# I need help with a natural log problem?

1. Dec 3, 2009

### empty.soul

1. The problem statement, all variables and given/known data
This is my first time ever posting anything on here.....but we just started working with ln? I know that it's the base e? or something like that...
but the problem is this....

1/3=ln(x^2/x-4)

2. Relevant equations

3. The attempt at a solution
I haven't attempted..i don't know how to do it o.o

Last edited: Dec 3, 2009
2. Dec 3, 2009

### ideasrule

ln is the opposite of "e to the power of", just like how addition is the opposite of subtraction. That is, if you raise e to the power of a certain number, taking the ln of the result gives you back the original number.

I presume you have to solve for x in 1/3=ln(x^2/x-4). I'll get you started: what happens when you raise e to the power of both sides? That is, e^(1/3)=e^(ln(x^2/x-4)), which simplifies to....