## Precalculus (6th Edition) Blitzer

a) The exponential growth function is $A=679{{e}^{0.00235t}}$ b) The population of Europe will be 800 million in the year $2045$
(a) Let us consider the exponential growth function as shown below: $A={{A}_{o}}{{e}^{kt}}$ (I) Here, $t$ is the number of years after 1975. In 1975, the population of Europe was 679 million. So, for $t=0,\ A=679.$ Put the above values in equation (1): \begin{align} & A={{A}_{o}}{{e}^{kt}} \\ & 679={{A}_{o}}{{e}^{k\times 0}} \\ & {{A}_{o}}=679 \end{align} Substitute the value of ${{A}_{o}}$ in equation (I). So, the exponential growth function after putting the value of ${{A}_{0}}$ is as shown below: $A=679{{e}^{kt}}$ (II) In the year 2015, the population of Europe was 746 million. So \begin{align} & t=2015-1975 \\ & =40 \end{align} And $A=746$ Put the values of $A$ and $t$ in equation (II), and simplify as shown below: \begin{align} & 746=679{{e}^{k\times 40}} \\ & {{e}^{k\times 40}}=\frac{746}{679} \\ & \ln \left( {{e}^{k\times 40}} \right)=\ln \left( \frac{746}{679} \right) \\ & 40k\times \ln \ e=\ln \left( \frac{746}{679} \right) \\ \end{align} Simplify it further: \begin{align} & 40k\times 1=0.094104 \\ & k=\frac{0.094104}{40} \\ & =0.002352 \\ & \approx 0.00235 \end{align} Put the value of $k$ in equation (II), so the exponential growth function is $A=679{{e}^{0.00235t}}$ (b) In the above part (a), the exponential growth function is as shown below: $A=679{{e}^{0.00235t}}$ It is stated in the problem that $A=800.$ Put the value of $A$ in the above equation, and simplify as shown below: \begin{align} & 800=679{{e}^{0.00235t}} \\ & {{e}^{0.00235t}}=\frac{800}{679} \\ & \ln \left( {{e}^{0.00235t}} \right)=\ln \left( \frac{800}{679} \right) \\ & 0.00235t=0.16399 \end{align} \begin{align} & t=\frac{0.16399}{0.00235} \\ & =69.7832 \\ & \approx 70 \end{align} Hence, the required year will be $1975+70=2045$ in which the population of Europe will be 800 million.