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# Exponential Growth and Decay - Problem 2

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

I want to talk about exponential decay. I’ve written on the board the exponential decay formula, and you may notice it looks exactly like the exponential growth formula. The only difference is the value of the base, b.

If b is between 0 and 1 you’ll get decay the function will be decreasing. If b is bigger than 1, you’ll get growth. That’s really the only difference, they both have the same equation. So here’s a problem; the radioactive isotope unobtainium decays at a rate 13% per hour. What is the half life of the substance? If there are 80 grams of it initially, how much will remain after 10 hours, after a day?

I’ve made a table down here to show you what the amount looks like over time, after 0 hours, 1 hour, 2 hours, 3 hours. The initial amount I don’t know. But I’m going to call it n sub 0. After an hour, it will be n sub 0 times .87. This number comes from the 13% rate of decrease. When something decreases at 13%, it’s like multiplying by .87, 1 minus .13.

And that going to happen every 3 hours. .87, .87², .87³ and that gives us the, equation n of t equals n sub 0 times .87 to the t. Now I’m asked to find the half life of the substance, that’s the amount of time it takes for the initial amount to decay to half.

Half the initial amount is 1/2 n sub 0. So I don’t really need to know what the initial amount is because the initial amount is going to cancel. So I’ll cancel that and I get 1/2 equals .87 to the t. Again we y have an exponential equation, solving an exponential equation means taking the log of both sides. We’ll take the natural log of both sides.

The reason this works is, the properties logarithms allow us to take exponents out in the front. The exponents we were trying to solve for. So I’m going to use the power property, the log of a power to write t times ln of .87. That equals ln of 1/2.

And then it’s just a linear equation. I divide both sides by ln.87 and I get ln of 1/2 over ln of .87. So that’s approximately; I need a calculator for this; natural log of .5 divided by natural log of .87. So my calculator gives me 4.97729 approximately. This is in hours because the decay rate was given to me in percent per hour. And this means it takes almost 5 hours for the amount to decay to half of what it originally was.

Now the question also asks me to calculate how much will remain after 10 hours. Again my formula is n sub 0 times .87 to the t. The initial amount was given to me as 80 kilograms. After 10 days, I’ll have 80 times .87 to the 10th. So I’ll need my calculator for that. I get approximately 19.874 kilograms.

Does this make sense? Yes this is roughly two half-lives. After the first half-life, the 80 will decay to about 40. After the second half-life the 40 will decay to about 20.

And this is ever so slightly more than two half-lives. And so we get a little bit less than 20 kilograms. And the last part asks, how much will remain after a day? Remember t is in hours so you don’t want to plug 1 in here, you want to plug in 24.

Again using the calculator, I already got the entry in so I’m just going to do a second entry, back up and enter 24. And I get 2.828 kilograms.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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