Answer
$x=\dfrac{-1+\sqrt{9+4e}}{2}$
Work Step by Step
$\ln(x-1)+\ln(x+2)=1$
Combine the logarithms on the left side of the equation as a product:
$\ln(x-1)(x+2)=1$
$\ln(x^{2}+x-2)=1$
Write this equation in exponential form:
$x^{2}+x-2=e^{1}$
$x^{2}+x-2=e$
Take the $e$ to the left side of the equation:
$x^{2}+x-2-e=0$
$x^{2}+x-(2+e)=0$
Solve using the quadratic formula, which is: $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
For this particular equation, $a=1$, $b=1$ and $c=-(2+e)$
$x=\dfrac{-1\pm\sqrt{1^{2}-4(1)[-(2+e)]}}{2(1)}=\dfrac{-1\pm\sqrt{1-(4)(-2-e)}}{2}=$
$...=\dfrac{-1\pm\sqrt{1+8+4e}}{2}=\dfrac{-1\pm\sqrt{9+4e}}{2}$
The initial equation is undefined for $x=\dfrac{-1-\sqrt{9+4e}}{2}$, so we can discard this solution. Our final answer is:
$x=\dfrac{-1+\sqrt{9+4e}}{2}$
