’t yet learned complex analysis, I guess the main takeaway is that the power of $-2$ being an integer saves the day.

Answer 2

The underlying issue here is simply that the power rule you wish to reference, $$ \log_a(x^p) = p \log_a(x) $$ only holds when everything on both sides is well-defined, i.e. when $x>0$.

But obviously if $x \le 0$, the right-hand side isn't even well-defined.

This is not unlike a multitude of fake proofs involving $i$ that follow the flavor of $$ 1 = \sqrt{1} = \sqrt{-1 \cdot (-1)} \stackrel{?}{=} \sqrt{-1} \sqrt{-1} = i^2 = -1 $$ but this assumes that $\sqrt{xy} = \sqrt{x} \sqrt{y}$ for all $x,y$, not just the positive ones.

Answer 3

Why take logarithms? Surely if you are beginning with $x$ you know what kind of thing $x$ is, and you are constrained by the nature of $x$ as to what you can do. If you are beginning with $y=\log x$ you need to specify with care what kind of thing $y$ is, and that may constrain what $x$ can be.

If $x$ is the kind of thing which doesn't have a logarithm, why try to take the logarithm?

Of course as Brian Moehring notes in a comment, there is the possibility of something interesting happening if you allow it, and such are the explorations which have enriched mathematics.

So are you beginning with $x$ or $y=\log x$? Your question is not clear on the point, and that is why you run into confusion.