To solve this problem, we need to determine which function correctly represents the number of days required for the stock of fruit flies to grow to [tex]\( n \)[/tex] given an initial stock of 200 fruit flies and a daily growth rate of 38%.
First, let's restate some key information:
- Initial stock of fruit flies, [tex]\( S_0 = 200 \)[/tex].
- Daily growth rate, [tex]\( r = 1.38 \)[/tex] (since a 38% increase is equivalent to multiplying by 1.38 each day).
The formula to find the number of fruit flies [tex]\( S \)[/tex] after [tex]\( t \)[/tex] days can be expressed exponentially as:
[tex]\[ S = S_0 \times r^t \][/tex]
[tex]\[ S = 200 \times 1.38^t \][/tex]
We want to find [tex]\( t \)[/tex] such that [tex]\( S = n \)[/tex], where [tex]\( n \)[/tex] is the desired number of fruit flies. Therefore, we set up the equation:
[tex]\[ n = 200 \times 1.38^t \][/tex]
To solve for [tex]\( t \)[/tex], we first isolate [tex]\( 1.38^t \)[/tex]:
[tex]\[ \frac{n}{200} = 1.38^t \][/tex]
Next, we take the logarithm of both sides. For convenience, we use the logarithm base [tex]\( 1.38 \)[/tex] (since [tex]\( t \)[/tex] will appear as a coefficient):
[tex]\[ \log_{1.38}\left(\frac{n}{200}\right) = t \][/tex]
Therefore, the function [tex]\( f(n) \)[/tex] that calculates the number of days [tex]\( t \)[/tex] required for the stock to grow to [tex]\( n \)[/tex] fruit flies can be expressed as:
[tex]\[ f(n) = \log_{1.38}\left(\frac{n}{200}\right) \][/tex]
Thus, the correct option is:
C. [tex]\( f(n)= \log_{1.38} \left(\frac{n}{200}\right) \)[/tex]
