Exact Solution for ln(x-2) = -2+ln(x) - Graphing Method

[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.

Please HELP!

 

[rocomath]

any work? got to show some b4 we can help.

and exactly what are you solving for?

 

[bob1182006]

Hm..without giving you the answer.

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.

 

[Dick]

Take e to the power of both sides.

 

[kdpointer]

you can raise e to both sides?

 

[kdpointer]

then is it simply just solving for x?

 

[kdpointer]

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

 

[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.

 

[Dick]

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

 

[rocomath]

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

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

 

[kdpointer]

e^a * e^b ?

 

[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.

 

[Dick]

e^a * e^b ?

You did it again. YES!

 

[kdpointer]

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

 

[Dick]

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

 

[kdpointer]

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