# Why isn't $-2$ solution for $x$?

I came across an logarithm problem recently. I don't know why solution to this problem cannot be $-2$. Now, don't downvote now because you don't know why I'm asking this. I know that logarithms' domains must be greater than $0$, can't take a negative number. Just read the whole post and wait until I get to the point to see why I'm asking this "ridiculous question".

This is the problem: $$\log_{10}x=1-\log_{10}(x-3)$$ You can solve it like this: $$\log_{10}x=\log_{10}10-\log_{10}(x-3)$$ $$\log_{10}x=\log_{10}\frac{10}{x-3}$$ $$x=\frac{10}{x-3}$$ $$x(x-3)=10$$ $$x^2-3x-10=0$$ $$(x+2)(x-5)=0$$ $$x\stackrel{?}{=}-2,\,\,x\stackrel{?}{=}5$$ If you plug in $5$, it will work just fine. You will get when simplified $\log_{10}5=\log_{10}5$. But if you plug in $-2$, you get something like this: $$\log_{10}-2=\log_{10}\frac{10}{-2-3}$$ $$\log_{10}-2=\log_{10}-2$$ I know that negative logarithms with any base don't exist, but I thought of something. There's a logarithm property which states: $$\log_by=\log_bx \Rightarrow y=x$$ Using this property, we can say: $$\log_{10}-2=\log_{10}-2 \Rightarrow -2=-2$$ And, now (I hope) everyone will agree that $-2$ is indeed equal to $-2$ (itself).

So why besides $5$, isn't $-2$ solution to this problem? Even if the property that I stated doesn't apply to this case, could I theoretically invent a new imaginary number unit of $v_b=\log_b(-1)$ like someone back in the days did with $i=\sqrt{-1}$ and state that $\log_{10}(-2)=v_{10}+\log_{10}2$ ?

• I would suggest you read a bit about the complex logarithm...logarithms are defined for complex bases, but it gets a whole lot more complicated. – KR136 Apr 2 '16 at 2:53
• In fact, going from second step to the third step imply that $x>0$. – user164550 Apr 2 '16 at 2:57
• Well you see, that property isn't quite valid here because you are equating two undefined quantities. You could similarly argue that $(\frac{1}{0})^2=(\frac{1}{0})^2$ – KR136 Apr 2 '16 at 2:57
• @Αδριανός is referring to the complex logarithm, which is what indeed applies if we wish to take this problem into the complex field and discuss logarithms of negative inputs. You can get a great introduction from the first couple chapters of Saff and Snider's text Fundamentals of Complex Analysis. – zahbaz Apr 2 '16 at 2:58
• It may spark some more interest to know that $\log_{10}(-2)=(\ln(2)+i\pi) / \ln(10)$, for the principal value of the complex logarithm. – Bonnaduck Apr 2 '16 at 3:08

Okay we've talked a lot about complex logarithms, let's try and solve the problem over the complex numbers. (I think that is what the OP is really interested in; it has been reiterated that the only solution over the reals is 5)

We have:

$\log_{10} (-2)$

The change of base formula is consistent for all complex numbers, of which 10 is one. ($0i+10$). But we must be careful about branch selection. In this exercise we will denote $L(x)$ as the principal branch of the natural complex logarithm.

Therefore:

$\log_{10} (-2)$ =$\frac{L(-2)}{L(10)}$

By the definition of the principle branch:

$\log_{10} (-2)$ =$\frac{\ln(2)+iπ}{\ln(10)}$

As @uqtredd1 has noted, this solution is not one that should be submitted in the context of the presented problem, but is completely extraneous. The complex logarithm is a multivalued function, which is why branch selection was so important. There are many more possible "answers" to what $\log_{10} (-2)$ is in the complex sense.

• All answers are great but this answer elaborated on the other answers, so I accepted it. – KKZiomek Apr 2 '16 at 3:29

You are assuming that the following equation holds for all $a,b$ when you say $\log_{10}-2=\log_{10}\frac{10}{-2-3}$

$$\log a+\log b=\log ab$$

It does not. It only holds when $a,b>0$. When $a,b<0$ it's different.

It's like how

$$\sqrt{a}\times\sqrt{b}=\sqrt{ab}$$

Is not true for $a,b<0$.

Thus, $\log_{10}-2+\log_{10}-5 \neq 1$.

It's a matter of the context of the problem, in real analysis $\text{log}_ax$ is only defined for $x>0$. In complex analysis you can use the complex logarithm.

If you were to define a "complex log base 10" as

$$\text{log}_{10} x =\frac{\text{log}\ x}{\text{log} \ 10}$$

Where $\text{log}$ is the complex logarithm, then you can show that $x=-2$ is indeed a valid solution. However these solutions likely make no sense in the context of the problem.

This is a great question and very well written. The property that you refer to near the end is called injectivity. It says that if a function $f$ satisfies the property that $f(x) = f(y) \implies x = y$ then the function is injective. It turns out that the logarithm function satisfies this property and also another property, namely "surjectivity", which says that I can assign to any $x \in X$ an output, called $f(y) \in Y$, where $X$ and $Y$ are called the domain and codomain respectively. If a function is both injective and surjective, we call it bijective and can in essence move backwards and forwards between inputs and outputs. This is what we mean when we talk about inverse functions. The logarithm is the inverse of the exponential function. That is to say, if

$$f(x) = e^x$$

then

$$f^{-1}(x) = \log_{e}(x) = \ln(x).$$

The problem is that over $\mathbb{R}$, the exponential function is not bijective. That is to say, there is no $x \in \mathbb{R}$ such that, for instance, $f(x) = -1$. For $e^x$ to be bijective over the reals (and, as such, for it to have an inverse), we restrict its domain to the positive reals, denoted $\mathbb{R}^+$. This is why we cannot have negative arguments for the logarithm over the real numbers.

The only answer here is $x=5$, specifically because this problem has an unstated assumption: we're working over the field of real numbers $\mathbb{R}$. Over the reals, the logarithm is defined only for $x>0$. If we were working over $\mathbb{C}$, we would indeed find (many) other solutions.

The fact that $\log_b a = \log_b c \implies a=c$ on $\mathbb{R}$ does not imply that $a$ is a solution to your equation. It only implies that $a=c$.

If $-2$ is a solution, you must calculate $\log_{10}-2$, but doesn't exist.