Algebra 2 Exponential Growth
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Watch It Grow
Exponential Growth


 Sometimes how fast something grows

 depends on how much we have.
 When bacteria grows, 
 the amount of new bacteria you get every minute
 depends on how much you already have.
 The way this looks in an equation is:

 Amount You Get = Amount You Start With econstant time

 That is, the amount that you have after some time has passed
 is equal to the amount you started with
 times "e" raised to a power made up of
 the amount of time that has passed times some constant (a number)
 The constant is there to make sure the units work out right.
 Usually in the equation:
 We use the letter k for the constant (this will be a different
 We use the letter t for the time
 We use the letter Q for the quantity of stuff we have at the end
 We use the letters Qo for the original quantity of stuff we start with.
 So put it all together, the equation look like this:

 Q = Qoekt

 The rate that mystery bacteria Q grows depends on the amount present.
 After 3 minutes, Q has grown from 100 to 500 units
 How much is there after 10 minutes?
 First we need to find out what k is.
 k will be different for each problem
 so we have to find k for each problem we do.
 Set up the problem.
 The amount we start with (Qo) is 100
 The amount we end up with (Q) is 500
 The time (t) is 3 minutes
 e is 2.718 ...
 That means we have:

 Q = Qoekt

500 = 100ek3

 Now we solve this problem for k.
 First we get rid of the big numbers ...


 Since k is an exponent we need our special log trick
 to turn it into just a factor.
 Take the log of each side. 
 It's easier in this case to use the ln type of log ...

 ln 5 = ln e3k

 Now use the log trick to bring down the exponents ...

 ln 5 = 3k ln e

 Now remember what ln e means
 It means "the power that e must be raised to, to get e"
 That gives us:

 ln 5 = 3k

 Now divide each side by 3 ...


 ln 5 is just a number, so we can solve this one ...


 Now we can put .536 in for k and solve the rest of the problem.
 Remember the rest of the problem?
 How much is there after 10 minutes?
 That means:
 Set up the problem.
 The amount we start with (Qo) is 100
 The amount we end up with (Q) is THE UNKNOWN
 The time (t) is 10 minutes
 The constant (k) is .536
 So we have:

 Q = Qoekt

 Q = 100e10 .536

 Simplify this a bit:

 Q = 100e5.36

 Now we need a calculator.
 On my calculator, this is how you punch this one in ...


 So after 10 minutes we have over 21,000 units of bacteria Q
 The rate that the population of "Our Town" grows
 depends on the number of people in the town.
 In 1900 there were 1000 people.
 In 1950 there were 2000 people.
 When will there be 10,000 people in "Our Town"?
 Since we're looking for total people, 
 let's use P instead of Q.
 So we have:
  Po = 1000 
  P = 2000 
  t = 50 (from 1900 to 1950)
  k = the first thing to solve for ...

P = Poekt 

 2000 = 1000e50k

 Divide each side by 1000 ...


 Now take the natural log (ln) of each side ...

 ln 2 = ln e50k

 and use the log trick (don't forget, ln e = 1)
  ln 2 = 50k   (ln 2 = .693)
  .693 = 50k
 Divide both sides by 50 to get k by itself ...


 That's great. Now we have k.
 But we're not done yet.
 The problem asked:
 'When will there be 10,000 people in "Our Town"?'
 Finding out WHEN means we need to solve for t.
 We have:
  Po = 1000 
  P = 10,000 
  k = .0139 
 Let's do it!

P = Poekt 

10,000 = 1,000e.0139t 

 Divide both sides by 1000 ...


 Now we need our log trick (OK, our ln trick)


 We need to solve for t so divide both sides by .0139 ...


 So it takes 165.7 years for the population of "Our Town"
 to grow from 1000 to 10,000 people.
 But the question was WHEN will there be 10,000 people
 in Our Town.
 That means WHAT YEAR.
 We started with 1000 people in 1900.
 165.7 years later we will have 10,000 people.
 The year will be 1900 + 165.7 = 2065.7.
 So Our Town will have 10,000 people in about the year 2065 or 2066
 When stuff grows like this it is called EXPONENTIAL GROWTH.

   copyright 2005 Bruce Kirkpatrick

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