Suppose the population in San Luis Obispo was 42,000 in the year 1990 (at the beginning of the year)
A) if back then the population was predicated to grow continuously at a rate of 0.8% what should it have been at the end of 2012?
B) in what year will the population reach 50,000 if it continues to grow at this rate?

Question

Suppose the population in San Luis Obispo was 42,000 in the year 1990 (at the beginning of the year) 
A) if back then the population was predicated to grow continuously at a rate of 0.8% what should it have been at the end of 2012? 
B) in what year will the population reach 50,000 if it continues to grow at this rate?

anonymous · Answer

@Directrix and @ranga

jdoe0001 · Answer

I'm thinking you need to use the compounded interest formula, for a yearly period

so  you have the principal amount, the years, the rate and the period, per year, so plug them in to find A)

directrix · Answer

@ninab731 I am with @jdoe001 on the continuously compounded interest formula. It is essentially the same as the continued growth formula which is attached. (Read more about it here: http://math.ucsd.edu/~wgarner/math4c/textbook/chapter4/expgrowthdecay.htm )

directrix · Answer

So, the population at the end of 2012 (22 years) is given by:

N = (42,000) * e ^(.008 * 22)  --> @ninab731   Crank this out.

 The .008 is the .8% growth rate expressed as a decimal and t is the number of years since 1990 until 2012.

So, get on your calculator and crank that out. You can approximate e but your calculator has an e button on it somewhere. 

Post what you can and we will compare answers, okay? And, we'll decide what to do about part b.