A math problem stuck in my head

[megalomania]

I hate thee deceptively simple calculus problems...

F(x) = A ln x + B where the x,y points 1,1 and 3,0 are on the graph. What are A and B? The closest I could get is taking the derivitave giving f'(x) = A1/x
I still don't know how to solve for A

Anyone have more math smarts than I?

[The_Duke]

If (1,1) and (3,0) are points on the graph, then this means

F(1) = 1,

and

F(3) = 0.

Since F(1) = A ln 1 + B, and ln 1 = 0 (remember that ln x = n is
equivalent to e^n = x, so ln 1 = 0 because e^0 = 1), we then see

1 = A ln 1 + B = B.

Therefore, B = 1. Now use the fact that F(3) = 0 to find A. ;)

[telkanuru]

To say the above in another way, you are basically given two equations to solve two unknowns, so you find one in terms of the other and substitute. Happy mathing, and may all your equations be Homeric (ie. long, descriptive, and full of Greek letters...)!

[megalomania]

Hmm, it all seems so easy once you can do it :) I figured out the B=1 part, but A took some doing. I eventually found it by plugging in 1 for B and using the 0,3 point. Solving for A is then 0 = A ln 3 + 1 and A = -1/ln 3