Question

Question #a5024

Answer 1

`3200`

Explanation:

According to the law of exponential grow

`{(y_1 = c e^(alpha t_1)),(y_2=c e^(alpha t_2)):}`

dividing side by side

`y_2/y_1 = e^(alpha(t_2-t_1))` or

`log(y_2/y_1) = alpha(t_2-t_1)` and then

`alpha = log(y_2/y_1)/(t_2-t_1)`

and substituting into `y_1 = c e^(alpha t_1)` we get

`y_1 = c (y_2/y_1)^(t_1/(t:2-t_1))` and then

`c = y_1 (y_2/y_1)^(t_1/(t_1 - t_2))` and now calling

`(t_1, y_1) = (2,100)`
`(t_2,y_2) = (5,800)`
`t_3 = 7`

then

`alpha = (log 8)/3 = 0.693147`
`c = 25`

`y_3 = 0.693147 e^(25 t_3) = 3200`

Answer 2

`3200`

Explanation:

`y(t) = a * e^(kt)`

value at time `t` = start value * e^(rate of growth * time)

value after `2`h = `100`
`a = 100`

`5` hours - `2` hours = `3` hours
`t = 3`

value after `5` hours = `800`
`y(3) = 800`

substituting these values into the equation:

`800 = 100 * e^(3k)`

`e^(3k) = 800/100 = 8`

`e^(3k) = 8`

take `ln` of both sides:

`ln(e^(3k)) = ln(8)`

`ln(e^x) = x -> ln(e^(3k)) = 3k`

`3k = ln8`

`k = ln8/3`

after `7` hours: `y(t) = 100 * e^(5k)`

`7`h-`2`h=`5`h

`5k = 5ln8/3`

`y(t) = 100 *e^((5ln8)/3)`

`y(5) = 100 * e^((5ln8)/3) = 100*31.9999...`

`= 3200` (nearest 1)

the number of bacteria is `3200` after 7 hours.