Question

Bt= 8500 *(8/27)^t/3After a special medicine is introduced into a Petri dish full of bacteria, the number of bacteria remaining in the dish decreases rapidly.

Answer 1

Answer:

Every 1.71 seconds, the bacteria loses $\frac{1}{2}$

Step-by-step explanation:

Given

$B(t) = 8500 * (\frac{8}{27})^\frac{t}{3}$

Required [Missing from the question]

Every __ seconds, the bacteria loses $\frac{1}{2}$

First, we model the function from t/3 to t.

$B(t) = 8500 * (\frac{8}{27})^\frac{t}{3}$

Apply law of indices

$B(t) = 8500 * (\frac{8^\frac{1}{3}}{27^\frac{1}{3}})^t$

Evaluate each exponent

$B(t) = 8500 * (\frac{2}{3})^t$ --- This gives the number of bacteria at time t

At time 0, we have:

$B(0) = 8500 * (\frac{2}{3})^0$

$B(0) = 8500 * 1$

$B(0) = 8500$

Let r be the time 1/2 disappears.

When 1/2 disappears, we have:

$B(r) = \frac{B(0)}{2}$

$B(r) = \frac{8500}{2}$

$B(r) = 4250$

So, we have:

$B(t) = 8500 * (\frac{2}{3})^t$

Substitute r for t

$B(r) = 8500 * (\frac{2}{3})^r$

Substitute $B(r) = 4250$

$4250 = 8500 * (\frac{2}{3})^r$

Divide both sides by 8500

$\frac{4250}{8500} = (\frac{2}{3})^r$

$\frac{1}{2} = (\frac{2}{3})^r$

Take log of both sides

$log(\frac{1}{2}) = log (\frac{2}{3})^r$

Apply law of logarithm

$log(\frac{1}{2}) = r\ log (\frac{2}{3})$

Make r the subject

$r = log(\frac{1}{2}) / log (\frac{2}{3})$

$r = \frac{-0.3010}{-0.1761}$

$r = 1.71$

Hence, it reduces by 1/2 after every 1.71 seconds