# Math Help - Solving for x in this natural log inequality

e^2-3x > 5; solve for x

So I went forward and figured that ln(5) < 2-3x

I don't know where to go from there, I thought I had a couple of answers but they weren't right. I had the other parts of similar questions right but this ones stumping me. Any help?

2. Originally Posted by BugzLooney
e^2-3x > 5; solve for x

So I went forward and figured that ln(5) < 2-3x

I don't know where to go from there, I thought I had a couple of answers but they weren't right. I had the other parts of similar questions right but this ones stumping me. Any help?
$e^{2-3x} > 5$

$2-3x > \ln{5}$

$-3x > \ln{5} - 2$

$3x < 2 - \ln{5}$

$x < \frac{2-\ln{5}}{3}$

3. Hey thanks that is easy I think I got something close but thanks for clearing that up! Maybe when I brush up my skills I'll be helping some guys around here.