Word problem about Chemical Half-Lives

**mankvill** · June 29th 2008, 01:21 PM

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!)

**Jhevon** · June 29th 2008, 01:26 PM

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 $A_0$ , after 38 seconds, you will have $\frac 12 A_0$ . thus, the following equation must hold:

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

now just solve for $a$

**mankvill** · June 29th 2008, 01:27 PM

a = 38?

**Jhevon** · June 29th 2008, 01:30 PM

Originally Posted by mankvill

a = 38?

ok, so lets say a = 38, that would mean $\frac 12 A_0 = A_0 (38)^{38} \implies \frac 12 = (38)^{38}$ . obviously that makes no sense

. how did you come up with that answer? do you have problems solving exponential equations?

**mankvill** · June 29th 2008, 01:41 PM

Originally Posted by Jhevon

ok, so lets say a = 38, that would mean $\frac 12 A_0 = A_0 (38)^{38} \implies \frac 12 = (38)^{38}$ . 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?

**Jhevon** · June 29th 2008, 01:51 PM

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

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

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

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

**mankvill** · June 29th 2008, 01:52 PM

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

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

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

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

Take the 38th root of $\frac 12$ ?

**Jhevon** · June 29th 2008, 01:58 PM

Originally Posted by mankvill

Take the 38th root of $\frac 12$ ?

yes!

and your answer is? (to 6 decimal places)

**mankvill** · June 29th 2008, 02:01 PM

Originally Posted by Jhevon

yes!

and your answer is? (to 6 decimal places)

I got:

.981925 after rounding.

**Jhevon** · June 29th 2008, 02:10 PM

Originally Posted by mankvill

I got:

.981925 after rounding.

yup

**mankvill** · June 29th 2008, 02:11 PM

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

**Jhevon** · June 29th 2008, 02:13 PM

Originally Posted by mankvill

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

Plug in $A_0 = 4$ and $t = 10$ and solve for A(t) (remember, you just found the value for a)

**mankvill** · June 29th 2008, 02:17 PM

Originally Posted by Jhevon

Plug in $A_0 = 4$ and $t = 10$ 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

?

**Jhevon** · June 29th 2008, 02:21 PM

Originally Posted by mankvill

To set it up, would that be:

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

?

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

**mankvill** · June 29th 2008, 02:23 PM

Originally Posted by Jhevon

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

So A(10) = 4(a)^10?

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