radium disintegrates at a rate proportional to the amount present. if 100 mg are set aside now, 96 mg will be left 100 years later. find how much will be left 2.56 centuries later and the half life too.

Question

lgbasallote · Answer

do i use $$100 - 4^t?$$

$$\implies 100 - 4^{2.56} = 65.22 = 66$$
lol what did i do wrong @satellite73 ?

anonymous · Answer

$$\frac{96}{100}=.96$$ so model as 
$$100\times(.96)^{\frac{t}{100}}$$

anonymous · Answer

you need the proportion it decreases by, not the absolute number
decreases by 4% retains 96% of its current amount

lgbasallote · Answer

this is pretty tricky haha

anonymous · Answer

half live is easy enough 
solve 
$$.96^{\frac{t}{100}}=\frac{1}{2}$$ for $t$
of course you will need logs for this solution

lgbasallote · Answer

hmm you dont use 0,5 instead of 0,96?

anonymous · Answer

but it is a on step solution 
$$t=100\times \frac{\log(.5)}{\log(.96)}$$

lgbasallote · Answer

it's lacking one year :O

t = 1697.9

it should be 1699

anonymous · Answer

ok i have confused you let me back up a step
you want the half life. you start with 100 so when half is gone you will have 50
you want to solve 
$$100\times (.96)^{\frac{t}{100}}=50$$ for $t$ but the first step is to divide both sides by 100 and write 
$$(.96)^{\frac{t}{100}}=\frac{1}{2}$$

lgbasallote · Answer

oh i see what happened

anonymous · Answer

i get 1698 rounded

anonymous · Answer

of course it doesn't matter what you start with, half is still half so you always set it equal to $\frac{1}{2}$

anonymous · Answer

i am not saying you shouldn't learn how to do it the other way, i am just saying this method is intuitive, snappy, and more accurate because if you solve for  $k$ and use  $p_0e^{kt}$ 
then since $k$ will be a log of some sort,  you will no doubt truncate the decimal when you solve and therefore be off by a little

lgbasallote · Answer

hmm i tried using diff eq but it gave that too

anonymous · Answer

other method you need to solve 
$$e^{100k}=.96$$ for $k$ 
not hard, you get 
$$k=\frac{\log(.96)}{100}=-0.0004082199452025517$$
then write the equation as $100e^{-0.0004082199452025517t}$

lgbasallote · Answer

how come when i do ln2/0.0408 then it gives the right answer?

lgbasallote · Answer

the more i add decimal places the lower the value

anonymous · Answer

i do not know where you got those numbers from

lgbasallote · Answer

$$\ln (\frac{100}{96}) = 0.0408$$

anonymous · Answer

you have it upside down

anonymous · Answer

$$100\times e^{100t}=96$$
$$e^{100t}=.96$$
$$100t=\log(.96)$$
$$t=\frac{\log(.96)}{100}$$

anonymous · Answer

and the reason  you got the right answer is that you had the numerator upside down as well
should be $\log(\frac{1}{2})$

anonymous · Answer

are you using centuries for your unit, or years?  looks like you might be using centuries, which is fine
in that case you solve 
$$e^k=.96$$ and get $$k=\log(.96)=-.0408$$

anonymous · Answer

then to solve for the half life put 
$$e^{-0.408t}=\frac{1}{2}$$
$$t=\frac{\ln(\frac{1}{2})}{-.0408}$$

anonymous · Answer

now your answer is in centuries instead of years

anonymous · Answer

oops typo there, first line should have been 
$$e^{-0.0408t}=\frac{1}{2}$$

anonymous · Answer

http://www.wolframalpha.com/input/?i=log%28.5%29%2Flog%28.96%29 http://www.wolframalpha.com/input/?i=log%28.5%29%2Flog%28.96%29 compare and constrast

lgbasallote · Answer

but if i use e^-0.0408 = 1/2 then wont it be 1.92?

anonymous · Answer

not according to wolfram http://www.wolframalpha.com/input/?i=e^%28-.0408t%29%3D.5

anonymous · Answer

should be $t=\frac{\log(.5)}{-.0408}$