# Thread: find half life, simple problem

1. ## find half life, simple problem

A 10 g sample of tritium decays to 8.5 grams in 2.93 years. What is the half life of tritium

the formula is y= A (0.5)^t/h

so A is the original amound, t is the time, and h is the half life, y is the changed amount...

i tried doing it but i didnt know...can anybody help? Does it involve logarithms?

this is what i got:

8.5 = 10 (0.5) ^2.93/h

log 8.5 = log (10/2) ^ 2.93/h

h = 2.93 log 8.5 / log 10/2

2. Originally Posted by magmagod
A 10 g sample of tritium decays to 8.5 grams in 2.93 years. What is the half life of tritium

the formula is y= A (0.5)^t/h

so A is the original amound, t is the time, and h is the half life, y is the changed amount...

i tried doing it but i didnt know...can anybody help? Does it involve logarithms?
Set t = 2.93, y = 8.5, and solve for h. And yes, there are logarithms involved. And of course set A = 10.

Edit: fixed typo

3. i know that...i meant for somebody to answer it cause i tried and failed in doing so.. i edited my question..you can check it again if you want

4. Originally Posted by magmagod
i know that...i meant for somebody to answer it cause i tried and failed in doing so.. i edited my question..you can check it again if you want
You are close.

Here is my work:

$\displaystyle 8.5 = 10\left(0.5\right)^{\frac{2.93}{h}}$

$\displaystyle 0.85 = \left(0.5\right)^{\frac{2.93}{h}}$

$\displaystyle log(0.85) = log\left((0.5)^{\frac{2.93}{h}}\right)$

$\displaystyle log(0.85) = \frac{2.93}{h}log(0.5)$

$\displaystyle \frac{2.93}{h} = \frac{log(0.85)}{log(0.5)}$

$\displaystyle h*log(0.85) = 2.93*log(0.5)$

$\displaystyle h = \frac{2.93*log(0.5)}{log(0.85)}$

5. ohhh...i get it so you just need to divide by 10 in the begining, then log both sides..thanks