How do you solve #ln(2x-3)+ln(x-2) = 2lnx#?

2 Answers
Jun 15, 2016

Answer:

Solution is #x=6#

Explanation:

#ln(2x-3)+in(x-2)=2lnx#

or #ln(2x-3)(x-2)=lnx^2#

or #(2x-3)(x-2)=x^2#

or #2x^2-4x-3x+6-x^2=0#

or #x^2-7x+6=0#

or #x^2-6x-x+6=0#

or #x(x-6)-1(x-6)=0#

or #(x-1)(x-6)=0#

i.e. #x=1# or #x=6#

But #x=1# does not lie in domain as this makes #2x-3# and #x-2# both negative, hence

Solution is #x=6#

Jun 15, 2016

Answer:

#:. x=6.#

Explanation:

#ln(2x-3)+ln(x-2)=2lnx=lnx^2,#... [since #mlna=lna^m#]
#:. ln {(2x-3)(x-2)}=lnx^2,#... [as #lna+lnb=ln(ab)#]
#:.ln(2x^2-3x-4x+6)=lnx^2.#
#:. ln(2x^2-7x+6)=lnx^2.#
#:. 2x^2-7x+6=x^2.#...[because #ln# is #1 to 1# fun.]
#:. x^2-7x+6=0.#
#:. (x-6)(x-1)=0.#
#:. x=6, x=1.#

But #x=1# is not permissible, as it makes both #ln(x-2)# & #ln(2x-3) [=ln(-1)]# undefined.# :. x!=1.#

We verify #x=6# : #LHS=ln(12-3)+ln(6-2)=ln9+ln4=ln(4*9)=ln36=ln6^2=2ln6=RHS.#

#:. x=6,# is the only soln.