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How Do You Solve a Word Problem Using an Exponential Function?

Tina is playing with her jump rope. At first, she can only jump 4 times before she gets tangled in the rope. She decides to practice, and each day, Tina triples her number of successful jumps. Assuming she continues at this rate, make an equation for her progress and use it to determine how many jumps she will be able to do on the sixth day.

Summary

  1. 'x' is the independent variable: the number of days Tina has practiced
  2. 'y' is the dependent variable: the number of jumps she can do
  3. We're letting 'Day 0' be her first try, so 'Day 1' is her first day of practice
  4. The number of jumps each day triples, so it is the result from the previous day times three
  5. The exponent of '3' in red tells us how many times the common factor '3' in black is multiplied by itself
  6. 30 = 1
  7. Since we have a variable 'x' in the exponent, 'y = f(x)' is an exponential function

Notes

    1. We want our equation to tell us how many jumps Tina can do on any day we pick
    2. So the two things in our equation that can vary are the day we pick and the number of jumps
    3. 'x' is often used as the independent variable
    4. Since the day is the thing we're picking, that will be the independent factor
    5. '#' means 'number'
    1. We have to calculate the number of jumps she can make each day
    2. Since the number of jumps depends on which day we pick, the number of jumps is the dependent factor
    3. 'y' is often used as the dependent variable
    4. '#' means 'number'
    1. A table is a good way to organize the information
    2. After just a few entries, we should be able to see a pattern and write an equation that describes the function
    3. Values for the independent variable 'x' are in the left column
    4. Values for the dependent variable 'y' are in the right column
    5. We're going to say that 'Day 0' is the day before she begins to practice
    1. Values for the independent variable 'x' are in the left column
    2. Values for the dependent variable 'y' are in the right column
    3. We're going to say that 'Day 0' is the day before she begins to practice
    4. 'Triple' means to multiply by 3
    1. After just a few entries in the table, we can detect a pattern
    1. We can replace 12 with 4•3
    2. Then since we can see that 36 equals 4•3•3, we can replace 36 with that expanded version in 36•3=108
    3. So we have that 108 equals 4•3•3•3
    1. 'Expanded form' means we're writing numbers out as their factors multiplied together
    2. Each time we go down a row, we're multiplying by 3 one more time
    3. When we have the same number multiplied together several times, we can rewrite it using exponents
    4. So 3•3•3 can be rewritten as 33, 3•3 can be rewritten as 32, and 3 can be rewritten as 31
    5. We can even extend our pattern to the first row, since 30 = 1
    1. We don't need to add any more table entries because the pattern is clear
    2. 'x' is the independent variable: the number of days Tina has practiced
    3. We can see from our table that the exponent on the 3 matches the value for 'x' each time
    4. So to find the number of jumps for ANY day, we can replace the exponent with 'x'
    5. That will give us the function f(x) = 4•3x
    1. For the sixth day, we plug '6' in for 'x' and evaluate the exponential function
    2. We need to address 36 first
    3. 36 just means we're multiplying 3 together 6 times
    4. So after 6 days of practice, when x = 6, Tina will be able to make 2916 successful jumps!