single cell tumor doubles every 4.5 months, how many cells will be present after 4 years?

Question

The number of cells in a tumor doubles every 4.5 months. If the tumor begins with a single? cell, how many cells will there be after 4 ?years?

Answer 1

 

Let x = number of months

Let y = number of cells

 

    This is an exponential relationship, not a linear one.

   

    Base equation is

 

        y = (b)^t

 

          Since the crlls are doubling every cycle, b = 2

     Since the period is every 4.5 months ...  9/2 months ... the growth rate each month is 1÷(9/2) = 2/9

 

      Now we can write our function

 

      y = 2^(2/9)t      where t is the number of months

 

We can perform some checks to confirm

 

      when t = 0     months, cell count should be 1 ... the initial value ...  2^(2/9)0 = 2⁰ = 1

      when t = 4.5  mpnths, cell count should be 2 ... 2^(2/9)(4.5) = 2^(9/9) = 2¹ = 2

      when t = 9     months, cell count should be 4 ... 2^(2/9)(9) = 2^18/9 = 2² = 4

      when t = 13.5 months, cell count should be 8 ... 2^(2/9)(13.5)= 2^27/9= 2³= 8

 

So it is clear to see that the function is accurate.

 

 

Since 4 years = 4(12) or 48 months, we substitute 48 into our function to determine the cell count after 4 years.

 

     y = 2^(2/9)(48) = 2^96/9 = 2^32/3 =  1625.5

 

After 4 years, the single cell will grow to become 1626 cells