Question

The population P of a town decreases at a rate proportional to the number by which the population exceeds $100$, the proportionality constant being $k>0$. If the initial population is $2500,$ and $10$ years later the population has fallen to $1900,$ and the time when the population will be $1500$ is expressed as $t=10{\log }_{5/3}\left(b\right)$, Find the value of b.

Answer 1

$\frac{dP}{dt}\alpha (P-100)\Rightarrow \frac{-dP}{dt}=K(P-1000)\int -dPP-1000=\int kt\Rightarrow ln(P-1000)=-kt+c$

At $t=0$

$P=2500C=ln1500P=1000+1500e^{-kt}$

At $t=10$

$P=19001900=1000+1500e^{-10k}{e}^{-10k}=\frac{900}{1500}=\frac{3}{5}\Rightarrow k=\frac{1}{10}ln\left(\frac{5}{3}\right)$

When $P=1500$

$1500=1000+1500e^{-kt}\Rightarrow e^{-kt}=\frac{500}{1500}=\frac{1}{3}\Rightarrow kt=ln3\Rightarrow \frac{1}{10}ln\left(\frac{5}{3}\right)t=ln3\Rightarrow ln\left(\frac{5}{3}\right)t=10ln\left(3\right)=ln\left(3^{10}\right)t=\frac{ln\left(3^{10}\right)}{ln\left(5/3\right)}=10{\log }_{5/3}\left(3\right)$