Question

☆15 POINTS AND MARKED BRAINLIEST IF CORRECT☆look at the image above to view the question!​

Answer 1

Answer:

We can express the question in a exponential function

The current population is 100,000.

The population doubles each day.

The exponential function is given by: P(t)=a(r)^t

The current population is 100,000, a = 100000.

The rate is doubling, r = 2.

P(t)=100000(2)^t

As we know that the bacterial population had been doubling for 5 days. Let A represent the initial population. So, the function is:

P(t)=A(2)^t

After 5 days, the population reaches 100,000. So, when t = 5, P(t) = 100000:

100000=A(2)⁵

Now solving for A, we get

A=(100000)/(2⁵)=3125

So, the function in terms of the original day is:

P (t) = 3125 (2)^t

Hence, the initial population is 3125 bacteria.

3125 is the right answer.

Answer 2

Answer:

3125 bacteria.

Step-by-step explanation:

We can write an exponential function to represent the situation.

We know that the current population is 100,000.

The population doubles each day.

The standard exponential function is given by:

$P(t)=a(r)^t$

Since our current population is 100,000, a = 100000.

Since our rate is doubling, r = 2.

So:

$P(t)=100000(2)^t$

We want to find the population five days ago.

So, we can say that t = -5. The negative represent the number of days that has passed.

Therefore:

$\displaystyle P(-5)=100000(2)^{-5} = 100000 \Big( \frac{1}{32}\Big) = 3125 \text{ bacteria}$

However, we dealing within this context, we really can't have negative days. Although it works in this case, it can cause some confusion. So, let's write a function based on the original population.

We know that the bacterial population had been doubling for 5 days. Let A represent the initial population. So, our function is:

$P(t)=A(2)^t$

After 5 days, we reach the 100,000 population. So, when t = 5, P(t) = 100000:

$100000=A(2)^5$

And solving for A, we acquire:

$\displaystyle A=\frac{100000}{2^5}=3125$

So, our function in terms of the original day is:

$P (t) = 3125 (2)^t$

So, it becomes apparent that the initial population (or the population 5 days ago) is 3125 bacteria.