Can't figure out exponential decay, can anyone help?

Question

Here's the problem: The concentration of the active ingredient in a chemical used to control insects decays exponentially with time. Twelve hours after application, 50% of the active ingredient decays. What percent of the active ingredient decays after 36 hours?

Thanks for the help...final preps for teacher certification test tomorrow!

Answer 1

(x).5 (in 12 hours)
(x).5^2 (in 24 hours)
(x).5^3 (in 36 hours)

12.5%

Answer 2

First, let's compare an exponential function to a linear function, a straight line. In a linear function (say, y = 3x + 2), every single "step" you take in the x-direction *adds* some amount to the line. In that example, a one-unit step in the x-direction results in a 3-unit addition to the y-value.

In an exponential function, instead of adding each step, you multiply. For each step in the x-direction, the y-value is *multiplied* by a constant ratio. For example, in the function y = (3)*(5^x), each one-unit step in the x-direction *multiplies* the y-value by 5.

Now, your chemstry problem. Remember that time is always the "x" axis in chemistry and physics. The problem tells us that if we take a "12 hour" step in the x-direction, the y-value is multiplied by 1/2, by 0.5. We know that for each "12 hour" step we take in the x-direction, we will multiply by that same fixed ratio, 1/2.

If I move 36 hours in the x-direction, that's three 12-hour steps --- for each one, I multiply by 1/2. That's (1/2)x(1/2)x(1/2) = (1/8). Thus, after 36 hours, you will be left with 1/8, which is equal to 12.5%. If you're left with 12.5%, how much decayed? 100% - 12.5% = 87.5% (that is, 7/8 of it) decayed.

The preceding was a very conceptual approach to exponential decay. Here's a formula that might also help you.

L = A*(r ^(t/P)), or L/A = r ^(t/P)

L = "left", remaining amount
A = starting amount
(any problem that ask for percent remaining wants L/A)
r = rate/ratio of decay given in problem
t = time
P = period given in problem.

Here, the problem gave us a decay rate of .5 in a period of 12 hours.

Thus L/A = 0.5^(t/12)

If I plug in t = 36, I get

L/A = 0.5^(36/12) = 0.5^3 = 0.125

That's what's left, so what decayed is 1 - 0.125 = 0.875, which is 87.5 percent.

Answer 3

For every 12 hours that elapse, half of the starting amount decays. So in the first 12 hours, 50% decays, leaving 50% remaining. In the next 12 hours, 50% of the 50% decays, leaving 25% remaining. In the next 12 hours, 50% of the 25% decays, leaving 12.5%. And so on. . . . . . .

Answer 4

In 12 hours 50% is decayed, so 50% is left
In 36 hours (50%)^(36/12)=(0.5)^3=0.125=12.5% is left
100% -12.5%=87.5% is decayed

Answer 5

y=(Initial)(1/2)^h/12 This equation is created based on the fact that 12 hours after the starting period, 1/2 of the initial value remains
y=initial(1/2)^36/12
y=initial(1/2)^3
y=initial(1/8)
After 36 hours, 1/8th of the initial ingredient remains.

Answer 6

C=C1e-kt, where c1 is the amount of concentration at t = 0 and C is amount at any time t.

So .5C1 = C1 e-12k
e-12k = .5
ln e-12k = ln.5
-12k = ln.5
k = ln .5/-12 = .05776
So C=C1e-.05776t
C= C1(e-.05776^36) = .125C1
So 12.5% remains after 36 hours

Answer 7

Ok, what that problem is saying is that for every twelve hours, the chemical is halved. So, twelve hours::one-half; twenty-four hours::one-fourth; and finally, thirty-six hours::one-eighth. This is also 12.5%.