growth rate and decay question

[borinbuscadecuartos]

Hello everyone, i was having trouble with this one problem on my homework, i just wanted to know if anyone could help me. The question is .... sixty percent of a radioactive substance decays in 15 years. By how much does the substance decay each year? what is the annual growth rate if the quantity grows by 7% in 8 months? thanks in advanced

 

[Deleted member 4993]

Hello everyone, i was having trouble with this one problem on my homework, i just wanted to know if anyone could help me. The question is .... sixty percent of a radioactive substance decays in 15 years. By how much does the substance decay each year? what is the annual growth rate if the quantity grows by 7% in 8 months? thanks in advanced

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

 

[borinbuscadecuartos]

I am just having trouble setting up the equation

 

[stapel]

I am just having trouble setting up the equation

To learn how to set up exponential equations (something they were supposed to have covered back in algebra, and then reviewed in your current class), try here.

Once you have refreshed yourself on the background material, please attempt the exercise (which does not appear to require any calculus, by the way). If you get stuck, you can then reply with a clear listing of what you have tried so far.

Thank you! :wink:

 

[BigGlenntheHeavy]

First, understand the formula for growth and/or decay.

$\displaystyle y(t) \ = \ Ae^{kt}, \ y(0) \ = \ A, \ the \ initial \ amount \ of \ mass \ at \ time \ t \ equals \ zero.$

in 15 years 60% of the mass decays, leaving 40% of the mass still intact, so;

$\displaystyle y(15) \ = \ .4A \ = \ Ae^{15k}, \ .4 \ = \ e^{15k}, \ 15k \ = \ ln(.4), \ k \ = \ \frac{ln(.4)}{15} \ = \ ln(.4)^{1/15}$

$\displaystyle Therefore, \ y(t) \ = \ Ae^{ln(.4)^{t/15}} \ = \ A(.4)^{t/15}$

$\displaystyle Halflife: \ \frac{A}{2} \ = \ A(.4)^{t/15}, \ \frac{1}{2} \ = \ (.4)^{t/15}, \ t \ = \ \frac{-15ln(2)}{ln(.4)} \ = \ 11.3470619605 \ years.$

$\displaystyle So \ this \ particular \ mass \ has \ a \ halflife \ of \ 11.3470619605 \ years, \ which \ means \ that \ 50 \ per \ cent$

$\displaystyle of \ the \ mass \ will \ have \ decayed \ at \ this \ time.$

$\displaystyle Now, \ I \ assume \ you \ can \ take \ it \ from \ here.$