Solving Exponential Equations Grade 11 Math?

👨🏻‍🏫 學術台數學

[匿名]

2007-12-03 4:00 pm

a town with a population of 12 000 has been growing at as average rate of 2.5% for the last 10 years. Suppose the growth rate will be maintained in the future. The function that models the town's growth is P(n)=12(1.025^n) where P(n) represents the population (in thousands) and n is the number of years from now.

a) Determine the population of the town in 10 years.

b) Determine the number of years until the population doubles.

c) Use the equation to determine the number of years age that the population was 800. Answer to the nearest year.

更新1:

PLEASE SHOW ALL WORK =D

更新2:

There is a mistake in the last question (c). is says 800 when it should be 8000 i am sorry for the mistake thank you to all who helped me so far

回答 (5)

[匿名]

2007-12-03 4:06 pm

✔ 最佳答案

a) plug 10 in for n and use a caculator:
P(10) = 12(1.025^10)

b) solve for n using logarithms:

24 = 12(1.025^n)
2 = 1.025^n
n = log 2 / log 1.025

c) solve for n using logarthims:

12 = .8(1.025^n)

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[匿名]

2016-10-10 3:42 pm

end fixing for x 3^(3x+a million) = 3/27 take logs (3x+a million)log(3) = log(a million/9) (3x+a million)log3 = log1 - log9 (3x+a million)log3 = 0 - log(3^2) (3x+a million)log3 = -2log3 (3x+a million) = -2log3/log3 3x+a million = -2 3x = -3 x= -a million does this make it seem simplier?

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pyrodude1031

2007-12-03 4:16 pm

I dont have calculator with me, please pluck in the expression in the final answer. SOrry.

a.
subsitute n=10,
P(10) = 12(1.025^10)

b.
the initial population is 12000 , the double of that is 24000, but notice the expression for P(n) is in thousands. Hence P(n) is actually just 24
therefore, 24 = 12(1.025^n)
2 = 1.025^n
ln(2) = ln(1.025^n)= n* ln(1.025)
n = ln2 / ln1.025 which is positivive

c.
population now is 800, and require to find n
once again P(n) is in thousands hence P(n) = 0.8
0.8 = 12(1.025^n)
0.8/12 = 1.025^n
n = [ln (0.8/12)] / ln1.025
this answer is negative , hence the answer would be so so many years ago

參考: Hope everything is right... Not quite sure about the last one though .

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[匿名]

2007-12-03 4:10 pm

a) let n = 10
P(10) = 12(1.025)^10 = 15.361 = 15,361 people

b) population double will be 24000

24= 12(1.025)^n divide by 12
2 = (1.025)^n

take the log of both sides

log2 = log(1.025)^n
log2 = nlog(1.025)
log2/log(1.025) = n
28.07 = n

about 28 years to double

c) 800 in terms of thousands = 800/1000 = 0.8

0.8 = 12(1.025)^n
divide b 12

1/15 = (1.025)^n
take the log

log(1/15) = n log (1.025)
log(1/15)/log(1.025) = n
-109.7

about 110 years ago

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[匿名]

2007-12-03 4:09 pm

P(0) = 12000 {Current population}

P(n) = population, in thousands, in n years

a) We need P(10) = 12 (1.025^10) * 1000

b) We need to find when P(n) = 24
12 (1.025^n) = 24
1.025 ^n = 2
n = log (base 1.025) 2

n = ln 2 / ln 1.025

ln 2 = approx. 0.69315
ln 1.025 = approx 0.02469
n = about 28 years.

c) P(n) = 800
12000 * 1.025^n = 800
1.025 ^n = 800/12000
n = ln (800/12000) / ln 1.025
n = approx. -110
The population was 800 about 110 years ago.

參考: Definition of logarithm, and the study of logarithmic operations, as well as consistent practice

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