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

1 answer:

3 0

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

Using properties of logarithms

</span> $\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 ,


$
<span>
so x2 is not a solution of this logarithmic equation.

Only x1 is a solution.

</span>

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