Answer:
Every 1.71 seconds, the bacteria loses [tex]\frac{1}{2}[/tex]
Step-by-step explanation:
Given
[tex]B(t) = 8500 * (\frac{8}{27})^\frac{t}{3}\\[/tex]
Required [Missing from the question]
Every __ seconds, the bacteria loses [tex]\frac{1}{2}[/tex]
First, we model the function from t/3 to t.
[tex]B(t) = 8500 * (\frac{8}{27})^\frac{t}{3}\\[/tex]
Apply law of indices
[tex]B(t) = 8500 * (\frac{8^\frac{1}{3}}{27^\frac{1}{3}})^t[/tex]
Evaluate each exponent
[tex]B(t) = 8500 * (\frac{2}{3})^t[/tex] --- This gives the number of bacteria at time t
At time 0, we have:
[tex]B(0) = 8500 * (\frac{2}{3})^0[/tex]
[tex]B(0) = 8500 * 1[/tex]
[tex]B(0) = 8500[/tex]
Let r be the time 1/2 disappears.
When 1/2 disappears, we have:
[tex]B(r) = \frac{B(0)}{2}[/tex]
[tex]B(r) = \frac{8500}{2}[/tex]
[tex]B(r) = 4250[/tex]
So, we have:
[tex]B(t) = 8500 * (\frac{2}{3})^t[/tex]
Substitute r for t
[tex]B(r) = 8500 * (\frac{2}{3})^r[/tex]
Substitute [tex]B(r) = 4250[/tex]
[tex]4250 = 8500 * (\frac{2}{3})^r[/tex]
Divide both sides by 8500
[tex]\frac{4250}{8500} = (\frac{2}{3})^r[/tex]
[tex]\frac{1}{2} = (\frac{2}{3})^r[/tex]
Take log of both sides
[tex]log(\frac{1}{2}) = log (\frac{2}{3})^r[/tex]
Apply law of logarithm
[tex]log(\frac{1}{2}) = r\ log (\frac{2}{3})[/tex]
Make r the subject
[tex]r = log(\frac{1}{2}) / log (\frac{2}{3})[/tex]
[tex]r = \frac{-0.3010}{-0.1761}[/tex]
[tex]r = 1.71[/tex]
Hence, it reduces by 1/2 after every 1.71 seconds
Answer:
Every second, the number of bacteria is multiplied by a factor of 0.67
Step-by-step explanation:
Let's rewrite the base so that the exponent is just t.
(8/27)^t/3=((8/27)^1/3)t=(2/3)^t
Therefore, we can rewrite the modeling function as follows.
B(t)=8500⋅(2/3)t
According to this model, the number of bacteria is multiplied by 2/3 every second. Rounding this to two decimal places, we get 2/3 ≈0.67
