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

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Answer 1 · 2018-05-25T08:56:32

#therefore x = \frac{e^2}{1 - e^2}#

Use the property of logarithms

#log(a)-log(b) = log(a/b)#

to write

#ln(x)-ln(x+1)=2 \iff ln(x/(x+1)) = 2#

Consider both sides as exponents of #e#:

#e^{ln(x/(x+1))} = e^2#

By definition, #e^ln(z)=z#, so

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

#therefore x = (x+1)e^2#

#therefore x = e^2x+e^2#

#therefore x - e^2x = e^2#

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

#therefore x = \frac{e^2}{1 - e^2}#

Answer 2 · 2018-05-25T08:57:39

#lnx-ln(x+1)=2#

Using logarithmic rules we have #ln(x/(x+1))=2=lne^2#

Then #x/(x+1)=e^2#

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