The population p(t) at time t of a certain mouse species satisfies the differential equation `(d p(t))/(dt)=0. 5 p(t)-450` If `p(0)""=""850` , then the time at which the population becomes zero is (1) 2 ln 18 (2) ln 9 (3) `1/2` In 18 (4) ln 18

Options

A. `2 log 18`
B. `log 9`
C. `1/2 log 18`
D. `log 18`

Answer

a

Explanation

Given , ` p'(t) = (dp(t))/(dt) = 0.5 p(t) - 450`
` rArr (2dp(t))/(p(t)-900)=(dt)`
On integrating , we get
` rArr (2dp(t))/(p(t)-900)= int dt rArr 2 log |p (t) - 900| = t +C`
To find the value of C, let's substitute t = 0
` rArr 2 log | p (0) - 900 | = 0+C`
` rArr C = 2 log | 850 - 900 |" " [ :' p (0) = 850]`
` rArr C = 2 log 50 `
` :. 2 log | p (t) - 900 | = t +2 log 50 `
Now, put `p(t) = 0` , then
` 2 log |0 - 900 | = t+2 log 50`
` rArr t = 2 log |(900)/(50)| = 2 log 18`