Quick math question

  • Thread starter joejo
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Hi guys...can someone help me out with this problem...

The population of Newfoundland and Labrador was 533,800 in 2001. At that time the population of Newfoundland was decreasing at a rate of 0.6% per year. The population of PEI was 138,500 in 2001. At that time the population was increasing at a rate of 0.2% per year. In what year will the population of PEI be more than Newfoundland? Explain?


I have this so far: y=#of years

533,800(1-0.006)^y

138,500(1-0.002)^y


Now what?
 
the equation of pop. growth is y(t) = y0 * e ^(kt)

where y0 is begining pop
k is the growth rate
and t is the time

plug in the values and make them equal to each other and solve for t
 
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we havent done that yet..any other way...
 
138,500*(1+0.002)^y = 533,800*(1-0.006)^y
138,500*(1.002)^y = 533,800*(0.994)^y
ln(138,500*(1.002)^y) = ln(533,800*(0.994)^y)
ln(138,500) + ln((1.002)^y) = ln(533,800) + ln((0.994)^y)
ln(138,500) + y*ln(1.002) = ln(533,800) + y*ln(0.994)
y*ln(1.002) - y*ln(0.994) = ln(533,800) - ln(138,500)
y = ln(533,800/138,500) / ln(1.002/0.994) = 168.3

2001 + 168.3 = 2169.3

... theoretical
 
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Thx faruk but wats ln?
 
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Natural log
 
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i haven't done logs yet...this is suppose to be grade 11 math...
 
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It's been many years since I've done stuff like this but isn't it the same sort of calculation as working out compound interests?

Create a graph with the decreasing population of Newfoundland represented using one colour and then do the same for Labrador. Where the line cross is where they equal each other.
 
joejo said:
i haven't done logs yet...this is suppose to be grade 11 math...
*joejo, logarithms are used to find exponents, which would be the years in your problem. For example, solving for [itex] c [/itex] in the equation
[tex] a = b^c ,\left( {a,b,c} \right) \in \mathbb{R} [/tex]
will give you
[tex] c = \log _b a \Leftrightarrow \frac{{\log a}}{{\log b}} \Leftrightarrow \frac{{\ln a}}{{\ln b}} [/tex]
*What's left is just plugging in values (*hint:smile:*)
---------------------------------------------------------------------
*P.S., I am a high school junior too (Grade 11), and the mathematics
at my school can seem at times quite slow as well :shy:;
[tex] \downarrow [/tex] For example, I just finished second-semester calculus this year, but I
could have studied it as a sophomore last year (i.e., in 10th grade :rolleyes:~)
 
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11th grade: that's 16yrs old? Is that the equivalent of O levels (GCSE's)
 
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Daminc said:
Is that the equivalent of O levels (GCSE's)
No-one calls them O Levels any more :smile:. Edit: A Levels still remain A Levels, though!
 
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You can also do it by the trial and error method

For Newfoundland : 533800*(1-0.006)^t => 533800*(0.994)^t
For PEI: 138500*(1+0.002)^t => 138500*(1.002)^t

Start with t+0 (t equals the number of years elapsed) and start jumping:

when t=20 Newfoundland=476123.8 & PEI=143858.8 which is no good

t=120 gives Newfoundland=259265.5 & PEI=176025.8 which is closer

t=150 gives Newfoundland=216439.4 & PEI=186899.4 which is closer still

t=170 gives Newfoundland=191895.1 & PEI=194519.2 has overshot a little

t=168 gives Newfoundland=194218.7 & PEI=193743.4 nearly there

t=169 gives Newfoundland=193053.4 & PEI=194130.9 which is the number you want

2001+169=2170 which is the year PEI overtakes Newfoundland :)
 
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No-one calls them O Levels any more
Hah, I'm old. Leave me alone :cry:

t=169 gives Newfoundland=193053.4 & PEI=194130.9 which is the number you want

2001+169=2170 which is the year PEI overtakes Newfoundland :)
Mmmm, the crossover occurs in the later part of 168 so the year would be 2169 (I think)
 
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BobG

Science Advisor
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joejo said:
i haven't done logs yet...this is suppose to be grade 11 math...
I would think they would have at least introduced common logs (base 10). Or maybe I'm old, too. When I was in high school, you had to have some understanding of logs or you wouldn't understand your calculator. But then, our calculators were three pieces of bamboo lashed together.
 
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Actually, in Belgium logs are introduced in grade 12, but that may not be a great reference... :rolleyes:
 
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thanks guys.....you all helped out a bit...Damincs answer was the way we did it in the book...Thanks Daminc!
 

saltydog

Science Advisor
Homework Helper
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This is what I have JoeJo:

Consider differential equations for both places:

[tex]\frac{dN}{dt}=-0.006N[/tex]

[tex]\frac{dP}{dt}=0.002P[/tex]

Solving them with the initial conditions given and equating them for when the populations just become equal we get:

[tex]533800e^{-0.006t}=138500e^{-0.002t}[/tex]

little of this, little of that, we get t=169 so in 2170 the populations become equal as PEI starts to overtake the other one.

Edit: I'm such a slow-poke. :yuck:
 
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I would think they would have at least introduced common logs (base 10). Or maybe I'm old, too. When I was in high school, you had to have some understanding of logs or you wouldn't understand your calculator.
When I was younger I think we were introduced to logs by having a small blue book filled with numbers. Can anyone else remember that book?

p.s. I'm sure we did logs when we were about 12-13 or so.
 

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