<s>[INST] Solve for $x$ - Logarithm Equation $\ln x+\ln(x+1)=\ln 2$ My attempt:
$\ln x(x+1)=\ln 2$
$e^{\ln x(x+1)}=e^{\ln 2}$
$x(x+1)=2$
$x^2+x-2=0$
$(x-1)(x+2)=0$
therefore $x=1, -2$
 [/INST] $$\ln { x+\ln { \left( x+1 \right) =\ln { 2 }  }  } \\ \ln { x\left( x+1 \right) =\ln { 2 }  } \\ x\left( x+1 \right) =2\\ x^{ 2 }+x-2=0\\ \left( x-1 \right) \left( x+2 \right) =0\\ { x }_{ 1 }=-2,{ x }_{ 2 }=1\\ $$
x should be $x>0$ hence ${ x }_{ 2 }=1$ is root
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