A COLONY OF BACTERIA IS GROWN UNDER IDEAL CONDITIONS IN A LAB SO THAT THE POPULATION INCREASES EXPONENTIALLY WITH TIME. At the end of the three hours, there are 10,000 bacteria. At the end of the 5 hours, there are 40,000 bacteria. How many bacteria were present initially?

I was planning to use Y=P(O)e^kt
but I'm not sure how to set it up.

Y(5)=40,000=Pe^5k

y(3)=10,000=Pe^3k

divide one equation by the other

4=e^2k solve for k. Put it back in, and solve for P(O)

There is an easier way. From the data, one can see that population doubles each hour.

Sure, use y = Po e^kt

when t = 3:
10,000 = Po e^3k
ln 10,000 = ln Po + 3k ln e but ln e is 1
so
ln 10,000 = ln Po + 3 k
similarly
ln 40,000 = ln Po + 5 k

subtract
ln 10,000 -ln 40,000 = -2 k

ln (10,000/40,000) = -2k
-1.386 = -2k
k = .693

then back
10,000 = Po e^(2.0794)
Po = 1250

To solve this problem, you can set up an exponential growth equation using the given information. The equation you mentioned, Y = P(0) e^(kt), can be used here.

In this equation:
- Y represents the final population count.
- P(0) represents the initial population count.
- e is Euler's number, approximately 2.71828 (a mathematical constant).
- k is the growth rate constant.
- t represents the time (in this case, in hours).

Let's plug in the given information to find the values of P(0) and k:

At the end of 3 hours, Y = 10,000.
At the end of 5 hours, Y = 40,000.

We can set up two equations using this data:

Equation 1: 10,000 = P(0) e^(3k)
Equation 2: 40,000 = P(0) e^(5k)

Now, we need to solve these two equations simultaneously to find the values of P(0) and k. We can do this by dividing Equation 2 by Equation 1:

40,000 / 10,000 = (P(0) e^(5k)) / (P(0) e^(3k))
4 = e^(5k - 3k)
4 = e^(2k)

Next, take the natural logarithm (ln) of both sides of the equation to isolate the exponent:

ln(4) = ln(e^(2k))
ln(4) = 2k

Now, solve for k by dividing both sides of the equation by 2:

k = ln(4) / 2

Substitute the value of k back into Equation 1 to find P(0):

10,000 = P(0) e^(3 * (ln(4) / 2))

Simplify the equation:

10,000 = P(0) e^(3ln(4) / 2)
10,000 = P(0) e^(ln(4^3) / 2)
10,000 = P(0) e^(ln(64) / 2)
10,000 = P(0) e^(3/2 ln(2))
10,000 = P(0) e^(1.03972) [approximately]

Now, solve for P(0) by dividing both sides of the equation by e^(1.03972):

P(0) = 10,000 / e^(1.03972)

Calculate the approximate value:

P(0) ≈ 10,000 / 2.82536
P(0) ≈ 3,542

Therefore, the initial population of bacteria was approximately 3,542.

To solve this exponential growth problem, you can use the formula for exponential growth:

N(t) = N₀ * e^(k * t)

In the given problem, we have two data points:

At t = 3 hours, N(3) = 10,000 bacteria.
At t = 5 hours, N(5) = 40,000 bacteria.

To find the initial population of bacteria (N₀), you need to solve the system of equations formed by substituting these values into the formula:

Equation 1: 10,000 = N₀ * e^(3k)
Equation 2: 40,000 = N₀ * e^(5k)

Now, we can solve this system step by step:

Step 1: Divide Equation 2 by Equation 1:

40,000 / 10,000 = (N₀ * e^(5k)) / (N₀ * e^(3k))

4 = e^(5k - 3k)

Step 2: Simplify:

4 = e^(2k)

Step 3: Take the natural logarithm (ln) of both sides:

ln(4) = ln(e^(2k))

Step 4: Use the property of logarithms:

ln(4) = 2k * ln(e)

Since ln(e) = 1, we have:

ln(4) = 2k

Step 5: Solve for k:

k = ln(4) / 2

Step 6: Substitute k back into Equation 1 to find N₀:

10,000 = N₀ * e^(3 * (ln(4) / 2))

Simplifying:

10,000 = N₀ * e^(3 * ln(2))

10,000 = N₀ * e^(ln(8))

10,000 = N₀ * 8

Dividing both sides by 8:

N₀ = 10,000 / 8

N₀ = 1250

Therefore, the initial population of bacteria was 1250.