Question

the doubling period of a bacterial population is 10 minutes. at time $t = 110$ minutes, the bacterial population was 90000.
what was the initial population at time $t = 0?$
find the size of the bacterial population after 3 hours.
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Explanation:

Step1: Define exponential growth formula

Exponential growth for doubling time: $P(t) = P_0 \times 2^{\frac{t}{d}}$, where $P_0$ = initial population, $d=10$ min (doubling time), $P(t)$ = population at time $t$.

Step2: Plug in known values for $t=110$

Substitute $P(110)=90000$, $t=110$, $d=10$:
$90000 = P_0 \times 2^{\frac{110}{10}}$
$90000 = P_0 \times 2^{11}$

Step3: Solve for initial population $P_0$

Calculate $2^{11}=2048$, then:
$P_0 = \frac{90000}{2048} \approx 43.95$
Round to nearest whole number (population is integer): $P_0 \approx 44$

Step4: Convert 3 hours to minutes

3 hours = $3 \times 60 = 180$ minutes

Step5: Calculate population at $t=180$

Use $P(t) = 44 \times 2^{\frac{180}{10}}$
$P(180) = 44 \times 2^{18}$
$2^{18}=262144$, so:
$P(180) = 44 \times 262144 = 11534336$

Answer:

Initial population at $t=0$: 44
Population after 3 hours: 11534336