exponential decay

[stefunny]

Problem: A substance X decreases by a third every 4 minutes. If 1/9 of the substance remains at 12:05am, when will only 1/3^103 remain? Give an exact time. Will the substance ever be zero? Why or why not?

I am quite clueless as to how I should start this problem. I know the formula for decay is M(t)=Ae^kt but I don't know how to use it here.
(I know that it will never be zero, I think because although e gets very small it never reaches zero)

 

[mmm4444bot]

… I know that it will never be zero, I think because although the ratio of e over an increasing denominator gets very small it never reaches zero …

Hello Stefunny:

There are two approaches to finding the time-of-day when (1/3)^103 units of substance X remain.

We can use the general function for exponential decay (the one that you listed), or we can use base 1/3 (instead of e). Personally, I like the latter, since the base in (1/3)^103 is also 1/3.

But, I'll get you started on using the natural exponential function, as you must have posted it for a reason; you can ask me about the other approach later, if you're interested.

M(t) is a variable that represents the mass of substance X after t minutes have elapsed.

Since we're told that M(t) is 1/9 units of mass at 12:05 AM, let's use this time-of-day as the very instant that variable t starts to count elapsed time.

In other words, let t = 0 correspond to 12:05 AM.

M(0) = 1/9

You should know that the constant A in the general function represents the mass at time 0. Therefore, A = 1/9.

If you forget what A represents, then you can also calculate it's value by substituting 1/9 for variable M(t) and substituting 0 for variable t.

1/9 = A e^(k*0)

1/9 = A e^0

1/9 = A

Next, we need to find the value of k, in our definition for function M:

M(t) = (1/9) e^(kt)

We know that the 1/9 unit at 12:05 AM will be decreased by one-third at 12:09 AM. In other words, when t = 4, M(t) = (1/9)(2/3) = 2/27

With these values for t and M(t), we can solve for k.

2/27 = (1/9) e^(4k)

Solve this equation for k, and you'll have the complete function definition for M(t).

Is this enough information for you to know how to use function M to answer the exercise?

Substitute (1/3)^103 for M(t), and solve the resulting equation for t.

Finally, determine how many hours and minutes your solution for t represents, and add that to 12:05 AM.

If you need more help, then please show whatever work that you can, or explain what you're thinking, so that we might determine where to continue helping you.

If I wrote anything that you do not understand, then please ask specific questions, and somebody will provide clarifications.

Cheers ~ Mark

MY EDIT: Corrected misread of amount of decrease

 

[stefunny]

Thank you, Mark!

I'm stuck. I solved for e^k=(1/3)^1/4, giving me M(t)=1/9(1/3)^1/4 and finally t=-4log3(9/3^103). But I am not sure how to get my time, I don't think this I ended up with the right answer. Will you please show me your other method, base 1/3? Thank you!

 

[mmm4444bot]

Hey there:

I apologize. Subhotosh pointed out that I've misread the amount of decrease every four minutes. I initially read it to mean one-third of the starting amount remains after four minutes. That's wrong.

The starting amount is reduced BY one-third, not TO one-third, which means 2/3rds remains, instead of 1/3rd.

In other words, we need to multiply the initial amount 1/9 by 2/3 after four minutes, not by 1/3. (I've corrected my previous post.)

So, with the correct amount of substance X remaining when t = 4, we are trying to solve the following equation, in order to find the value of parameter k.

2/27 = (1/9) e^(4k)

?

… I solved for e^k=(1/3)^1/4, giving me M(t)=1/9(1/3)^1/4 …

Our goal is not to substitute an expression for e^k into function M's definition. We're trying to find the value of k.

You started with some okay steps.

(9/1)(2/27) = (9/1)(1/9) e^(4k)

(2/3) = e^(4k)

(2/3)^(1/4) = [e^(4k)]^(1/4)

(2/3)^(1/4) = e^k

I'm going to start using a calculator, at this point.

e^k = 0.9036

We solve for k by taking the natural logarithm of both sides.

ln(e^k) = ln(0.9036)

k = -0.1014

Now we have the complete definition for function M. It allows us to find the number of elapsed minutes t required to reduce M(t) to any number of units, like (1/3)^103 units, for examle. :wink:

M(t) = (1/9) e^(-0.1014t)

Can you continue wtih the exericise?

PS: Since the factor is 2/3, instead of 1/3, my alternate method no longer applies. But, if you're interested in seeing what I was thinking (in the case where substance X is reduced by a factor of 1/3), let me know.