SOLUTION: Substances X and Y decompose at a rate proportional to the amount present. Tests shows that substance X loses one half of its mass every 16 hours and substance Y loses one half o

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Substances X and Y decompose at a rate proportional to the amount present. Tests shows that substance X loses one half of its mass every 16 hours and substance Y loses one half o      Log On


   



Question 1077705: Substances X and Y decompose at a rate proportional to the amount present.
Tests shows that substance X loses one half of its mass every 16 hours and
substance Y loses one half of its mass every 21 hours. At this moment there are
3.0 kg of X and 3.0 kg of Y. When will there be three times as much of Y remaining
as there is of X?

Found 2 solutions by Boreal, jorel1380:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
half the mass lost in 16 hours is (1/2)=e^(-16k)
-ln2=-16k
k=ln2/16=0.04332 for X
k=ln2/21=0.03301 for Y
We want e^(-0.04332t)/e^(-0.03301t)=1/3
3e^(-0.04332t)=e^(-0.03301t)
do ln both sides
ln3-0.04332t=-0.03301t
ln3=0.01031t
t=106.56 hours
In that time, X will have 3e^(-4.616)=0.02967 kg
Y will have 3e^(-3.517)=0.0890 kg
That ratio is 3.00 to 1.

Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
Half the mass lost in 16 hours is (1/2)=e^(-16k)
-ln2=-16k
k=ln2/16=0.04332 for X
k=ln2/21=0.03301 for Y
We want e^(-0.04332t)/e^(-0.03301t)=1/3
3e^(-0.04332t)=e^(-0.03301t)
So:
ln 3-0.04332t=-0.03301t
1.0986122886681096913952452369225=0.01031t
t=106.55793 hours. ☺☺☺☺