You need to convert the sum of logarithms, to the left side, into a product, using the identity `ln a + ln b = ln(a*b)` , such that:

`ln((x - 2)(2x - 3)) = 2ln x`

You need to use the following logarithmic ientity to the right side:

`a*ln b = ln b^a`

Reasoning by analogy yields:

`ln((x - 2)(2x - 3)) = ln x^2 => ((x - 2)(2x - 3)) = x^2`

You need to open the brackets to the left side, such that:

`x^2 - 3x - 4x + 6 = x^2`

Reducing duplicate terms yields:

-`7x + 6 = 0 => -7x = -6 => x = 6/7`

You need to test the value `x = 6/7` in equation, such that:

`ln(6/7 - 2) + ln(12/7 - 3) = 2ln(6/7)`

`ln((6-14)/7) + ln((12-21)/7) = 2ln(6/7)`

`ln(-8/7) + ln(-9/7) = 2ln(6/7)`

You should notice that `ln(-8/7)` and `ln(-9/7)` do not exist, hence, x = `6/7` does not satisfy the logarithms.

**Hence, evaluating the solution to the given equation yields that there is no solution to this equation.**