Respuesta :

ln(x+2)-ln(4x+3)=ln(1/2*x)

Using properties of  logarithms

[tex] \frac{ln(x+2)}{ln(4x+3)} = ln \frac{x}{2} \\ \\ \frac{x+2}{4x+3} = \frac{x}{2} \\ \\2(x+2)=x(4x+3) 2x+4=4x^2+3x \\ \\ 4 x^{2} +x-4=0 \\ \\ x= \frac{-b+/- \sqrt{b^{2}-4ac} }{2a} \\ \\ x= \frac{ -1+/-\sqrt{1+64} }{8} \\ \\ x_{1} = \frac{-1+ \sqrt{65} }{8} \\ \\ x_{2} = \frac{-1- \sqrt{65} }{8} Check: When you substitute x_{2} into \\ \\ ln(4x+3)=ln(4* \frac{-1- \sqrt{65} }{8} ) =ln( \frac{-1- \sqrt{65} }{2} ) you will get negative number under ln, that is impossible , [/tex]

so x2 is not a solution of this logarithmic equation.

Only x1 is a solution.

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