The population of smartsville in 1950 was 9618 . if the population was growing 3.25% per year. determine the population in 1960

Question

The population of smartsville in 1950 was 9618 . if the population was growing 3.25% per year. determine the population in 1960 and in 2003. and when will the populatioin reach 20,000

Answer 1

You need to make a basic exponential growth equation of the form P(t) = ab^x , where a is your initial conditions at some start time, and b is the growth factor.

 

For your problem:

 

a=9618 if we consider 1950 year 0

b=1+3.25%, or 1.0325

 

So P(t)=9618(1.0325)^t, where t is years after 1950

 

• P(1960) really is P(10), as 1960 is 10 years after 1950
• P(10)= (calculator required) = 13, 243 (rounded to the nearest person)
• For the second part, you want to know when P=20,000.  So we have to work backwards.  This will required logs
• 20,000=9618(1.0325)^t
• 2.0794=(1.0325)^t
• Log(2.0794)=t(log(1.0325)) 
• t = Log(2.0794)/Log1.0325)
• t = 22.89 years, bur remember this is years after 1950, so the year would be late 1972.

Answer 2

P(t) = (9618)*(1.0325)^t

 

Where:

• P(t) = the population in year t
• t = years since 1950

 

In 1960, t = 10:

 

P(10) = (9618)*(1.0325)¹⁰ = 13,243

 

In 2003, t = 53:

 

P(53) = (9618)*(1.0325)⁵³ = _______

 

The population will reach 20,000:

 

20,000 = (9618)*(1.0325)^t

 

20,000/9618 = (1.0325)t

 

log(2.08) = log(1.0325)^t = t*log(1.0325)

 

log(2.08)/log(1.0325) = t = __________