To demonstrate that the Product Rule of Logarithms holds true for the expression \( \log_4{16} + \log_4{4} \), you want to show that this sum is equal to another logarithm expression based on the product of the numbers.
The Product Rule of logarithms states that:
\[ \log_b{m} + \log_b{n} = \log_b{(m \cdot n)} \]
In your case, we have:
- \( m = 16 \)
- \( n = 4 \)
Now, we can calculate:
\[ m \cdot n = 16 \cdot 4 = 64 \]
Therefore, according to the Product Rule:
\[ \log_4{16} + \log_4{4} = \log_4{(16 \cdot 4)} = \log_4{64} \]
Thus, to verify the statement, you can evaluate \( \log_4{16} \) and \( \log_4{4} \) separately:
- \( \log_4{16} = 2 \) (since \( 4^2 = 16 \))
- \( \log_4{4} = 1 \) (since \( 4^1 = 4 \))
Now, add these two results together:
\[ \log_4{16} + \log_4{4} = 2 + 1 = 3 \]
Next, calculate \( \log_4{64} \):
Since \( 64 = 4^3 \), we have:
\[ \log_4{64} = 3 \]
Since \( \log_4{16} + \log_4{4} = \log_4{64} \) yields the same value of \( 3 \), this confirms the Product Rule of Logarithms is indeed valid in this case.
Thus, the best option is: Evaluate both logarithm subscript 4 baseline 16 and logarithm subscript 4 baseline 4 and show that their sum is equal to the value of logarithm subscript 4 baseline 64.