How would I go about isolating for x in this case?
1 = -ln(x+1)
Hmm... that's what I was doing and I got the same answer, but that's not what it's supposed to be. It should be 1.7 instead of -0.6.
EDIT: I see why I'm wrong. I checked my previous work and I made a mistake. The ln should NOT be negative. Now it works and I get the answer I want.
$\displaystyle 1 =ln(x+1)$
$\displaystyle e^{1}=x+1 $
$\displaystyle e^{1}-1 = x$
$\displaystyle 1.7 = x $