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The population of a city in 1910 was 50,000, and it doubles every 10 years. What will it be (a) in 1970 (b) in 1990 (c) in 2,000?

Im trying to brush up my algebra skills and found this question.

I couldnt think of a purely algebraic way to find the answer, this is what I did.

There are 60 years between 1970 and 1910. Since the population doubles every 10 years, it will double 6 times in period of 60 years.

So if P = 50,000, 2P = 100000, 4P = 200000 8P = 400000, 16P = 800000, 32P = 16,00000 64P = 32,00000.

Is it done incorrectly/is there a algebraic method?

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You have answered the question perfectly but if you want to find the population after $n$ decades note that

$$P_0=50K$$ $$P_1=2(50K)$$ $$P_2=2(2(50K))=2^2(50K)$$ $$P_3=2(2^2(50K))=2^3(50K)$$ $$\vdots$$ $$P_n=2^n(50K)$$

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  • $\begingroup$ sorry to ask but does 50K represent 50,000? Could this be written as $$P_n = 2^n(50,000)$$? $\endgroup$ – salman Aug 26 '13 at 17:02
  • $\begingroup$ @user90771 Yes it does and yes it could. $\endgroup$ – JP McCarthy Aug 26 '13 at 17:05
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Your answer is correct, I don't think there is much a problem the way you did it.

So after 10 years, the population would be 2P, after 20 years it would be $2\times2\times P$, after 30 years it would be $2 \times 2\times2\times P$.

So after 10k years population $P_k= 2^k P$

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