Aus Online Mathematik Brückenkurs 1

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

\displaystyle \ln x+\ln (x+4) = \ln (x(x+4))\,,

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

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

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(x+4) and \displaystyle 2x+3 are equal to each other and positive, i.e.

\displaystyle x(x+4) = 2x+3\,\textrm{.}

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

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

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

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