We probably all know that

$$\log (x y)=\log x + \log y$$

However the expression on the left needs only $x y >0$ to be defined, whereas the expression on the right requires $x>0$ and $y>0$.

For example suppose you were maximizing $$\max \log(x)+\log(y), s.t. constraints$$


$$\max \log(xy) s.t. constraints$$

The two problems seem to be entirely different, although a naive substitution would imply that the two are identical.

My question is: Is this point discussed anywhere? I haven't come across a discussion of this. Are there any references to find more on that?

  • 1
    $\begingroup$ In this context, the correct statement is "for all positive numbers $x$ and $y$, $\log(xy) = \log(x) + \log(y)$." The statement "for all real numbers $x$ and $y$, $\log(xy) = \log(x) + \log(y)$" is ill-defined (assuming we only define $\log$ on positive numbers). $\endgroup$
    – L. F.
    Nov 12, 2022 at 2:50
  • $\begingroup$ If $\log (x y)$ is defined opportunistically (for anything that makes sense) then it does not equal $\log x+\log y$ defined opportunistically (for anything that makes sense, ignoring imaginary numbers) $\endgroup$
    – Cris
    Nov 12, 2022 at 2:55
  • $\begingroup$ Yes, only when $x,y>0$. $\endgroup$
    – Z Ahmed
    Nov 12, 2022 at 2:58
  • 2
    $\begingroup$ It is not technically appropriate to directly talk about the truth value of an open formula (i.e., a formula that includes unquantified variables). For example, it is technically inappropriate to say that the formula "$x - x = 0$" is true. The formula "for all real numbers $x$, $x - x = 0$", however, is a true closed formula. In practice, the latter is usually implied when referring to the former. $\endgroup$
    – L. F.
    Nov 12, 2022 at 3:00

3 Answers 3


You are right: Formulas and theorems come with conditions on the variables they are about. $\log(x)+\log(y)=\log(xy)$ is false. What is true is

For every positive real numbers $x$ and $y$, $\log(x)+\log(y)=\log(xy)$

I guess that some people would consider that, when writing $\log(x)+\log(y)=\log(xy)$, it is implicitly assumed that each number exists (here, $x$ and $y$ are positive), but I think it is a bad habit, especially for students.

This has practical implications. As an example, the equation $2\log(x+1)-\log(x+3)=0$ is NOT equivalent to $\log\left( \frac{(x+1)^2}{x+3} \right)=0$ since $-2$ is a solution of the latter but not of the former ($\log(-1)$ is not defined).

Your optimization problem is another example. In maximizing $\log(xy)$, you have the constraint that the product $xy$ should be positive (so $x$ and $y$ are nonzero and have the same sign), while in maximizing $\log(x)+\log(y)$, the constraints are stronger ($x$ and $y$ should be positive).


If we work with real valued logarithm and if we only have the constraint $xy>0$, you can derive the similar logarithm rule:

$$ \begin{align}\log(xy)&=\log \left(|xy|\right)\\ &=\log \left(|x|×|y|\right)\\ &=\log |x|+\log |y|.\end{align} $$

In general, note that

$$\log (xy)=\log x+\log y$$

holds, iff $x>0\wedge y>0$.


If you define it via integral formula i.e. $\displaystyle \ln(x)=\int_1^x \dfrac{dt}t$

You can see that you need the interval $[1,x]$ (or $[x,1]$ if $x<1$) to not cross $0$ else the integral would be divergent, which consequently means that $x$ must be strictly positive.

The multiplication to addition formula can be proven this way ($a,b>0$):

First by substituting $u=\frac 1t$ then $$\ln(\tfrac 1x)=\int_1^{\frac 1x}\dfrac{dt}{t}=\int_1^x \dfrac{-u\mathop{du}}{u^2}=-\int_1^x du=-\ln(x)$$

Then by substituting $t=bu$ then $$\ln(ab)=\int_1^{ab}\dfrac{dt}{t}=\int_{\frac 1b}^a \dfrac{b\mathop{du}}{bu}=\int_{\frac 1b}^a \dfrac{du}{u}\ \overset{(*)}{=}\ \ln(a)-\ln(\tfrac 1b)=\ln(a)+\ln(b)$$

Justification of (*):

  • It is straightforward when $a,b>0$ since we can rewrite $\displaystyle\int_{\frac 1b}^1 \frac{du}{u}+\int_1^a \frac{du}{u}\ $ and then use the definition.

  • On the contrary when $a,b<0$ this split is not possible since each individual integral is divergent. But we can substitute $v=-u$ and get

$$\int_{\frac 1b}^a \dfrac{du}{u}=\int_{-\frac 1b}^{-a} \dfrac{-dv}{-v}=\int_{\frac 1{|b|}}^{|a|} \dfrac{dv}{v}=\ln(|a|)+\ln(|b|)$$

Now with $|a|,|b|>0$ the split is possible.

In the end the "true" formula is rather:

  • if $\ln(a),\ln(b)$ are defined (i.e. then $a,b>0$) then their sum is equal to $\ln(ab)$
  • if $\ln(ab)$ is defined (i.e. $ab>0\iff\{a,b>0\}\lor\{a,b<0\}$) then it is equal to $\ln(|a|)+\ln(|b|)$

Yet commonly the case where $a,b$ are both negative is not very frequent, and the usage settled for the omission of the absolute values.


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