find the exact value of x for which
ln(x^2+4x+3)-ln(x^2+x)=4
$\displaystyle \ln (a) - \ln (b) = \ln \left(\frac{a}{b}\right)$
$\displaystyle \ln (x^2+4x+3) - \ln (x^2+x) = 4$
$\displaystyle \ln \left(\frac{x^{2} + 4x + 3}{x^{2} + x}\right) = 4$
Simplify the inside by factoring and canceling the common factor.
This should help you right after: $\displaystyle \ln (a) = b \iff \log_{e}a = b \iff a^{b} = e$ where e is the constant 2.718...
Then it should be a matter of simple algebra to solve for x.