# How do you do this math problem?The number of bacteria b on a Petri dish is given by `b(t) = 500e^0.04t` , where t is measured in minutes. a. how many bacteria are there at t=0? b. How many...

How do you do this math problem?

The number of bacteria b on a Petri dish is given by `b(t) = 500e^0.04t` , where t is measured in minutes.

a. how many bacteria are there at t=0?

b. How many bacteria will there be when t=30 minutes?

c. When will the bacteria double?

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For each portion of the question, we simply are plugging in values to the funtion b(t):

`b(t) = 500e^(0.04t)`

**a) How many bacteria are there at t=0?**

To find this, we simply substitute 0 for t

`b(0) = 500e^(0.04*0)`

To solve this, we just need to recognize that e^0 = 1:

`b(0)=500*1 = 500`

Therefore, we have **500 bacteria as our starting point**.

**b) How many bacteria will there be at t=30 minutes?**

Again, we will simply substitute 30 for t in b(t):

`b(30) = 500e^(0.04*30)`

`b(30) = 500e^1.2`

Now, we simply evaluate the above expression on a calculator to get:

`b(30) = 1660`

Wow, so apparently we have **1660 bacteria in just half an hour**. This is reasonable, though, based on the equation and the fact that bacteria multiply fairly quickly. Now, we'll move on to the last part.

**c) When will the bacteria double?**

This question requires our answer from the first question. Our starting point was 500 bacteria (at time t=0). To determine when the number of bacteria will double, we'll need to determine when the number of bacteria reaches 1000 (2*500). To do this, we'll substitute 1000 into the b(t) term in our function and we'll solve for t:

`b(t) = 500e^(0.04t)`

`1000 = 500e^(0.04t)`

Now, we can divide both sides by 500:

`1000/500 = 500/500e^(0.04t)`

`2 = e^(0.04t)`

Now, to bring the exponent down and allow us to find t directly, we'll need to take the natural logarithm of both sides:

`ln(2) = ln(e^(0.04t))`

Recall, now, the property of natural logarithm that `ln(e^x) = x` because the natural logarithm is the inverse function of `e^x` (see link).``

In our case, we'll use this property to bring down the 0.04t:

`ln 2 =0.04t`

Now, we'll simply divide both sides by 0.04t:

`ln2/0.04 = (0.04t)/0.04`

`ln2/0.04 = t`

Now, depending on your teacher, you can leave it like that or just solve using a calculator to get the following result:

**t = 17.3 minutes**

Hope that helps!