# A couple of Exponential Functions Questions

• Jul 30th 2007, 04:31 PM
usm_67
A couple of Exponential Functions Questions
8). The oldest rock knows to data found in Minnesota contains 2g of carbon 14(C^14). Using geological methods toestimate the original size of the rock, scentists estimate it to originally have contained 4.096kg of C^14 .The Half-life of c^14 is 5760 years.What is the approximate age of rock?

9). Land developers had to delay a housein development because the land had been previously used as a nucleur waste pit for Radium-D, an isotope has decayed to 1/√2 of its original. When could they build ifRadium-D has a half life of 22 years.

both use the y=ka^x, i just cant seem to get the answer right when i plugin the values its prol cuz i subed it in wrong,if some1 cud kindly help me it wud be apreciatted.
• Jul 30th 2007, 04:47 PM
topsquark
Quote:

Originally Posted by usm_67
8). The oldest rock knows to data found in Minnesota contains 2g of carbon 14(C^14). Using geological methods toestimate the original size of the rock, scentists estimate it to originally have contained 4.096kg of C^14 .The Half-life of c^14 is 5760 years.What is the approximate age of rock?

9). Land developers had to delay a housein development because the land had been previously used as a nucleur waste pit for Radium-D, an isotope has decayed to 1/√2 of its original. When could they build ifRadium-D has a half life of 22 years.

both use the y=ka^x, i just cant seem to get the answer right when i plugin the values its prol cuz i subed it in wrong,if some1 cud kindly help me it wud be apreciatted.

The decay equation is indeed
$\displaystyle y = ke^{bt}$

How do we use this?

Well we know the half-life in each case, so what does that mean? It means that the when the time is equal to the half-life only half of the sample is left.

When t = 0
$\displaystyle y = ke^{(0)} = k$

So k is the amount of sample initially, before it started to decay. So when t is equal to the half-life $\displaystyle \tau$ we only have (1/2)k of the substance:
$\displaystyle \frac{1}{2}k = ke^{b \tau}$

$\displaystyle \frac{1}{2} = e^{b \tau}$

$\displaystyle ln \left ( \frac{1}{2} \right ) = b \tau$

$\displaystyle b =\frac{ln \left ( \frac{1}{2} \right )}{\tau} = - \frac{ln(2)}{\tau}$

So the decay equation is:
$\displaystyle y = ke^{-(ln(2) \cdot t)/\tau} = k2^{-t/\tau}$

Try using this form for your questions. If you still have problems go ahead and let us know.

-Dan
• Jul 30th 2007, 04:58 PM
usm_67
well bro u totally lost me with the equation because he din teach us anythin like that, can u please solve it so the answer for first one is 63360 years and second one is 11 years
• Jul 30th 2007, 05:05 PM
topsquark
Quote:

Originally Posted by usm_67
8). The oldest rock knows to data found in Minnesota contains 2g of carbon 14(C^14). Using geological methods toestimate the original size of the rock, scentists estimate it to originally have contained 4.096kg of C^14 .The Half-life of c^14 is 5760 years.What is the approximate age of rock?

$\displaystyle y = k2^{-t/\tau}$

$\displaystyle \tau = 5760~yr$
$\displaystyle k = 4096~g$
$\displaystyle y = 2~g$

Thus
$\displaystyle 2 = 4096 \cdot 2^{-t/5760}$

Solve for t.

-Dan
• Jul 30th 2007, 05:10 PM
usm_67
what du u mean when u say t/5760 how wud i go by solvin that i get the rest of it
• Jul 30th 2007, 05:45 PM
galactus
Quote:

Originally Posted by usm_67
what du u mean when u say t/5760 how wud i go by solvin that i get the rest of it

You might want to find an English forum also.
• Jul 30th 2007, 05:46 PM
usm_67
lol my bad i was in a hurry, got the letters mixed up
• Jul 30th 2007, 05:53 PM
usm_67
• Jul 31st 2007, 04:33 AM
topsquark
Quote:

Originally Posted by topsquark
$\displaystyle 2 = 4096 \cdot 2^{-t/5760}$

(I apparently left of the "-" in the exponent. Sorry about that!)

$\displaystyle \frac{2}{4096} = 2^{-t/5760}$

$\displaystyle 2^{-t/5760} = \frac{1}{2048}$

$\displaystyle -\frac{t}{5760} = log_2 \left ( \frac{1}{2048} \right )$

$\displaystyle t = -5760 \cdot log_2 \left ( \frac{1}{2048} \right )$

$\displaystyle t = -5760 \cdot log_2 \left ( \frac{1}{2^{11}} \right )$
$\displaystyle t = -5760 \cdot -11 = 63360$