Year population

[Dr_Suess]

Im doing a program that gets the amount of years after 2003 to output the year and population beside or after it. The population in 2003 is 350000 and it increases 5% every year.

Can you help?
Thanx

[Scrapyard]

Just some simple maths...

Code:

INPUT "Year you want to know the population of ",Year%

Population! = INT(350000 + 1.05 ^ (Year%-2003))

PRINT "Population in";Year%;"is";Population!

[Meg]

Double check that math.

If the population in 2003 = 350,000 and increases 5% each year, then you need to multiply, not add:

Code:

input "Enter year you wish to estimate population: ", year
population = 350000 * 1.05 ^ (year - 2003)
print "Population will be: "+ STR$(population)

*peace*

Meg.

[toonski84]

I think the formula goes:

f(x) = c*e^(kt)

where e is log(1), c is the initial population, k is .05 and t is the number of years. some dumb algebra 2 formula i keep forgetting... i think the way they give might work too, i dont know (i was never much of a mathematician).

[Scrapyard]

Woops my mistake :wink:

[oracle]

That's just the compound interest formula... nothing hard about it (but toonski your method is just weird and confusing)

[toonski84]

yeah, it's used for interest problems, but my calc book uses it mostly for population problems, and pretty much any kind of exponential problem.

[Agamemnus]

Toonski, e?
Depends on if it is a continual growth or not.