Question

A bacteria population is 6000 at time t = 0 and its rate of growth is 1000 · 9t bacteria per hour after t hours. what is the population after one hour? (round your answer to the nearest whole number.)

Answer 1

Answer:

14384

Step-by-step explanation:

We are given that

P(0)=6000

$\frac{dP}{dt}=1000\cdot 9^t$

We have to find the population after 1 hour.

$dP=1000(9^t)dt$

Taking integration on both sides

$P=1000\int 9^t dt$

$P(t)=\frac{1000}{log 9}9^t$+C

Using formula:$\int a^t=\frac{1}{log a}a^t+C$

Substitute t=0 and P(0)=6000

$6000=\frac{1000}{log9}+C$

$C=6000-\frac{1000}{log 9}=6000-\frac{1000}{0.954}$

$C=4952$

Substitute the values then we get

$P(t)=1047.95(9^t)+4952$

Substitute t=1

Then, we get

$P(1)=1047.95(9)+4952$

$P(1)=14384$

Hence, the population after one hour=14384

Answer 2

 dP/dt = 1000 * 9^t is the bacteria population's growth rate.where:P(t) is the population at time t, andP(0) = 6000. 
The population after one hour would be: 
P(1) = P(0) + <Integral of dP/dt from 0 to 1> 
P(1) = P(0) + [ 1000 * 9^1 / ln(9) - 1000 * 9^0 / ln(9) ] 
P(1) = 6000 + 3641 = 9461 would be the answer