Question

The half-time for the elimination of a drug is h, show that k=ln2/h. Where A=De^(kt).

Answer 1

y-2x-6=0 and y/2-3=x solve the two variables

Answer 2

A friend of mine did it like this

D=De^kh
D/2=De^kh
1/2=e^kh
ln(1/2)=kh
(ln(1/2))/h=k

And somehow got k=(ln2)/ h

Answer 3

Wait, k is supposed to be negative.

Answer 4

please help me solve y-2x-6=and y/2-3=x

Answer 5

wtf? Make your own question haha

Answer 6

I think it is A=De^(-kt) in the question?

Answer 7

Yeah

Answer 8

But how would you get k=ln2/h?

Answer 9

Now,that makes sense
A=De^(-kt), where A = D/2, t=h
A=De^(kt)
D/2 =De^(-kh)
1/2 = e^(-kh)
Take ln on both sides
ln (1/2) = ln e^(-kh)
-ln2 = (-kh) lne , note that lne =1
-ln2 = -kh
k = (ln2)/h
Does it make sense to you?

Answer 10

The step where ln(1/2) goes to -ln2 is confusing me

Answer 11

Oh wait ln(1/2) is equal to -ln2?