Solve: ln(2x-3)+ln(x-2) = 2lnx

please solve the question above :) thanks

ln(2x-3) + ln(x-2) = 2lnx
ln(2x-3) + ln(x-2) - 2lnx = 0
ln{[(2x-3)(x-2)] / ln(x^2)} = 0
[(2x^2 - 7x + 6) / x^2] = e^0
2x^2 - 7x + 6 = x^2
x^2 - 7x + 6 = 0
(x - 1)(x - 6) = 0
x = 1 or 6

Note however, that we cannot use x = 1 because we'd get ln(-1) for the first quantity, which is not in the domain. So the only answer is x = 6.

Hope this helped.d

ln(2x-3)(x-2)=lnx^2
multiply the whole thing by e, the ln's cancel out
(2x-3)(x-2)-x^2=0
then distribute and solve for x

Heck no. Do your homework yourself.