# Thread: Exponential growth + decay

1. ## Exponential growth + decay

I'm having some trouble with this hwrk problem, and help would be greatly appreciated. Thanks!

If a substance decomposes at a rate proportional to the amount of substance present, and if the amount decreases from 40gm to 10gm in 2 hours, then the constant of proportionality is:

2. Originally Posted by wombat1
I'm having some trouble with this hwrk problem, and help would be greatly appreciated. Thanks!

If a substance decomposes at a rate proportional to the amount of substance present, and if the amount decreases from 40gm to 10gm in 2 hours, then the constant of proportionality is:
Hello,

the process could be best described by an exponential funktion: $\displaystyle f(t)=a \cdot b^t$

where a is the value at the beginning
b (the base) is the constant of proportionality you are looking for
t is the time.

Put in the values you know: $\displaystyle 10=40 \cdot b^2$

Solve for b: $\displaystyle \frac{10}{40}= b^2$ and you'll get: b = 0.5.

Greetings

EB

3. Originally Posted by wombat1
I'm having some trouble with this hwrk problem, and help would be greatly appreciated. Thanks!

If a substance decomposes at a rate proportional to the amount of substance present, and if the amount decreases from 40gm to 10gm in 2 hours, then the constant of proportionality is:
Here is one way.
"Decomposes at a rate proportional to amount of substance at present"
Let present be when time is zero.
So,
amount of substance at present = "P at t=0" = Po
amount of substance at time t = "P at t=t" = P of t = P(t)
P(t) = (Po)*e^(kt) -------------exponential function

Given:
at time t=2hrs,
P(t) = P(2) = 10gm
Po = 40gm at anytime.

So,
10 = 40*e^(k*2)
10/40 = e^(2k)
1/4 = e^(2k)
ln(1/4) = ln[e^(2k)] -----(i)
ln(0.25) = (2k)*ln(e)
ln(0.25) = 2k
k = (1/2)*ln(0.25)

Or, from (i),
ln(1/4) = ln[e^(2k)] -------(i)
ln(1) -ln(4) = (2k)*ln(e)
0 -ln(2^2) = (2k)*1
-2ln(2) = 2k
k = -ln(2) ---------------(ii)
k = -(0.693147181)
k = -0.693147181
Same as above, hence, k = -ln(2).

Then,
k = -0.693147181 = -ln(2) ----------if you are using the "e".

If you are not using the "e",
P(t) = (Po)*e^(kt)
P(t) = (Po)*[e^(k*t)]
P(t) = (Po)*[(e^K)^t]

To not to use "e", let (e^k) = a = constant of proportionality.
So,
P(t) = (Po)*[a^t]
Plug in the known values,
10 = 40[a^2]
10/40 = a^2
1/4 = a^2
a = 1/2