Question

The number of bacteria in a culture is increasing according to the law of exponential growth. The initial population is 250 bacteria, and the population after 10 hours is double the population after 1 hour. How many bacteria will there be after 6 hours?

Answer 1

Answer:

The correct answer is 396.85.

Step-by-step explanation:

The number of bacteria in a culture is increasing according to the law of exponential growth.

Let A = $A_{0}$ × $e^{kt}$ where A is the final count of bacteria after t time and k is the rate of increase and $A_{0}$ is the initial count of bacteria.

Given the initial population is 250 bacteria, or $A_{0}$ = 250.

The population after 10 hours say M is M = 250 × $e^{10k}$

The population after 1 hour say N is N = 250 × $e^{k}$.

Given M = 2× N

⇒ 250 × $e^{10k}$ = 500 × $e^{k}$

⇒ $e^{10k}$ = 2× $e^{k}$

Taking logarithm to the base e both sides gives the value of k as

10k = ㏑ 2 + k

⇒ k = $\frac{ln2}{9}$ = 0.07701

Amount of bacteria present after 6 hours is 250 × $e^{6k}$ = 396.85.

Answer 2

The population increase of the bacteria can be found by using the

exponential growth formula.

• The population of the bacteria after 6 hours is approximately 396 bacteria

Reason:

Given:

The initial population of the of bacteria = 250

The formula for population growth is P = P₀ × $e^{r \cdot t}$

Where;

r = The growth rate

t = The time (in hours)

P₀ = The initial population

The growth parameters are;

P₁₀ = 2·P₁

Therefore;

250 × $e^{r \times 10}$  = 2 × 250 × $e^{r \times 1}$

$e^{r \times 10}$ = 2 × $e^{r \times 1}$

$e^{9 \cdot r}$ = 2

9·r = ㏑2

$r= \dfrac{ln(2)}{9}$

When the time, t = 6 hours, we have;

$P_6 = 250 \times e^{6\times r}$

Therefore;

$P_6 = 250 \times e^{6\times \dfrac{ln(2)}{9}} \approx 396.85$

The approximation of population to a whole number is given by rounding

off down, therefore, we have;

• The population of the bacteria after 6 hours ≈ 396 bacteria

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