# Thread: help with a question

1. ## help with a question

the active chemical in a certain medicine breaks down according to an exponential law. if the chemical has a half life of 20 years find the shelf life of the medicine if it is considered to be expired when less than 90% of the active chemical is still available...

pls can someone show me how to go about these type of questions as i have many similar questions to go through like this...

2. Originally Posted by jhbguy
the active chemical in a certain medicine breaks down according to an exponential law. if the chemical has a half life of 20 years find the shelf life of the medicine if it is considered to be expired when less than 90% of the active chemical is still available...

pls can someone show me how to go about these type of questions as i have many similar questions to go through like this...
$
y = y_0 e^{kt}
$

$0.5 = 1 \cdot e^{k \cdot 20}$

solve for $k$ , the determine the value of $t$ when $y = 0.9$

3. Originally Posted by skeeter
$
y = y_0 e^{kt}
$

$0.5 = 1 \cdot e^{k \cdot 20}$

solve for $k$ , the determine the value of $t$ when $y = 0.9$
thanks for the reply, but wat exactly does y = y_0 e^{kt}[/tex]
mean and how do i calculate it...im still really new with maths and im forced to do this coarse for university
thank you

4. $
(\frac{1}{2})^\frac{x}{20}
$

you use the 1/2 as your base because its a half life, and then it is to the x power, but since the half life only occurs every 20 years the x is over 20.
Now to find when there is less than 90% of the chemical you simply set the equation equal to .1, aka 10%, and solve.

$
(\frac{1}{2})^\frac{x}{20}=0.1
$

$
\log_\frac{1}{2} 0.1=\frac{x}{20}
$

$
\frac{\log 0.1}{\log \frac{1}{2}}=\frac{x}{20}
$

$
x=20(\frac{\log 0.1}{\log \frac{1}{2}})
$

$
x=66.439ish
$

5. Originally Posted by jhbguy
thanks for the reply, but wat exactly does y = y_0 e^{kt}[/tex]
mean and how do i calculate it...im still really new with maths and im forced to do this coarse for university
thank you
It is the exponential decay law. Sometimes it's written as $y = y_0 e^{-kt}$ but it's the same equation, only difference is the sign of the decay constant k. This is the way I'd do it as it means one less formula to remember but drewbears is quicker

6. Originally Posted by drewbear
$
(\frac{1}{2})^\frac{x}{20}
$

you use the 1/2 as your base because its a half life, and then it is to the x power, but since the half life only occurs every 20 years the x is over 20.
Now to find when there is less than 90% of the chemical you simply set the equation equal to .1, aka 10%, and solve.

$
(\frac{1}{2})^\frac{x}{20}=0.1
$

$
\log_\frac{1}{2} 0.1=\frac{x}{20}
$

$
\frac{\log 0.1}{\log \frac{1}{2}}=\frac{x}{20}
$

$
x=20(\frac{\log 0.1}{\log \frac{1}{2}})
$

$
x=66.439ish
$
hi there thanks for the help with that question....are these the type of questions that logs are needed for?? bcuz i got a question like in 1990 west germany population was 60 million and decreasing at rate of 0.1% per year, and east germany population was 16 million and increasing at 0.06% per yr...if this continued when would the populations be equal? my friend tells me the answer is definately 2816AD but he uses a really complex formuale. is it possible to work this out with logs as they seem really handy. thank you

7. hi there thanks for the help with that question....are these the type of questions that logs are needed for?? bcuz i got a question like in 1990 west germany population was 60 million and decreasing at rate of 0.1% per year, and east germany population was 16 million and increasing at 0.06% per yr...if this continued when would the populations be equal? my friend tells me the answer is definately 2816AD but he uses a really complex formuale. is it possible to work this out with logs as they seem really handy. thank you
to solve that i would use the equations y=60(.999)^x for west germany and y=16(1.006)^x for east germany and then just set them equal and solve..you could probably use logs to get it but it would be overly complicated and i dont think they would really work. and i just did it on my calculator real quick and would have agree with your friends answer