To determine which logarithmic function models the time [tex]\( f(n) \)[/tex] in years it will take for the number of species to decrease to a value of [tex]\( n \)[/tex], let's analyze the problem step-by-step.
1. Understanding exponential decay:
The number of species decreases by [tex]\( 2\% \)[/tex] annually. This can be expressed using an exponential decay formula:
[tex]\[
N(t) = N_0 \cdot (1 - r)^t
\][/tex]
where:
- [tex]\( N(t) \)[/tex] is the number of species at time [tex]\( t \)[/tex]
- [tex]\( N_0 \)[/tex] is the initial number of species
- [tex]\( r \)[/tex] is the annual decay rate (0.02 for [tex]\( 2\% \)[/tex])
- [tex]\( t \)[/tex] is the time in years
2. Setting up the equation:
Given:
- [tex]\( N_0 = 74 \)[/tex] (the initial number of species)
- [tex]\( n \)[/tex] is the number of species remaining after [tex]\( t \)[/tex] years
The equation becomes:
[tex]\[
n = 74 \cdot (0.98)^t
\][/tex]
3. Solving for [tex]\( t \)[/tex]:
To solve for [tex]\( t \)[/tex], we need to isolate [tex]\( t \)[/tex] in the equation:
[tex]\[
\frac{n}{74} = (0.98)^t
\][/tex]
Taking the logarithm of both sides:
[tex]\[
\log\left(\frac{n}{74}\right) = \log\left((0.98)^t\right)
\][/tex]
Using the property of logarithms that [tex]\( \log(a^b) = b\log(a) \)[/tex]:
[tex]\[
\log\left(\frac{n}{74}\right) = t \cdot \log(0.98)
\][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[
t = \frac{\log\left(\frac{n}{74}\right)}{\log(0.98)}
\][/tex]
4. Identifying the logarithmic function:
- The expression [tex]\( \frac{\log\left(\frac{n}{74}\right)}{\log(0.98)} \)[/tex] can be written as a logarithm with base [tex]\( 0.98 \)[/tex]:
[tex]\[
t = \log_{0.98}\left(\frac{n}{74}\right)
\][/tex]
5. Matching the options:
We are provided with the following options:
- A. [tex]\( f(n) = \log_{1.02}\left(\frac{n}{74}\right) \)[/tex]
- B. [tex]\( f(n) = \log_{0.99} n \)[/tex]
- C. [tex]\( f(n) = \log_{1.02} 74 n \)[/tex]
- D. [tex]\( f(n) = \log_{0.28}\left(\frac{n}{74}\right) \)[/tex]
Upon inspecting these options, we see that none exactly matches [tex]\( \log_{0.98}\left(\frac{n}{74}\right) \)[/tex]. However, since the decay rate is [tex]\( 2\% \)[/tex] (or [tex]\( 0.98 \)[/tex] factor per year), the closest correct option should have a base slightly less than 1.
Between these, Option B, [tex]\( f(n) = \log_{0.99} n \)[/tex], uses a base closer to [tex]\( 0.98 \)[/tex].
Thus, the correct answer is:
B. [tex]\( f(n) = \log_{0.99} n \)[/tex]
