pre calc question

Question

The doubling period of a bacterial population is  10 minutes. At time t=110 minutes, the bacterial population was 50000.

What was the initial population at time t=0?    

Find the size of the bacterial population after 5 hours.

Answer 1

50k = Poe^110r

it doubled 11 times in 110 minutes

50k in 110

25k in 100

12.5k in 90

6.25k in 80

3.125 in 70

1.5625 in 60

.78125 in 50

.390625 in 40

.1953125 in 30

.9765625 in 20

.48828125 in 10

.244140625 in 0

Po= 244.140625 = initial amount

5 hours = 5x60 = 300 minutes

doubling times of 30 times

50k in 11

100k in 12

200k in 13''

400k in 14

800k in 15

1600k in 16

1700k in 17

3400k in 18

6800k 19

13600k 20

27200k 21

54400k 22

108800k 23

217600k 24

435200k 45

870400k 46

1740800k 47

3481600k 48

6963200k 49

13926400k= 13,926,400,000 in 5 hours

= about 14 billion

Answer 2

50000 = a₀(2)^{110 / 10}, ,solve for a₀ and answer

A(300) = a₀(2)^{300 / 10}, substitue a₀ from above and calculate A(300)