Is there some reason why $\ln(x^2+x) - \ln x = \ln (x+1)$ would not always be true?

Question

I ask because although I've learned in calculus class that it is a valid algebraic rule to work with logarithms, wolfram alpha doesn't unequivocally say that it is always true, which leaves me wondering if there is some condition I'm missing somewhere.

Thank you for any help!

Check for $x \in (-1,0]$. In that interval $\ln(x+1)$ is defined but not $\ln(x)$. However your equality is true for all $x>0$. — Axel, Jun 24 '21 at 08:17
It's true where everything is defined, but it's not always defined — Stephen Donovan, Jun 24 '21 at 08:19

Lalit Tolani · Accepted Answer · 2021-06-24T09:13:15.237

5

The domain of LHS is $(0,\infty)$ while that of RHS is $(-1,\infty)$

So yes , LHS=RHS would not be true in the case of $x\in (-1,0]$ because if $x\in(-1,0]$ then LHS will be undefined and RHS will be defined and since they are equal it will lead to a contradiction.

They both will be equal to each other in their common domain i.e. $x\in(0,\infty)$

edited Jun 24 '21 at 09:13

answered Jun 24 '21 at 08:32

Lalit Tolani

3,426

Is there some reason why $\ln(x^2+x) - \ln x = \ln (x+1)$ would not always be true?

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