The Quotient Rule for logarithms states that for any logarithm with base \( b \), the following holds:
\[ \log_b \frac{a}{c} = \log_b a - \log_b c \]
We need to demonstrate that this rule applies to the expression \( \log_4 64 - \log_4 16 \).
First, we can individually calculate \( \log_4 64 \) and \( \log_4 16 \):
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Calculating \( \log_4 64 \): We know that \( 64 = 4^3 \), since \( 4^3 = 64 \). Therefore, \[ \log_4 64 = \log_4 (4^3) = 3. \]
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Calculating \( \log_4 16 \): Similarly, \( 16 = 4^2 \), so: \[ \log_4 16 = \log_4 (4^2) = 2. \]
Now we can evaluate \( \log_4 64 - \log_4 16 \): \[ log_4 64 - log_4 16 = 3 - 2 = 1. \]
Next, let's use the Quotient Rule to see if it yields the same result. We can rewrite our original expression \( \log_4 64 - \log_4 16 \) as: \[ \log_4 \frac{64}{16}. \]
Now, we need to simplify \( \frac{64}{16} \): \[ \frac{64}{16} = 4. \]
Now we can calculate \( \log_4 4 \): Since \( 4 = 4^1 \), we have: \[ \log_4 4 = 1. \]
Both methods yield the same result: \[ log_4 64 - log_4 16 = 1, \] and \[ log_4 \frac{64}{16} = 1. \]
Thus, we have shown that the Quotient Rule of Logarithms holds for this base 4 expression, confirming that:
\[ \log_4 64 - \log_4 16 = \log_4 \frac{64}{16}. \]
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