To evaluate the given logarithms, we will use the change of base formula:
[tex]\[
\log_a x = \frac{\log_b x}{\log_b a}
\][/tex]
where [tex]\( \log_b \)[/tex] denotes the logarithm with base [tex]\( b \)[/tex]. We will use the natural logarithm (base [tex]\( e \)[/tex]) for convenience.
### 1. Evaluating [tex]\( \log_3 6 \)[/tex]:
Using the change of base formula, we get:
[tex]\[
\log_3 6 = \frac{\log 6}{\log 3}
\][/tex]
Note that [tex]\( \log \)[/tex] denotes the natural logarithm here.
The result of calculating [tex]\( \log_3 6 \)[/tex] rounded to the nearest thousandth is:
[tex]\[
\log_3 6 \approx 1.631
\][/tex]
### 2. Evaluating [tex]\( \log_5 20 \)[/tex]:
Similarly, using the change of base formula, we get:
[tex]\[
\log_5 20 = \frac{\log 20}{\log 5}
\][/tex]
The result of calculating [tex]\( \log_5 20 \)[/tex] rounded to the nearest thousandth is:
[tex]\[
\log_5 20 \approx 1.861
\][/tex]
### 3. Evaluating [tex]\( \log_2 \left( \frac{1}{5} \right) \)[/tex]:
Again, using the change of base formula, we have:
[tex]\[
\log_2 \left( \frac{1}{5} \right) = \frac{\log \left( \frac{1}{5} \right)}{\log 2}
\][/tex]
Since [tex]\( \frac{1}{5} \)[/tex] is a fraction, its logarithm is negative. The result of calculating [tex]\( \log_2 \left( \frac{1}{5} \right) \)[/tex] rounded to the nearest thousandth is:
[tex]\[
\log_2 \left( \frac{1}{5} \right) \approx -2.322
\][/tex]
Summarizing, the evaluated logarithms are:
[tex]\[
\begin{array}{l}
\log_3 6 \approx 1.631 \\
\log_5 20 \approx 1.861 \\
\log_2 \left( \frac{1}{5} \right) \approx -2.322
\end{array}
\][/tex]
