Solve this math problem for me!

[BD2003]

This might be an easy problem for some of you, but it left me totally mystified! Logarithmic equations are just not my thing. Anyways:

ln(x-3)=-4+ln5

Solve for x.

 

[BA]

ln(x-3)-ln(5)=4

ln((x-3)/5)=4

e^4=(x-3)/5

5e^4=x-3

x=5e^4+3

 

[jhu]

e^(4+ln5) + 3

 

[DAM]

im on crack



x = e^-4 + e^5 + 3



dam()

 

[jhu]

eh, mine wurx 2

 

[BA]

DAM, I don't know what you're doing...jhu and I got equivalent answers

 

[BD2003]

Well I figured some of it out during the test. I got it down to ln((x-3)/5)=-4. I thin multiplied both sides by e to get (x-3)/5=-4e Is that where I went wrong? I shouldnt have multipled by e, but put both sides to the power of e?

 

[BD2003]

Ack! F*ck it! I hate math!

 

[BA]

yup, to get rid of the ln, it's e^whatever

 

[Mday]

yeah, to solve a problem, it's a combination of arithmetic, and the proper use of inverses...

 

[thEnEuRoMancER]

Don't forget the minus: x=5e^(-4)+3

 

[Ranger X]

So funny how you people jump at the opportunity to answer this question. Hell I would have but you guys beat me to it.