Question

e population of the city was 80000 in the year 2008.the population was increasing at the rate of 3% per annum.what was the population in 1995?

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{population}\mathrm{in}1995=\mathrm{P} \\ \mathrm{Rate}\mathrm{of}\mathrm{increase},\mathrm{R}=3\%\mathrm{p}.\mathrm{a}. \\ \mathrm{Time},\mathrm{n}=13\mathrm{yrs} \\ \mathrm{Now},\mathrm{population}\mathrm{in}2008,\mathrm{A}=80000 \\ \mathrm{Now},\mathrm{A}=\mathrm{P}{\left[1+\frac{\mathrm{R}}{100}\right]}^{\mathrm{n}} \\ \Rightarrow 80000=\mathrm{P}{\left[1+\frac{3}{100}\right]}^{13} \\ \Rightarrow \mathrm{P}=\frac{80000}{{\left(1.03\right)}^{13}}......\left(1\right) \\ \mathrm{Let}\mathrm{x}={\left(1.03\right)}^{13} \\ \Rightarrow \log \mathrm{x}=13\log \left(1.03\right) \\ \Rightarrow \log \mathrm{x}=13\times 0.0128 \\ \Rightarrow \log \mathrm{x}=0.1664 \\ \Rightarrow \mathrm{x}=\mathrm{antilog}0.1664=1.1467 \\ \mathrm{Now},\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get} \\ \mathrm{P}=\frac{80000}{1.1467}=\approx 69765 \end{gathered}$$