Let's solve each part step-by-step.
Part 1: \( \log_2 \frac{1}{16} \)
We can express \( \frac{1}{16} \) as \( 16^{-1} \) and since \( 16 = 2^4 \), we have:
\[ \frac{1}{16} = 16^{-1} = (2^4)^{-1} = 2^{-4} \]
Using the property of logarithms that states \( \log_b(a^n) = n \cdot \log_b(a) \):
\[ \log_2 \frac{1}{16} = \log_2(2^{-4}) = -4 \cdot \log_2(2) = -4 \cdot 1 = -4 \]
So,
\[ \log_2 \frac{1}{16} = -4 \]
Part 2: \( \log_2 1 \)
Any logarithm of 1, regardless of the base, is 0 because \( b^0 = 1 \) for any base \( b \):
\[ \log_2 1 = 0 \]
Part 3: \( \ln e^2 \)
Using the property of logarithms that states \( \ln(a^n) = n \cdot \ln(a) \):
\[ \ln e^2 = 2 \cdot \ln e \]
Since \( \ln e = 1 \):
\[ \ln e^2 = 2 \cdot 1 = 2 \]
Part 4: \( \log 0.001 \)
We can express \( 0.001 \) in exponential form. \( 0.001 = \frac{1}{1000} = 10^{-3} \):
\[ \log 0.001 = \log(10^{-3}) = -3 \cdot \log(10) \]
Since \( \log(10) = 1 \):
\[ \log 0.001 = -3 \cdot 1 = -3 \]
Final Answers
To summarize:
(a) \( \log_2 1/16 = -4 \)
(b) \( \log_2 1 = 0 \)
(c) \( \ln e^2 = 2 \)
(d) \( \log 0.001 = -3 \)