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 $x=1, -2$
$$\ln { x+\ln { \left( x+1 \right) =\ln { 2 }  }  } \\ \ln { x\left( x+1 \right) =\ln { 2 }  } \\ x\left( x+1 \right) =2\\ x^{ 2 }+x-2=0\\ \left( x-1 \right) \left( x+2 \right) =0\\ { x }_{ 1 }=-2,{ x }_{ 2 }=1\\ $$
x should be $x&gt;0$ hence ${ x }_{ 2 }=1$ is root
We have, $$\ln x+\ln x(x+1)=\ln 2$$ $$\implies \ln(x(x+1))=\ln 2$$
$$\implies \frac{\ln(x(x+1))}{\ln 2}=1$$ $$\implies \log_2(x(x+1))=1$$ $$\implies x(x+1)=2$$ $$\implies x^2+x-2=0$$ now, solving above quadratic equation for $x$ as follows $$\implies x=\frac{-1\pm\sqrt{(1)^2-4(1)(-2)}}{2(1)}$$   $$\implies x=\frac{-1\pm\sqrt{9}}{2}$$   $$\implies x=\frac{-1\pm 3}{2}$$ $$\implies x=\frac{-1+3}{2}=\color{}{1}$$ &amp;$$\implies x=\frac{-1-3}{2}=\color{}{-2}$$
Edit: Since log is defined for positive number i.e. $x&gt;0$ hence we have $x=\color{blue}{1}$