Solve this following

Question:

The population $\mathrm{P}=\mathrm{P}(\mathrm{t})$ at time ' $\mathrm{t}$ ' of a certain species follows the differential equation

$\frac{\mathrm{dP}}{\mathrm{dt}}=0.5 \mathrm{P}-450 .$ If $\mathrm{P}(0)=850$, then the time

at which population becomes zero is :

 

  1. $\log _{\mathrm{e}} 18$

  2. $\log _{\mathrm{e}} 9$

  3. $\frac{1}{2} \log _{c} 18$

  4. $2 \log _{\mathrm{e}} 18$


Correct Option: , 4

Solution:

$\frac{\mathrm{dP}}{\mathrm{dt}}=0.5 \mathrm{P}-450$

$\Rightarrow \int_{0}^{t} \frac{d p}{P-900}=\int_{0}^{t} \frac{d t}{2}$

$\Rightarrow[\ell \mathrm{n}|\mathrm{P}(\mathrm{t})-900|]_{0}^{\mathrm{t}}=\left[\frac{\mathrm{t}}{2}\right]_{0}^{\mathrm{t}}$

$\Rightarrow \quad \ell \mathrm{n}|\mathrm{P}(\mathrm{t})-900|-\ell \mathrm{n}|\mathrm{P}(0)-900|=\frac{\mathrm{t}}{2}$

$\Rightarrow \quad \ell \mathrm{n}|\mathrm{P}(\mathrm{t})-900|-\ell \mathrm{n}|50|=\frac{\mathrm{t}}{2}$

for $P(t)=0$

$\Rightarrow \quad \ell \mathrm{n}\left|\frac{900}{50}\right|=\frac{\mathrm{t}}{2} \Rightarrow \mathrm{t}=2 \ell \mathrm{n} 18$

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