Answer :

[tex]ln(x-2)+ln(2x-3)=2lnx\\\\D:x-2 > 0\ \wedge\ 2x-3 > 0\ \wedge\ x > 0\\x > 2\ \wedge\ x > 1.5\ \wedge\ x > 0\\x\in(2;\ \infty)[/tex]

[tex]ln[(x-2)(2x-3)]=lnx^2\\\\ln(2x^2-3x-4x+6)=lnx^2\\\\ln(2x^2-7x+6)=lnx^2\iff2x^2-7x+6=x^2\\\\2x^2-x^2-7x+6=0\\\\x^2-7x+6=0\\\\x^2-x-6x+6=0\\x(x-1)-6(x-1)=0\\(x-1)(x-6)=0\iff x-1=0\ \vee\ x-6=0\\\\x=1\notin D;\ x=6\in D\\\\Solution:x=6.[/tex]
konrad509
[tex]\ln(x-2)+\ln(2x-3)=2\ln x \\ D:x-2>0 \wedge 2x-3 >0 \wedge x>0\\ D:x>2 \wedge 2x>3 \wedge x>0\\ D:x>2 \wedge x>\frac{3}{2}\\ D:x>2\\ \ln(x-2)(2x-3)=\ln x^2\\ 2x^2-3x-4x+6=x^2\\ x^2-7x+6=0\\ x^2-x-6x+6=0\\ x(x-1)-6(x-1)=0\\ (x-6)(x-1)=0\\ x=6 \vee x=1\\ 1\not \in D\Rightarrow x=6 [/tex]

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