# Best Exponential Growth Functions Review

This video takes a look at an application of algebra in biology known as an exponential growth function by using logarithms and exponents.

Biology Application – Exponential Growth Function

Biology Application - Exponential Growth Function

In this video, we’re going to look at an application of algebra in biology. And so, this particular problem that we’re looking at, tells us that we have this bacteria that initially starts out at a mass 5mg and it doubles in size every 30 minutes. And what we’re asked to find is the amount of time that it takes to get up to 100 mg. And so, this is an

**exponential growth problem**, and so the basic form of the equation that we need is f of x equals a times e to the kx. Since we’re talking about growth over time, what the function is going to look like is f of t – because we’re doing this with respect of time – is equal to f not – or f at times zero – times e to the kt.

And so, we don’t know what k is yet, but we have all the other information we need to solve this. And so the first thing we want to do is figure out what k is in this equation, so we can figure out how long it will take to get up to 100 mg. And so what we’re told is that it doubles every 30 minutes. And so implicitly we’re told that at time t equals 30 minutes, we’ve gone up to 10 mg. And so we can create an equation from that information that will let us solve for k. And so let’s go ahead and plug those numbers in.

So, our time t equals 30 minutes, will give us the 10 mg – because at 30 minutes we were up to 10 mg – times the initial amount, 5 mg, times e raised to the k times 30 minutes. And so now we just have one equation with one variable, and we can solve it. And so let’s divide both sides by 5 and that gives us 2 equals e to the k times 30. Now we have an exponential here, so we need to take the natural log of both sides, and that gives us natural log of 2 is equal to 30 k – we took the natural log of something raised – e raised to something – and so the value is just going to be the value of the exponent.

And so the natural log of 2 is equal to 30 k, or k equals natural log of 2 divided by 30. And so we could calculate this value with a calculator, but I’m going to wait to do that until we get to the end. And so now that we have the value of k, we’ll go ahead and go back to our original equation here, and solve it for the time when we have 100 mg. And so, plugging this back in, we now have 100 mg is equal to the initial amount 5 times e raised to the k t. We’re looking for t this time. We have the value of k so we can treat it just like a number, so we have just one unknown variable in t. And so we can go ahead and solve this. We’ll divide both sides by 5. And that gives us 20 equals e raised to the kt. Take the natural log of both sides once again, that will give us the natural log of 20 is equal to k times t. And t is equal to the natural log of 20 divided by k or the natural log of 20 divided by log of 2 divided by 30 or since it’s on the bottom, times 30. And you can use a calculator to figure this out, but what it comes out to is 2.15 hours. That’s going to be the time at which the bacteria reaches a mass of 100 mg. And it’s usually a good idea in these situations to make sure that the number that you get from your calculation makes sense with what you have in the problem statement. And so we’re told that this doubles every half hour, and so after the first half hour, we’re going to have 10 mg, after the second half hour we’ll have 20, after the third we’ll have 40, after the 4th, 80, and after the 5th, 160. And so we’re looking for the time when we get to 100 mg, and so we know it’s going to be between 4 and 5 periods, or between 2 and 2.5 hours. And so, since our number, 2.15, falls between these two time periods, we know at least that it’s a reasonable number for this problem.

And so that’s how you can go about solving an exponential growth problem using logarithms and exponents.

Provided by:

*Mometrix Test Preparation*

Last updated: 04/13/2018