To demonstrate that the Quotient Rule of Logarithms works for the expression \( \log_4 64 - \log_4 16 \), we start by applying the Quotient Rule. The Quotient Rule states that:
\[ \log_b \frac{m}{n} = \log_b m - \log_b n \]
This means we can rewrite the expression:
\[ \log_4 64 - \log_4 16 = \log_4 \frac{64}{16} \]
Next, we calculate \( \frac{64}{16} \):
\[ \frac{64}{16} = 4 \]
Now substituting this back into our logarithm expression gives us:
\[ \log_4 \frac{64}{16} = \log_4 4 \]
Now, we find \( \log_4 4 \). Since \( 4 \) is the base of the logarithm, we have:
\[ \log_4 4 = 1 \]
Now, we will separately evaluate \( \log_4 64 \) and \( \log_4 16 \) to verify our calculations:
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To find \( \log_4 64 \), express \( 64 \) as a power of \( 4 \): \[ 64 = 4^3 \quad \text{(since \( 4^3 = 64 \))} \]
Therefore, \[ \log_4 64 = 3 \]
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To find \( \log_4 16 \), express \( 16 \) as a power of \( 4 \): \[ 16 = 4^2 \quad \text{(since \( 4^2 = 16 \))} \]
Therefore, \[ \log_4 16 = 2 \]
Now we can substitute these values back into our original expression \( \log_4 64 - \log_4 16 \) to confirm:
\[ \log_4 64 - \log_4 16 = 3 - 2 = 1 \]
Since we derived earlier that \( \log_4 \frac{64}{16} = 1 \), we conclude that both methods yield the same result.
Thus, we have shown that:
\[ \log_4 64 - \log_4 16 = \log_4 \frac{64}{16} \]
and verified it. The Quotient Rule of Logarithms holds true for the given expression.