|Start||Now we learn how to solve a problem involving exponential growth.
Exponential growth and decay is something that you will see a lot of in physics, biology, in chemistry, in all sorts of fields.
So, what does that mean? It means when you have something and it is growing,
|0:30||instead of growing linearly it grows exponentially or it can shrink exponentially.
The way you model it is you say well: y is a times b to the x (a*b^x) and that is an exponential equation.
For example: let's say if initially I have 200 bacteria, and 3 hours later I have
|1:27||500 bacteria, how much will I have after 10 hours? We are not that concerned with getting the numbers|
|1:48||all the way to the end, we just want to set the problem. So if initially I have 200 bacteria
this means at the beginning, at time zero, I have 200 bacteria, at time 3 I have 500 bacteria.
|2:04||How many am I going to have at time 10? Well you use my equation: y=a*b^x.
x is zero and y is 200, I know that 200 is a*b^0. b to the zero is 1 so this is 200=a times 1
|2:33||so a=200. In fact when you have these kinds of problems at zero, at the starting point this
number will always be a. Now, I can take my equation y=a*b^x and I have solved for a.
Now, if I figure out what b is, then I can do this equation with any number I need to.
Well I know that when x is 3 y is 500. So 500 is 200 times b to the 3 (b cubed) so
|3:08||if I divide by 500, I get 5/2 = b^3, b is 5/2 to the 1/3. Thats the cube root. Because any time I have the exponent on one side I want to isolate the base and just figure out what the other side is, I just have to flip the exponent. So now I can say that my equation|
|3:36||becomes y=200 times 5/2 to the 1/3 to the x or y is 200 times 5/2 to the x/3 because I can multiply the powers together. And now I just plug in 10 and we get y= 200 times|
|4:05||5/2 to the 10/3 and you can leave the answer like that, you don't have to actually need to figure out what it is, you can use a calculator, but this is not about the calculator, this is about getting to this set to coming up with an exponential growth equation. Let's do another example: Let's say initially I have 1000 grams of plutonium and after 10 minutes I have 950 grams. How|
|5:05||much will I have after 1 hour? Well we use the same equation. I know at time zero I have 1000. At time 10 minutes I have 950, so how much will I have after one hour? So one hour is 60 minutes. So I go to the equation. I know that 1000 is a times b to the zero and|
|5:38||as we saw on the last one, anything to the zero is one so a is 1000. So as I said before, you can get used to the idea that the initial amount is a. That means that this equation now becomes y= 1000 times b to the x. Now I know that when x is 10 y is 950. So 950 is|
|6:05||1000 times b^10. So I divide by 1000 and I get 95/100 or .95 is b^10. So 95/100 to the|
|6:31||1/10 equals b. So that means that my equation becomes y= 1000 times 95/100 to the 1/10 to the x or y= 1000 times 95/100 to the x/10. (that's a terrible 10!). So now I just need to plug|
|7:07||in 60 I get y is 1000 times 95/100 to the 60/10 or to the 6. Now we are going to do one last type, we are going to use this to find half life which is a kind of decay. That|
|7:31||was a decay problem, the second one. So now if initially I have 100 grams of uranium and 4 minutes later I have 80 grams, what is its half life? In|
|8:13||other words, when will I have 50 grams. So I have 0, at time zero I have 100. 4 minutes later I have 80. At what time will I have 50?. So Let's set this up. Now we know that|
|8:35||initially we have 100 so we can just say that this is going to be 100 times b^x. So 80 will be 100 times b^4, divide by 100 you get 8/10 is b^4 or b is 4/5 to the 1/4. So my equation|
|9:06||becomes y=100 times 4/5 to the 1/4 to the x or y is 100times 4/5 to the x/4. And now I want to find when will this get to 50. Well, 50 is 100 (4/5) to the x/4. Divide by 100|
|9:40||you get 1/2, that is why this is called half life. And how am I going to figure out x,
I need to use logarithms. You take the log on both sides and you can use any log you
want, you can use natural logs, you can use log base 2, log base 10, it does not matter.
|10:04||You take the log of both sides and you get log of 1/2 is x/4 times the log of 4/5. And
now you just have to solve for x. So multiply by 4, and divide by the log of 4/5.
|10:36||So you get 4 times the log of 1/2 divided by the log of 4/5 equals x.