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Answer:
the n-th root of the ratio of ending to starting values, less 1
Step-by-step explanation:
When I have starting and ending values over a time period, I usually use those directly in the exponential equation. If you're asked for a percentage growth rate, then you have to convert the growth factor to a growth rate.
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The exponential equation can be written ...
f(t) = (initial value) × ((final value)/(initial value))^(t/(time period))
In your example case, this would be ...
f(t) = 2 × (20/2)^(t/18)
f(t) = 2(10^(t/18)) . . . . . . simplified
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To convert this to the form ...
f(t) = 2(1 +(growth rate))^t
recognize that ...
(1 +(growth rate)^t = (10^(1/18))^t
So, you need to find the value of ...
10^(1/18) = 10^(1/18) ≈ 1.13646 = 1 + 13.646%
Then the growth rate is about 13.6% each week.
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Sometimes, you want to use this in the form ...
f(t) = (initial value)×(1 + (growth rate))^t
and sometimes you want the form ...
f(t) = (initial value)×e^(kt)
When you solve for k, you find it is ...
k = ln(1 + (growth rate)) . . . . . . . . where ln(x) is the natural logarithm of x
Here, that is ...
k = ln(10^(1/18)) = ln(10)/18 ≈ 0.12792
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Your final growth equation could be any of ...
f(t) = 2(10^(t/18)) . . . . . . . my personal favorite, as no rounding is involved
f(t) = 2(1.13646^t)
f(t) = 2e^(0.12792t)
