anonymous
  • anonymous
The function f(x) = 396(3)x represents the growth of a dragonfly population every year in a remote swamp. Pierre wants to manipulate the formula to an equivalent form that calculates every month, not every year. Which function is correct for Pierre's purposes?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
A. f(x)= 33(3)^x b. F(x)=396(3^12)x/12 C. f(x)=396(3^1/12)^12x D. f(x)=4752(3)^x
anonymous
  • anonymous
So, I got C?
IrishBoy123
  • IrishBoy123
**absent typos**: \(\large 396(3^{\frac{1}{12}})^{12x} = 396(3^{\frac{1}{12} \times 12x}) = 396(3)^{x}\) ie back to where you started. plug a number in and see. none of these look right https://gyazo.com/5b74d0c9b3753f89f9467f1bc3f2858c

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IrishBoy123
  • IrishBoy123
screenshot?
anonymous
  • anonymous
@campbell_st
anonymous
  • anonymous
f(x) = 33(3)x f(x) = 396(312)the x over 12 power f(x) = 396(3 to the one twelfth power)12x f(x) = 4752(3)x
campbell_st
  • campbell_st
this is the 2nd time I have seen one of these questions and they don't make much sense to me... I'd expect for this exponential model if the time changes to months the value of x needs to be divided by 12 so you can test by looking at the model \[f(1) = 396(3)^{\frac{1}{12}}\] then find f(2) and f(3) to see if there is growth hope it helps
anonymous
  • anonymous
That"s not an answer choice:( @campbell_st
campbell_st
  • campbell_st
well remember the index law for power of a power, multiply the powers \[(x^a)^b = x^{a \times b}\] so you have \[3^{\frac{x}{12}} = (3^?)^x\]
anonymous
  • anonymous
12?
anonymous
  • anonymous
So would I be right when I said, C?

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