Question 1062936: A scientist has 200 grams of a substance where half of it decays every hour. Write an equation to determine how long it takes until 3.125 grams are remaining
Found 2 solutions by Fombitz, Theo:
Answer by Fombitz(32388)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! f = p * e^(rt)
this is the formula for continuous growth or decay.
if r is negative, then decay.
if r is positive, then growth.
f = future value
p = present value
e = scientific constant of 2.718281828.....
r = rate per time period
t = number of time periods
you are given that p = 200
you are given that f = 100
you are given that t = 1 hour
you need to solve for r.
formula of f = p * e^(rt) becomes 100 = 200 * e^(r*1) which becomes 100 = 200 * e^(r)
divide both sides of this equation by 200 to get .5 = e^r
take the natural log of both sides of this equation to get ln(.5) = ln(e^r).
this is equivalent to ln(.5) = r * ln(e).
this is equivalent to ln(.5) = r because ln(e) is equal to 1.
solve for r to get r = ln(.5) = -.6931471806 per hour.
you started with 100 = 200 * e^r.
replace r with -.6931471806 to get 100 = 200 * e^-.6931471806.
simplify to get 100 = 100.
this confirms the solution for r is correct.
now that you know r, you want to find the number of hours for the substance to become equal to 3.125 grams.
same formula of f = p * e^(rt) is used, only this time we are looking for t.
formula becomes 3.125 = 200 * e^(-.6931471806 * t)
divide both sides of the equation by 200 to get 3.125/200 = e^(-.6931471806 * t)
take the natural log of both sides of the equation to get ln(3.125/200) = ln(e^(-.6931471806 * t))
this is equivalent to ln(3.125/200) = -.6931471806 * t * ln(e)
since ln(e) = 1, this becomes ln(3.125/200) = -.6931471806 * t
divide both sides of the equation by -.6931471806 and solve for t to get t = ln(3.125/200) / -.6931471806.
this makes t equal to 6 hours.
the mass of 200 grams will reduce to 3.125 grams in 6 hours.
formula becomes f = 200 * e^(-.6931471806 * 6) which results in f = 3.125 grams.
this was solved based on continuous compounding.
you could also have solved using discrete compounding.
the formula, in that case, would be f = p * (1+r)^n
f = 100
p = 200
n = 1
the time, in this case, is represented by n.
it doesn't really matter what variable name you use, as long as you are consistent within the confines of the formula you are using.
formula becomes 100 = 200 * (1+r)^1 which becomes 100 = 200 * (1+r).
divide both sides of the equation by 200 and simplify to get .5 = 1 + r
subtract 1 from both sides of the equation and solve for r to get r = -.5 per hour.
this says that the substance becomes half of what it was every hour.
now you need to use the same formula to solve for n.
formula of f = p * (1 + r)^n becomes 3.125 = 200 * (1 - .5)^n.
simplify to get 3.125 = 200 * .5^n
divide both sides of the equation by 200 to get 3.125/200 = .5^n.
take the log of both sides to get log(3.125/200) = log(.5^n).
this is equivalent to log(3.125/200) = n * log(.5)
divide both sides by log(.5) and solve for n to get n = log(3.125/200)/log(.5) = 6.
the substance will reduce from 200 grams to 3.125 grams in 6 hours.
you get the same answer whether you used continuous compounding formula or discrete compounding formula.
|
|
|
