Respuesta :

[tex]\bf 2ln(x)=ln(3x)-[ln(9)-2ln(3)] \\\\\\ ln(x^2)=ln(3x)-[ln(9)-ln(3^2)] \\\\\\ ln(x^2)=ln(3x)-[ln(9)-ln(9)] \\\\\\ ln(x^2)=ln(3x)-\stackrel{\stackrel{\textit{it happened right here}}{\downarrow }}{\left[ \cfrac{ln(9)}{ln(9)} \right]} \\\\\\ ln(x^2)=ln(3x)-[1]\impliedby \textit{recall }\frac{same}{same}=1\ne 0[/tex]

The answer is Since 0 in ln(3x) - 0 is not a logarithm, the property of logarithms cannot be used here.

The difference shown cannot be written as a quotient of logarithms.

The step ln(x2) = ln(3x) - (0) reduces to

ln(x2) = ln(3x).

The possible solutions are 0 and 3, with 0 being extraneous.