# Solving logarithmic equations

1. Nov 11, 2013

### Coco12

1. The problem statement, all variables and given/known data
0ops.. The title should read solving logs..
Mod note: Fixed.

the original dosage contains 280 MBq of Iodine-131. If none is lost from the body, then after 6 hr there are 274 MBq of iodine-131. What is the half life of iodine I-131?

2. Relevant equations

Logcx=y

3. The attempt at a solution

I tried to put
274=280(1/2)^t/6 in and solve for t. But now I know it can't be right because it is not halving every 6 hrs.. However I don't know how you would figure it out. Can someone give me a hint?

Last edited by a moderator: Nov 11, 2013
2. Nov 11, 2013

### Staff: Mentor

The concentration varies with time according to the equation $I=I_0e^{-kt}$, where k is a constant that you need to determine from the data: -kt = ln(I/I0). Once k is known, find the time at which I is equal to half of I0: -kt1/2=ln(1/2)

3. Nov 11, 2013

### Ray Vickson

Put $D = D_0 (1/2)^{t/c},$ where $D =$ dosage after t hours, $D_0 =$ initial dosage (at t = 0) and $c$ is a constant. You are given $D_0 = 280$, and at t = 6 you have $D = 274$. You have enough information to find $c$.

4. Nov 11, 2013

### Dick

The radioactivity will follow the decay law 280*(1/2)^(t/k). k is the half life. You want to solve for k, not put k=6.

5. Nov 12, 2013

Thank you!