How do you solve #Ln 12 - ln(x - 1) = ln(x-2)#?

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Answer 1 · 2016-08-28T10:43:05

#x=5#

#ln12-ln(x-1)=ln(x-2)#

#hArrln12/(x-1)=ln(x-2)# or

#12/(x-1)=(x-2)# or

#12=(x-1)(x-2)# or

#x(x-2)-1(x-2)=12# or

#x^2-2x-x+2=12# or

#x^2-3x-10=0# or

#x^2-5x+2x-10=0# or

#x(x-5)+2(x-5)=0# or

#(x-5)(x+2)=0# or

#x=5# or #x=-2#

But as #x=-2# is extraneous as we cannot have log of a negative number,

Answer is #x=5#.