Why is Desmos not showing $\ln(y)-\ln(y-1)$ as the same as $\ln(y/(y-1))$?

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Desmos link.

The first two equations are equivalent and I can see this since both lines perfectly overlap: enter image description here

According to my understanding and spreadsheet, the left side of the second equation, $\ln(y)-\ln(y-1)$ should be the same as $\ln(y/(y-1))$ However, if you see the third formula in the link, it is not the same, it both overlaps the first two and then has another line going from $0$ to $-\infty$..

Quick example from my spreadsheet:

y=5
y-1 = 4
ln(5)-ln(4)=0.2231435513
ln(5/4)=0.2231435513

They are the same then?

Why is the third formula not the same as the first two? enter image description here

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The reason is simple:

  • $f(x) := \ln(x) - \ln(x-1)$ is always positive, since $\ln(x-1) < \ln(x)$. Note that we cannot have $x \le 1$ for this equation to be well-defined.
  • $g(x) := \ln (x/(x-1))$ is allowed to take on negative values on the other hand, since while a logarithm is not allowed to have a negative argument, we have a positive one here when $x<0$.

The usual rule of logarithms you're thinking of -- $\log_b(x) - \log_b(y) = \log_b(x/y)$ -- implicitly assumes $x,y>0$ in order for the logarithms on the left-hand side to be well-defined. Of course, you could have $x,y < 0$ and thus $x/y > 0$ and thus the logarithm on the right-hand side is perfectly fine, but those on the left are not.

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The main reason is that the formula $\ln(a/b)=\ln(a)-\ln(b)$ is false in general!

The real statement is

If $a$ and $b$ are positive, then $\ln(a/b)=\ln(a)-\ln(b)$.

Indeed, if $a$ and $b$ are both negative, then $\ln(a)$ and $\ln(b)$, hence their difference, are undefined but, since $a/b>0$, $\ln(a/b)$ is well defined.

In your case, $\ln(y)-\ln(y-1)$ is defined if and only if both $y>0$ and $y-1>0$, that is for $y>1$. On the other hand, $\ln(y/(y-1))$ is defined if and only if the whole ratio $\frac{y}{y-1}$ is positive, which happens when $y>1$ or $y<0$.

For exemple, if $y=-1$, then $\ln(y/(y-1))=\ln(1/2)=-\ln(2)\simeq -0.6$.