1. ## Word problem about Chemical Half-Lives

Dunno how to write this one, so I used a screenshot.

http://img111.imageshack.us/img111/8039/mathwordux7.jpg

The thing that trips me up is the 0 seconds.

<(I hate Word Problems!)

2. Originally Posted by mankvill
Dunno how to write this one, so I used a screenshot.

http://img111.imageshack.us/img111/8039/mathwordux7.jpg

The thing that trips me up is the 0 seconds.

<(I hate Word Problems!)
the half-life is 38 seconds. thus, if you start with $\displaystyle A_0$, after 38 seconds, you will have $\displaystyle \frac 12 A_0$. thus, the following equation must hold:

$\displaystyle \frac 12 A_0 = A_0a^{38}$

now just solve for $\displaystyle a$

3. a = 38?

4. Originally Posted by mankvill
a = 38?
ok, so lets say a = 38, that would mean $\displaystyle \frac 12 A_0 = A_0 (38)^{38} \implies \frac 12 = (38)^{38}$. obviously that makes no sense . how did you come up with that answer? do you have problems solving exponential equations?

5. Originally Posted by Jhevon
ok, so lets say a = 38, that would mean $\displaystyle \frac 12 A_0 = A_0 (38)^{38} \implies \frac 12 = (38)^{38}$. obviously that makes no sense . how did you come up with that answer? do you have problems solving exponential equations?
I thought I understood it. D:

...a = 19?

Or 2?

6. Originally Posted by mankvill
I thought I understood it. D:

...a = 19?

Or 2?

haha, stop guessing, and solve it. a would have to be less than 1, of course, otherwise, if you raise it to the 38th power, you would get a number greater than or equal to 1, which is bad, since we have 1/2 on the left hand side

$\displaystyle \frac 12A_0 = A_0a^{38}$

$\displaystyle \Rightarrow \frac 12 = a^{38}$

Now, how would you solve for $\displaystyle a$ here. what must you do to get rid of that 38th power, and be left with just $\displaystyle a$ on the right side?

7. Originally Posted by Jhevon
haha, stop guessing, and solve it. a would have to be less than 1, of course, otherwise, if you raise it to the 38th power, you would get a number greater than or equal to 1, which is bad, since we have 1/2 on the left hand side

$\displaystyle \frac 12A_0 = A_0a^{38}$

$\displaystyle \Rightarrow \frac 12 = a^{38}$

Now, how would you solve for $\displaystyle a$ here. what must you do to get rid of that 38th power, and be left with just $\displaystyle a$ on the right side?
Take the 38th root of $\displaystyle \frac 12$?

8. Originally Posted by mankvill
Take the 38th root of $\displaystyle \frac 12$?
yes!

9. Originally Posted by Jhevon
yes!

I got:

.981925 after rounding.

10. Originally Posted by mankvill
I got:

.981925 after rounding.

yup

11. 2nd part: How much of a 4 gram sample would be left after 10 seconds?

12. Originally Posted by mankvill

2nd part: How much of a 4 gram sample would be left after 10 seconds?
Plug in $\displaystyle A_0 = 4$ and $\displaystyle t = 10$ and solve for A(t) (remember, you just found the value for a)

13. Originally Posted by Jhevon
Plug in $\displaystyle A_0 = 4$ and $\displaystyle t = 10$ and solve for A(t) (remember, you just found the value for a)
To set it up, would that be:

1/2 (4) = (4)a^10

?

14. Originally Posted by mankvill
To set it up, would that be:

1/2 (4) = (4)a^10

?
no, your original equation is $\displaystyle A(t) = A_0a^t$ now plug in the values you got for $\displaystyle A_0, ~a, \mbox{ and }t$

15. Originally Posted by Jhevon
no, your original equation is $\displaystyle A(t) = A_0a^t$ now plug in the values you got for $\displaystyle A_0, ~a, \mbox{ and }t$
So A(10) = 4(a)^10?

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