Question

find how much time passes before the Blood Alcohol level reaches .04, we made C(t)=.04, and then solved the original equation (not the derivative) for t.

This is the work that I've done for the problem, but I know it's not right, the right answer is t=5.66

.04=(.12t)(e^(-t/2)
ln(.04)=ln((.12t)(e^(-t/2))
ln(.04)=(.12t)(-t/2)
2ln(.04)=(.12t)(-t)
2ln(.04)=-.12t^2
(2ln(.04))/-.12=t^2
Sqrt(2ln(.04))/-.12) = t

Answer 1

What is the original equation?

C(t) = C(final)[1 - exp(-kt)] ? Or ?

Answer 2

C (t) = 0.12te^(-t/2)
and
c'(t)=.06(t-2)e^(-t/2)

Answer 3

0.04 = 0.12 t exp(-t/2)
ln(0.04) = ln(0.12) + ln(t) - t/2

solve for t.

t = 2[ln(3t)]

Plot y = t and y = 2[ln(3t)] vs. t

and see value of t where they intersect.

Answer 4

How did you simplify to t=2[ln(3t)]
for the problem that this is a part of, I need to show all of my steps for solving of t

Answer 5

ln(0.12) - ln(0.04) = ln(0.12/0.04) = ln(3)

ln(3) + ln(t) = ln(3t)

Answer 6

Thank you so so so much!