in this video we're going to focus on solving exponential growth and Decay word problems so let's get right into it in 2005 there were a thousand rabbits on an island the population grows 8% every year at this rate how many rabbits will there be on the island by 2020 so what should we do in this problem how can we find the answer so we need to write an equation now is this an exponential growth or Decay problem since the population is growing it's an exponential growth problem the equation that you want to use is y is equal to a * B raised to the X this is basically the generic formula for exponential growth but for this particular uh problem a is going to represent the initial population B is going to be 1 + r raised to the T since we're dealing with time so the initial population of rabbits is a thousand the it grows 8% every year 8% is basically 08 to convert a percentage into a decimal divide by 100 so 8id 100 is 8 now what should we plug in for T how many years will pass pass by between 2005 and 2020 if we subtract the two numbers it's going to be 15 years so let's type this in the calculator 1,00 * 1.08 rais to the 15th power you should get 3,172 rabbits round into the nearest whole number the value of a new car in 2015 was $40,000 it depreciates 7% each year how much will the car be worth in 2024 so go ahead take a minute pause the video and work on this problem see if you can do it and then unpause it when you're ready to see the solution so let's uh begin so what equation do we need what we have here is an exponential decay problem because the value of the car is decreasing over time so the equation is going to be similar to the last equation but instead of 1 + r it's going to be 1 minus r due to the fact that it's a Decay problem so in this case a is going to be the initial value of the car it's 40,000 and it depreciates by 7% 7 divid 100 is 07 and what is the value of T so we know we need to subtract 2024 and 2015 that will give us a difference of 9 years 1us 07 is93 so let's type this in the calculator so 40,000 * .93 raised to the 9th power should be about $2,816 44 so this is going to be the expected value of the car N9 years later John bought a new home in 2002 the value of the home increases 4% each year if the price of the house is 225,000 in 2015 how much did he pay for it in 2002 so is this an exponential growth or Decay problem since the value of the home is increas in every year it's an exponential growth problem so therefore the formula that we need is going to be 1+ r instead of 1 minus r now what are we looking for in this particular problem notice that the 225,000 occurs in 2015 that's the present value and we're looking for the price of the home in 2002 that's the value in the past so let's say if a represents the present value y represents the future value so I'm going to write PR for present and fut for future now in this problem y represents the present value which means a represents the value in the pass so a is always the value that occurs sometime before y y occurs Y is the value basically after sometime Beyond a so we have this situation for this problem hopefully I didn't confuse you on that so in 13 years the value of the home increases to 225,000 we need to find out how much it was worth 13 years ago R is going to be 04 and T is going to be 2015 minus 2002 which is 13 so let's find out what this value is equal to First 1.04 raed to the 13 power is about 1.6 65735 so let's divide both sides by this number so a is equal to 225,000 ID 1. 6650 0735 so John he purchased a home for 135,000 $129 17 according to this formula so that's how much it was worth in 2002 let's work on this problem a sample contains a, counts of bacteria the bacteria doubles every 20 minutes at this rate how many counts of bacteria will there be in 3 hours so if you get a question like this on a test what will you do so we know this is an exponential growth problem now we want to use this form of the equation Y is equal a * B raised to the X but we're going to modifying just as before a represents the initial amount in this case the 1,00 counts of bacteria B represents the growth the bacteria doubles so B is equal to 2 and a is 1,000 now the bacteria doubles every 20 minutes typically X is usually represented in hours you can make in minutes if you want but since I'm looking for the amount of bacteria in three hours I'm going to use hours so if x represents the time in hours or T and it doubles every 20 minutes then it doubles three times in an hour because there's three 20 minute periods in an hour an hour is 60 minutes if you take 60 Minutes divid by 20 this is going to be three so the equation is going to be n * T where n is the number of times it doubles an hour in this case n is so let's plug in what we have the equation is now Y is = 1,00 * 2 ra to 3 T and T is 3 hours so 3 * 3 is 9 so it's going to double nine times it doubles three times in an hour but nine times in three hours 2 to the KN power is 512 multiply by th000 so this going to be 512,000 counts of bacteria in 3 hours a sample contains 100 counts of bacteria the bacteria triples every 15 minutes so let's write the equation we can see that a is 100 that's the initial population now the bacteria triples so B is three and it does so every 15 minutes so how many 15 minute intervals are there in a single hour well 1 hour is 60 minutes and if we divide that by 15 this is going to equal four so the equation is 100 * 3 ra to 14 where T is in hours so let's answer part A how much bacteria will there be in 1 hour so T is 1 3 to 4th power is 81 and 100 * 81 is 8100 so that's how much bacteria there will be in a single hour now let's focus on the second part of the problem how long will it take for the sample to contain 500 million counts of bacteria so we need to solve for t t represents the time so in this case Y is 500 m ion we can write that as 500 * 10 6 now what's the first thing that we need to do in order to solve for T the first thing that we should do is divide by 100 500 ID 100 is five now what's our next step in order to solve for T we need to use logarithms let's take the log of both sides now a property of logs allows us to take the exponent and move it to the front so what we now have is Log 5 * 10 6 and that's = to 4T * log 3 so let's divide both sides by log 3 so log 5 * 10 6 well let's divided by 4 log3 let's get T by itself ID 4 log 3 is equal to T so if you have your calculator with you type in Log 5 * 10 6 you should get 6.69 897 log 3 is about 47712 if you multiply that by four on a bottom you should have 1.9 8485 now let's divide these two numbers this will give you 3.51 so it's going to take 3.51 hours for the sample to contain 500 million counts of bacteria