0
$\begingroup$

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!

$\endgroup$
2
  • 6
    $\begingroup$ 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 $x>0$. $\endgroup$
    – Axel
    Jun 24, 2021 at 8:17
  • 4
    $\begingroup$ It's true where everything is defined, but it's not always defined $\endgroup$
    – Stephen Donovan
    Jun 24, 2021 at 8:19

1 Answer 1

Reset to default
5
$\begingroup$

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. $x\in(0,\infty)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.