Question 446794
{{{ln((x+3)/(x+2)) = 5}}}
==> {{{(x+3)/(x+2) = e^5}}}
==> {{{x+3 = e^5x + 2e^5}}}
==> {{{3 - 2e^5 = (e^5 - 1)x}}}
==> {{{x = (3 - 2e^5)/(e^5 - 1)}}}, approximately -1.9932 to 4 decimal places.

Since this value would satisfy the original equation, this is the final answer.