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

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Answer 1 · 2016-07-23T07:03:11

no solution

Since #ln a - ln b=ln(a/b) and lne=1#

you have the equivalent equation:

#ln(x/(x+1))=lne#

Then:

#x/(x+1)=e#

You also must fix:

#x>0 and x+1>0#

that is #x>0#

and

#x=e(x+1)#

#x=ex+e#

#x-ex=e#

#x(1-e)=e#

#x=e/(1-e)#

but it is a negative number, against the domain of the solution fixed as a positive number, then the equation has no solution