The number of species in a sample is given by [tex]$ S(n)=a \ln \left(1+\frac{n}{a}\right) $[/tex]. Here, [tex]$ n $[/tex] is the number of individuals in the sample, and [tex]$ a $[/tex] is a constant that indicates the diversity of species in the community. If [tex]$ a=0.48 $[/tex], find [tex]$ S(n) $[/tex] for each value of [tex]$ n $[/tex].

1. Find [tex]$ S(n) $[/tex] for [tex]$ n=100 $[/tex].
[tex]\[
S(100)= \square \quad \text{(Round to the nearest whole number as needed.)}
\][/tex]

2. Find [tex]$ S(n) $[/tex] for [tex]$ n=200 $[/tex].
[tex]\[
S(200)= \square \quad \text{(Round to the nearest whole number as needed.)}
\][/tex]

3. Find [tex]$ S(n) $[/tex] for [tex]$ n=150 $[/tex].
[tex]\[
S(150)= \square \quad \text{(Round to the nearest whole number as needed.)}
\][/tex]

4. Find [tex]$ S(n) $[/tex] for [tex]$ n=10 $[/tex].
[tex]\[
S(10)= \square \quad \text{(Round to the nearest whole number as needed.)}
\][/tex]

To solve for the number of species in a sample using the given equation [tex]$ S(n) = a \ln \left( 1 + \frac{n}{a} \right) $[/tex] where [tex]$a = 0.48$[/tex], let's calculate [tex]$ S(n) $[/tex] for the specified values of [tex]$ n $[/tex] step by step.

### Step-by-Step Calculations

#### Find [tex]$ S(100) $[/tex]

1. Substitute [tex]$ n = 100 $[/tex] and [tex]$ a = 0.48 $[/tex] into the equation:
[tex]\[ S(100) = 0.48 \ln \left( 1 + \frac{100}{0.48} \right) \][/tex]
2. Calculate [tex]$ \frac{100}{0.48} $[/tex]:
[tex]\[ \frac{100}{0.48} \approx 208.33 \][/tex]
3. Add 1 to the result:
[tex]\[ 1 + 208.33 \approx 209.33 \][/tex]
4. Calculate the natural logarithm of 209.33:
[tex]\[ \ln(209.33) \approx 5.34 \][/tex]
5. Multiply by the constant [tex]$ 0.48 $[/tex]:
[tex]\[ 0.48 \times 5.34 \approx 2.56 \][/tex]
6. Round to the nearest whole number:
[tex]\[ S(100) \approx 3 \][/tex]

#### Find [tex]$ S(200) $[/tex]

1. Substitute [tex]$ n = 200 $[/tex] and [tex]$ a = 0.48 $[/tex] into the equation:
[tex]\[ S(200) = 0.48 \ln \left( 1 + \frac{200}{0.48} \right) \][/tex]
2. Calculate [tex]$ \frac{200}{0.48} $[/tex]:
[tex]\[ \frac{200}{0.48} \approx 416.67 \][/tex]
3. Add 1 to the result:
[tex]\[ 1 + 416.67 \approx 417.67 \][/tex]
4. Calculate the natural logarithm of 417.67:
[tex]\[ \ln(417.67) \approx 6.03 \][/tex]
5. Multiply by the constant [tex]$ 0.48 $[/tex]:
[tex]\[ 0.48 \times 6.03 \approx 2.89 \][/tex]
6. Round to the nearest whole number:
[tex]\[ S(200) \approx 3 \][/tex]

#### Find [tex]$ S(150) $[/tex]

1. Substitute [tex]$ n = 150 $[/tex] and [tex]$ a = 0.48 $[/tex] into the equation:
[tex]\[ S(150) = 0.48 \ln \left( 1 + \frac{150}{0.48} \right) \][/tex]
2. Calculate [tex]$ \frac{150}{0.48} $[/tex]:
[tex]\[ \frac{150}{0.48} \approx 312.5 \][/tex]
3. Add 1 to the result:
[tex]\[ 1 + 312.5 \approx 313.5 \][/tex]
4. Calculate the natural logarithm of 313.5:
[tex]\[ \ln(313.5) \approx 5.75 \][/tex]
5. Multiply by the constant [tex]$ 0.48 $[/tex]:
[tex]\[ 0.48 \times 5.75 \approx 2.76 \][/tex]
6. Round to the nearest whole number:
[tex]\[ S(150) \approx 3 \][/tex]

#### Find [tex]$ S(10) $[/tex]

1. Substitute [tex]$ n = 10 $[/tex] and [tex]$ a = 0.48 $[/tex] into the equation:
[tex]\[ S(10) = 0.48 \ln \left( 1 + \frac{10}{0.48} \right) \][/tex]
2. Calculate [tex]$ \frac{10}{0.48} $[/tex]:
[tex]\[ \frac{10}{0.48} \approx 20.83 \][/tex]
3. Add 1 to the result:
[tex]\[ 1 + 20.83 \approx 21.83 \][/tex]
4. Calculate the natural logarithm of 21.83:
[tex]\[ \ln(21.83) \approx 3.08 \][/tex]
5. Multiply by the constant [tex]$ 0.48 $[/tex]:
[tex]\[ 0.48 \times 3.08 \approx 1.48 \][/tex]
6. Round to the nearest whole number:
[tex]\[ S(10) \approx 1 \][/tex]

### Final Results

- [tex]$ S(100) = 3 $[/tex]
- [tex]$ S(200) = 3 $[/tex]
- [tex]$ S(150) = 3 $[/tex]
- [tex]$ S(10) = 1 $[/tex]