Question 1188359: When the population growth of a certain city was first studied, the population was 22,000. It was found that the population P grows with respect to time t in years by the formula P=22,000(10exponential of 0.163t)
How long will it take for the city to double its population?
How long will it take for the population of the city to triple?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the growth formula is either:
f = p * (1 + r) ^ n
or:
f = p * e ^ (r * t)
i don't understand p = 22,000 * 10 ^ .163t.
i suspect you want f = p * e ^ (r * t)
under that aswsumption, .....
for the population to double, the formula becomes:
44,000 = 22,000 * e ^ (.163 * t)
divide both sides of the formula by 22,000 to get:
2 = e ^ (.163 * t)
take the natural log of both sides of the formula to get:
ln(2) = ln(e ^ (.163 * t)) which becomes:
ln(2) = .163 * t * ln(e) which becomes:
ln(2) = .163 * t
solve for t to get:
t = ln(2) / .163 = 4.252436691
confirm by replacing t in the original requation of f = 22,000 * e ^ (.163 * t) with 4.252436691 to get:
f = 22,000 * e ^ (.163 * 4.252436691).
solve for f to get:
f = 44000.
to triple the growth, you'll go through the same steps from:
66,000 = 22,000 * e ^ (.163 * t)
and wind up with:
ln(3) = .163 * t
solve for t to get:
t = ln(3) / .163 = 6.739952691
confirm by replacing t in the original equation of f = 22,000 * e ^ (.163 * t) with that to get:
f = 22,000 * e ^ (.163 * 6.739952691)
solve for f to get:
f = 66000.
the equation can be graphed, as shown below:
the coordinate points are in (x,y) format.
x in the graph equation is the same as t in the algebra equation.
y in the graph equation is the same as f in the algebra equation.
what this means is:
f = 22000 * e ^ (.163 * t) and y = 22000 * e ^ (.163 * x) are equivalent equations and will give you the same results.
f and y represent the future value of the population.
t and x represent the number of years.
let me know if you have any questions.
theo
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