population math problem?

Question

If the population of a town doubles every 10 years, then the population in year "x + 100" will be how many times the population in year "x"?

Answer 1

The population after t years is P(t) = P0*e^(kt), where P0 is the initial population.

It doubles every 10 years, so P(t+10) = 2P(t). Therefore

P0*e^(k(t+10)) = 2P0*e^(kt).

Divide by P0, and use a^(x+y) = a^x a^y:

e^(kt)e^(10k) = 2e^(kt)

so

e^(10k) = 2.

P(t+100) / P(t) = P0*e^(k(t+100)) / [P0*e^(kt)]
= e^(k(t+100)) / e^(kt)
= e^(100k)
= [e^(10k)]^10
= 2^10
= 1024.

Answer 2

it would want to were effective to point the answer you've been getting so it would want to be checked, yet that aside it is the technique that i used to sparkling up for t = 0. N(0) = 2000/(a million + 24.3e^0.7(0)) N(0) = 2000/(a million + 24.3e^0) bear in mind once you've a nil in the amplify part of an exponential it would want to have a outcome of one million. N(0) = 2000/(a million + 24.3(a million)) N(0) = 2000/(25.3) N(0) = seventy 9...... because it has a decimal position you are able to both round up or down reckoning on the numbers. yet you ought to apply an finished style, yet you do not have too. i desire this helps and strong success.

Answer 3

every 10 years it doubles, so in 10 yrs, it will be 2 times
20 yrs, 2*2 times
30 yrs, 2*2*2 times
...
..
n*10 yrs 2^n times

so in x+100 yrs, it will be 2^10=1024 times...

Answer 4

Doubles in 10 years:
10 years: 2x
20 years: 4x
30 years: 8x
...
100 years: 1024x

This is called exponential growth.

Answer 5

population = 10x/2

if x = 100

population = 500