How do you solve #ln x - ln(x-1) = 2#?

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Answer 1 · 2018-04-04T22:05:18

#x=e^2/(e^2-1)#

First note that

#lnx-ln(x-1)=ln(x/(x-1))#

So for your problem we can write

#ln(x/(x-1))=2#.

Now take the exponential of both sides

#x/(x-1)=e^2#

(Another way to think of this is if #lna=b#, then #a=e^b#.)

#x=xe^2-e^2#

#x=e^2/(e^2-1)=1/(1-1/e^2)#