Algebra II Help

4) The function f(x)=2(3)^x represents the growth of a dragonfly population every year in a remote swamp. Rose wants to manipulate the formula to an equivalent form that calculates seven times a year, not just once a year. Which function is correct for Rose’s purpose and what is the new growth rate?
A) f(x)=2(3)^x, growth rate 200%
B) f(x)=2(1.17)^7x, growth rate 17%
C) f(x)=2(3)^x,growth rate 2%
D) f(x)=2(1.17)^x,growth rate 2%

I do not know what to do with this question. I am very confused. I'm thinking it is B or D, but maybe D?
Please explain this to me!! :)

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  1. If I understand your question correctly, this is the same as finding the equivalent of an annual rate with new compounding frequencies
    the way it stands, f(x) = 2(3)^x
    or f(x) = 2( 1 + 2)^x to me represents a growth of 200% compounded annually
    so you want
    f(x) = 2(1 + r)^7x to be equivalent

    2(3)^x = 2(1+r)^7x
    3^x = (1+r)^7x
    take log of both sides
    x log3 = 7x log (1+r)
    log 3 = 7log(1+r)
    log3 = log( (1+r)^7)
    3 = (1+r)^7
    take 7th root of both sides
    1.16993 = 1+r
    r = .16993 or 16.993%

    so f(x) = 2(1.16993)^(7x) should do it
    which looks like B

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    posted by Reiny
  2. I was going to check this:
    e.g how much after 3 years
    original:
    f(3) = 2(3)3 = 54
    new one:
    f(3) = 2(1.16993)^21 = 53.9992 , not bad
    According to B
    f(3) = 2(1.17)^21 = 54.067 ,

    so it is B

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    posted by Reiny

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