Question

The function f(x) = 200 × (1.098)x represents a village's population while it is growing at the rate of 9.8% per year. Create a table to show the village's population at 0, 2, 4, 6, 8, and 10 years from now. Use your table to create a graph that represents the village's population growth. When the population doubles from its current size, the village will need to dig a new water well. To the nearest half of a year, about how long before it is time for the village to dig the new well?

Answer 1

Solution:

$f(x)= 200 \times (1.098)^x$

 $f(x)=200 \times(1+\frac{9.8}{100})^x$

$I_{0}=200,$, R= 9.8%,

Time =x=0,2,4,6,8,10

At, x=0

$I_{0}=200 \times (1.098)^0=200$

At, x=2

$I_{2}=200 \times (1.098)^2=241.1208$

At, x=4

$I_{4}=200 \times (1.098)^4=290.696$

At, x=6

$I_{6}=200 \times (1.098)^6=350.464$  

At, x=8

$I_{8}=200 \times (1.098)^8=422.521$

At, x=10

$I_{10}=200 \times (1.098)^{10}=509.393$

We will draw the graph of , $f(x)= 200 \times (1.098)^x$.

So, New Population, when population of village doubles=400

$I_{P}=400\\\\ 400=200\times (1.098)^x\\\\ 2=(1.098)^x\\\\ x=\frac{log2}{log1.098}$

$x=\frac{0.30102}{0.04060}\\\\ x=7.414$

So, it take approximately, 7 years and approximately 146 days that is 7.414 years by the village to dig the new well.

Answer 2

A) See the attached for a table and graph.

b) It will be 7.5 years before it is time to dig a new well.