How do you solve #lnx=3+ln(x-5)#?

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Answer 1 · 2016-03-09T13:10:27

#x=5/(1-e^-3)#

Transposing term in #lnx=3+ln(x-5)#

#lnx-ln(x-5)=3# or

#ln(x/(x-5))=3# - as #lna-lnb=ln(a/b)#

Hence, as #lna=brArra=e^b#

#(x/(x-5))=e^3# or

#x=(x-5)e^3# or

#(e^3-1)x=5e^3# or

#x=(5e^3)/(e^3-1)# or #x=5/(1-e^-3)#