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Let's say we have a function with absolute value like:

$f(x) = \ln\vert x\vert$ where $x$ is any real number except 0

Now, when we get rid of the absolute value, we get this:

$f(x) = \ln(x)$ where $x$ is positive

$f(x) = \ln(-x)$ where $x$ is negative

But here is the thing, $\ln$ properties allow us to do something like

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

So we apply this to the second form of function

$f(x) = \ln(-x)$ means

$f(x) = \ln(-1) + \ln(x)$

But $\ln$ is undefined for negative real numbers so I don't get it, how does this work?

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    $\begingroup$ $\log(xy)=\log(x)+\log(y)$ only holds in real numbers when $x,y>0$. $\endgroup$ Jan 22, 2021 at 9:59
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    $\begingroup$ You are already assuming $x < 0$ when you write $f(x) = \ln (-x)$, and the $\ln$ properties are only valid for $x, y > 0$. $\endgroup$
    – macton
    Jan 22, 2021 at 10:00

3 Answers 3

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$\ln (xy)=\ln (x)+\ln (y)$ holds only for positive numbers $x$ and $y$. You cannot use negative numbers in this.

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The absolute value is present exactly to avoid this.

When you write $f(x) = ln|x|$ It means that a negative number cannot come inside the bracket.

so when you apply the second form of function, x is a negative number and negative and negative cancel out. What you have done is similar to

$ln(1)=ln(-1*-1)=ln(-1)+ln(-1)$, which cannot be done, as the property is defined only for positive real numbers :~)

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    $\begingroup$ An analogy can be made with the square root function. It is true to say that $2 = \sqrt{4} = \sqrt{(-2) \times (-2)}$ but one cannot conclude that $2 = \sqrt{-2}\times \sqrt{-2}$. $\endgroup$
    – Didier
    Jan 22, 2021 at 10:06
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    $\begingroup$ Yes quite true. $\endgroup$
    – Aatmaj
    Jan 22, 2021 at 10:07
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The equality $\ln (xy) = \ln x + \ln y$ only holds if all components are well defined, namely if $x,y >0$. The same happens in situations that do not involve the absolute value... For instance $\ln (x^2)$ is well defined for any $x \ne 0$, but the property $\ln(x^2) = 2 \ln x$ only holds for $x>0$.

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  • $\begingroup$ Nice viewpoint... $\endgroup$
    – Aatmaj
    Jan 22, 2021 at 10:13

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