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Exponential Growth and Decay - Concept

Teacher/Instructor 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.

Exponential growth refers to an amount of substance increasing exponentially. Exponential growth is a type of exponential function where instead of having a variable in the base of the function, it is in the exponent. Exponential decay and exponential growth are used in carbon dating and other real-life applications.

I want to talk about exponential growth, I have an example here the population of mice in the Duchy of Grand Fenwick grows at a rate of 6% per year. How long will it take for the population to double or quadruple? I have a table of values here I wanted to show you why this is exponential growth. Increasing at 6% per year means very year we're multiplying by 1.06 and so we get p sub 0 when t=0, p sub 0 times 1.06 when t=1 and p sub 0 times 1.06 squared when t=2 and so on.
This would suggest the formula p sub t equals p sub 0 times 1.06 to the t and that's an exponential growth formula. Now to find the doubling time I need to plug in twice the initial population here. I don't know what the initial population is but twice the initial population is 2 times p sub and after you plug in you can see that the actual initial population doesn't matter it's going to cancel out. So now I have the equation 2=1.06 to the t, this is an exponential equation and the way we solve exponential equations is to take the log of both sides. I'm going to take that natural log of both sides, it doesn't matter what log you use as long as it's on your calculator so you can use either the common log or the natural log.
Okay before I calculate I actually need to use the property of the logs, natural log of 1.06 to the t this is the log of a power so the exponent can come out in front t times ln of 1.06 and then we have a simple linear equation to solve this all we need to do is divide both sides b y natural log of 1.06. So ln 2 over ln of 1.06 now I'd like a numerical answer so I'm going to calculate this value ln2 divided by ln 1.06 enter, I get t, t is approximately 11.89566 and that would be in years. Because the population growth rate was given as 6% per year, I want to round this off let's say your teacher likes you to round off to the nearest hundredth then when I calculate quadrupling time a reason like this in order for a population to quadruple, it's got to double and then double again. So we're going to have 2 doubling times in a row. So it stands to reason that the quadrupling time is twice the doubling time. Twice this, but if I write 23.80 years my answer is not quite right let me multiply in my calculator answer times 2. it's actually 23.79 years.
You have to be really careful when you're using rounded values to do calculations. These values rounded to the nearest hundredth and when I double it I double whatever round off error there was. So the best way to get my final answer for quadrupling time is to double this value which is still stored in the calculator so multiply this times 2 and you get the correct value to the nearest hundredth. This is my answer.

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