**p(1.67)=2500(4)^1.67**

**p(1.67)=25315**

**p(4)=2500(4)^4**

**p(4)=640000**

Find the size of the bacterial population after 100 minutes.

Find the size of the bacterial population after 4 hours.

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Marked as Best Answer

To answer this questions we must first find the constant k. If we find k then we can calculate our population for anytime.

I am going to assume that time t is measured in hours as the problem statement doesn't specify.

The population doubles every half hour, therefore:

p=5000 when t=0.5 hours

Plug these values in the equation to obtain:

5000=2500e^(0.5k)

Simplify...

2=e^(0.5k)

Square both sides

4=e^k

k=ln(4)

Now we have k so we can plug it in the original equation and obtain p for any time

p(t)=2500e^(ln(4)t), which simplifies to

p(t)=2500(4)^t

Now evaluate for t=100 minutes (1.67 hours) and t=4 hours

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