A) 80 years
B) \[80\frac{\log 2}{\log 3}years\]
C) \[40\frac{\log 3}{\log 2}years\]
D) \[40\log 2\log 3\,years\]
Correct Answer: C
Solution :
| [c] Let the initial population be \[{{x}_{0}}\] and it is x in t years, then the differential equation is |
| \[\frac{dx}{dt}=kx,\] k is a constant |
| \[\Rightarrow \frac{dx}{x}=kdt.\]Integrating we get |
| \[\log x+kt+C\] ? (i) |
| When \[t=0,x={{x}_{0}}\Rightarrow c=\log {{x}_{0}}\] |
| Then from (i) \[\log x=kt+\log {{x}_{0}}\] |
| \[\Rightarrow \log \frac{x}{{{x}_{0}}}=kt\] ? (ii) |
| Now when |
| \[\Rightarrow \log \frac{x}{{{x}_{0}}}=2\Rightarrow \log 2=k.40\Rightarrow k=\frac{\log 2}{40}\] |
| \[\therefore \] (ii) becomes \[\log \frac{x}{{{x}_{0}}}=\frac{\log \,2}{40}\,\,.\,\,t\] |
| Next put \[\frac{x}{{{x}_{0}}}=3\Rightarrow t=40\frac{\log 3}{\log 2}\] |
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