Question

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

Answer 1

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`

Answer 2

`:. 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.