From Förberedande kurs i matematik 1

With the log laws, we can write the left-hand side as one logarithmic expression,

\displaystyle \ln x+\ln \left( x+4 \right)=\ln \left( x\left( x+4 \right) \right)

but this rewriting presupposes that the expressions \displaystyle \text{ln }x\text{ } and \displaystyle \text{ln}\left( x+\text{4} \right) are defined, i.e. \displaystyle x>0 and \displaystyle x+\text{4}>0. Therefore, if we choose to continue with the equation

\displaystyle \ln \left( x\left( x+4 \right) \right)=\ln \left( 2x+3 \right)

we must remember to permit only solutions that satisfy \displaystyle x>0 (the condition \displaystyle x+\text{4}>0 is then automatically satisfied).

The equation rewritten in this way is, in turn, only satisfied if the arguments \displaystyle x\left( x+\text{4} \right)\text{ } and \displaystyle \text{2}x+\text{3} are equal to each other and positive, i.e.

\displaystyle x\left( x+\text{4} \right)=\text{2}x+\text{3}

We rewrite this equation as \displaystyle x^{\text{2}}-\text{2}x-\text{3}=0 and completing the square gives

\displaystyle \begin{align} & \left( x+1 \right)^{2}-1^{2}-3=0 \\ & \left( x+1 \right)^{2}=4 \\ \end{align}

which means that \displaystyle x=-\text{1}\pm \text{2}, i.e. \displaystyle x=-\text{3} and \displaystyle x=\text{1}.

Because \displaystyle x=-\text{3} is negative, we neglect it, whilst for \displaystyle x=\text{1} we have both that \displaystyle x>0\text{ } and \displaystyle x\left( x+\text{4} \right)=\text{2}x+\text{3}>0. Therefore, the answer is \displaystyle x=\text{1}.