Answer:
Step-by-step explanation:
At the time t = 0, population of the town = 5000
Rate of population increase = 500 per year
Therefore, the equation that will represent the population will be

Where
= Population after t years
= Initial population
t = Time in years
a). For double once the population will be 500×2 = 10000
By plugging in the values in the equation,
10000 = 5000 + 500t
500t = 10000 - 5000
500t = 5000
t = 
t = 10 years
For Double twice,
Population will be = 10000×2 = 20000
Now we plug in the values in the equation again
20000 = 5000 + 500t
500t = 20000 - 5000
500t = 15000
t = 
t = 30 years
For double thrice,
Population of the town = 20000×2 = 40000
Now we plug in the values in the equation,
40000 = 5000 + 500t
500t = 40000 - 5000
500t = 35000
t = 
t = 70 years
b). If the population growth is 5%.
Then the growth will be exponential represented by

= Population after t years
= Initial population
t = time in years
For double once,
Population after t years = 10000



Take log on both the sides

tlog(1.05) = log2
t = 
t = 14.20 years
For double twice,
Population after t years = 20000



Take log on both the sides

tlog(1.05) = log4
t = 
t = 28.413 years
For double thrice
Population after t years = 40000



Take log on both the sides

tlog(1.05) = log8
t = 
t = 42.620 years