## anonymous 5 years ago What is the half life of a radioative substance that decays by 6% in 7 years?

1. anonymous

half life is the period of time it takes to decay 50% of its original value. so what ration of 50% is 6% and apply that ratio to 7

2. anonymous

ummmm 50/6?

3. anonymous

So I make it about 58.3333

4. anonymous

yep 50/6 * 7

5. anonymous

Ohh! Thanks!

6. anonymous

hmm that is not what i got. let me try again.

7. anonymous

i got 78.4

8. anonymous

i think this problem is a little more complicated that ratios. the general formula for decay is $A=A_0e^{rt}$ where r is the rate of decay (r will be negative) t is time, $A_0$ is the initial amount and $A$ is what you have after t years. we do not know r, but we know that when t = 7 you have decreased by 6%. decreasing by 6% means you have 94% =.94 of the original amount so set $.94=e^{7r}$ and solve for r $7r = ln(.94)$ $r=\frac{ln.94}{7}=-.008839$ rounded now half life means you have half the original amount so set $\frac{1}{2}=e^{-.00839t}$ or better still (to avoid rounding error) $\frac{1}{2}=e^{\frac{ln(.94)}{7}t}$ and solve for t $ln(\frac{1}{2})=\frac{ln(.94)t}{7}$ $t=\frac{7ln(\frac{1}{2})}{ln(.94)}=78.4$ rounded

9. anonymous

i could be wrong of course, but i am fairly sure this is correct.

10. anonymous

ah yes what was I thinking. but isn't decay $e^{-rt}$ and not $e^{rt}$ because that's growth not decay so taking the log of both sides should give $-7r=\ln(0.94)$ the rest follows?

11. anonymous

I agree with the answer given. here is a doc to clarify the above comment. sorry for the misunderstanding

12. anonymous

you can write $A=A_0e^{-rt}$ it is just a matter of preference. my 'r' was negative. if you write it your way r is positive. it makes no difference.

13. anonymous

there is another way to solve this problem that amounts to the same thing. since you know it decreases by 6% in 7 years you could write the formula is $A=A_0(.94)^{\frac{t}{7}}$ then set $\frac{1}{2}=(.94)^{\frac{t}{7}}$ and solve for t, via $ln(\frac{1}{2})=ln((.94)^{\frac{t}{7}})=\frac{t}{7}ln(.94)$ and still get $t=\frac{7ln(\frac{1}{2})}{ln(.94)}$

14. anonymous

Nice :D