A culture starts at 8600 bacteria. After one hour the count is 10,000.
Find a function that models the number of bacteria n(t) after t hours.
The answer is n(t) = 8600e^.1506t
Where does this 0.1506 come from?
Thanks.
n(t) = 8600 e^(kt), where k is an unknown constant and t is the number of hours. We know that n(1), or the population after one hour is 10000. So 10000 = 8600e^(k*1)
10000/8600 = e^k
natural log ln(10000/8600) = k = .1508
The general expression is n(t) = no* e^(t/T), where T is the time for an increase by a factor of e.
In this case,
10000 = 8600 e^(1/T)
That equation can be solved for T.
10000/8600 = 1.16279 = e^(1/T)
Take the natural log of both sides.
0.15082 = 1/T
n(t) = e^(0.15082 t)
The 0.1506 in your version is not quite right.
Well, it seems like there was a little miscalculation in the previous response. The correct value for the constant in the equation is actually approximately 0.1508, not 0.1506.
As for where this value comes from, it is determined by solving the equation when t = 1, which gives us the population of 10,000 bacteria. By plugging these values into the equation and solving for k, we find that k ≈ 0.1508.
So, the correct function that models the number of bacteria n(t) after t hours is:
n(t) = 8600e^(0.1508t)
There you have it!
The value of 0.1506 comes from solving the equation ln(10000/8600) = k, where k is the growth constant.
The value of 0.1506 is derived from solving the equation 10000 = 8600e^(k*1), where k is the unknown constant. To find the value of k, we need to isolate e^(k*1) on one side of the equation.
Dividing both sides of the equation by 8600 gives us:
10000/8600 = e^(k*1)
Simplifying the left side of the equation gives us:
1.16279 = e^(k*1)
To remove the exponential function e from the equation, we can take the natural logarithm (ln) of both sides:
ln(1.16279) = ln(e^(k*1))
Using the property of logarithms that ln(e^x) = x, the equation becomes:
0.15082 ≈ k*1
Simplifying further gives us:
k ≈ 0.15082
Therefore, the value of k, or the unknown constant, is approximately 0.15082. This leads us to the function n(t) = 8600e^(0.15082t) for modeling the number of bacteria after t hours.
a)n(t)= 8600e^0.1508t
b) around 11,600
c) 4.6 hrs.