# Math Help - Isolating for x

1. ## Isolating for x

How would I go about isolating for x in this case?

1 = -ln(x+1)

2. Originally Posted by NAPA55
How would I go about isolating for x in this case?

1 = -ln(x+1)

Write in expo form..

$1 =-ln(x+1) \iff -1=ln(x+1)$

$e^{-1}=x+1 \iff e^{-1}-1 = x$

3. 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.

$1 =ln(x+1)$

$e^{1}=x+1$

$e^{1}-1 = x$

$1.7 = x$