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Mathematics 24 Online
OpenStudy (boxman61):

solve: ln x - ln(x-1)=1

OpenStudy (dumbcow):

log(a) - log(b) = log(a/b) use this log property

OpenStudy (ash2326):

We have \[\ln x -\ln (x-1)=1\] Do you know the properties of logarithm? How it's defined?

OpenStudy (boxman61):

i have it to \[\ln (x/x-1)=1\]

OpenStudy (ash2326):

Now take antilogarithm both sides. Do you know how to do that?

OpenStudy (boxman61):

not ringing a bell currently

OpenStudy (ash2326):

Ok \[\ln x=y\] \[=>x=e^y\] Do you understand this?

OpenStudy (boxman61):

yes

OpenStudy (ash2326):

SO here we have \[\ln\frac{x}{x-1}=1\] Taking antilog \[\frac{x}{x-1}=e^1\] \[\frac{x}{x-1}=2.782\] I think you can find x now, Can't you?

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