# Differential Equation Rate of decay problem

1. Sep 29, 2008

### clope023

1. The problem statement, all variables and given/known data

A chemical reaction converts a certain chemical into another chemical, and the rate at which the 1st chemical is converted is proportional to the amount of of this chemical present at any time. At the end of 1 hour, 50g of the 1st chemical remain; while at the end of 3 hours only 25g remain.

a) how many grams of the 1st chemical were present initially?

2. Relevant equations

dx/dt = -kx

x(0) = x0
x(1) = 50g
x(3) = 25g

int[dx/x = -kdt] -> ln(x) = -kt + c

x = ce^(-kt)

x(0) = x0 -> x0 = C(1), x = x0e^(-kt)

3. The attempt at a solution

I'm pretty much stuck where I ended the relevant equations, I'm not sure what to do to get x0 by itself, I tried plugging in the 1st hour value which got me

50 = x0e^(-k), but I need to solve for k or the value of e^(-k) before I can solve for x0, which I could do easily by taking e^(-k)/50 and then I could solve the rest of the problem; any help would be greatly appreciated.

oh and if th3plan is reading this, thanks for the help in the other thread I made.