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 $x=1, -2$
 A: $$\ln { x+\ln { \left( x+1 \right) =\ln { 2 }  }  } \\ \ln { x\left( x+1 \right) =\ln { 2 }  } \\ x\left( x+1 \right) =2\\ x^{ 2 }+x-2=0\\ \left( x-1 \right) \left( x+2 \right) =0\\ { x }_{ 1 }=-2,{ x }_{ 2 }=1\\ $$
x should be $x>0$ hence ${ x }_{ 2 }=1$ is root
A: By inspection, $x=1$ is a solution. As the logarithm function is strictly increasing, it is the only one.
A: $\ln x+\ln (x+1)=\ln 2$
$e^{\ln x+\ln (x+1)}=e^{\ln 2}$
$e^{\ln x}. e^{\ln x+1}=2$
$x.(x+1)=2$
Now we have a quadratic equation    
A: 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 $$\implies x=\frac{-1\pm\sqrt{(1)^2-4(1)(-2)}}{2(1)}$$   $$\implies x=\frac{-1\pm\sqrt{9}}{2}$$   $$\implies x=\frac{-1\pm 3}{2}$$ $$\implies x=\frac{-1+3}{2}=\color{}{1}$$ &$$\implies x=\frac{-1-3}{2}=\color{}{-2}$$
Edit: Since log is defined for positive number i.e. $x>0$ hence we have $x=\color{blue}{1}$
