# Ln(x-2) = -2+ln(x)

1. Aug 30, 2007

### kdpointer

I have no idea where to start. I need to know how solve this.

I graphed both of the equations and then found the intersection, but it is a non repeating decimal and my answer has to be exact. I don't know how else to do it.

2. Aug 30, 2007

### rocomath

any work? gotta show some b4 we can help.

and exactly what are you solving for?

3. Aug 30, 2007

### bob1182006

Try to get rid of those logs, what can you do to get rid of the logs?

If that doesn't work try combining the logs to get 1 single log and see if that gives you the answer.

4. Aug 30, 2007

### Dick

Take e to the power of both sides.

5. Aug 30, 2007

### kdpointer

you can raise e to both sides?

6. Aug 30, 2007

### kdpointer

then is it simply just solving for x?

7. Aug 30, 2007

### kdpointer

i know that e^(ln(x-2)) is x-2 but i don't know what happens to the other side

8. Aug 30, 2007

### Dick

I thought you knew you were solving for x. What else is there to solve for? And sure you can raise e to both sides. If a=b, then e^a=e^b.

9. Aug 30, 2007

### Dick

What's e^(a+b). And how are your replies showing up so fast?

10. Aug 30, 2007

### rocomath

when you need help = constantly refreshing ;) and through e-mail notifications

11. Aug 30, 2007

### kdpointer

e^a * e^b ???

12. Aug 30, 2007

### Dick

My email notifications show up in a slug-like fashion. Like at least a minute or two between. I thought the forum was delaying for last minute edits. Guess I'm wrong, or the OP is anticipating the reply.

13. Aug 30, 2007

### Dick

You did it again. YES!

14. Aug 30, 2007

### kdpointer

oh dang.. i got it i think. -2/(e^(-2)-1) maybe?

15. Aug 30, 2007

### Dick

Yes!!!!!!! I know I've got to get this in quick.

16. Aug 30, 2007

### kdpointer

sweet.. thanks a lot! all i needed was a little push in the right direction haha