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.
