To demonstrate that the Quotient Rule of Logarithms holds for the expression \( \log_4 64 - \log_4 16 \), we can use the Quotient Rule itself, which states:
\[ \log_b \frac{a}{c} = \log_b a - \log_b c \]
This means that the left-hand side is equivalent to taking the logarithm of the quotient of \( a \) and \( c \).
Let's first calculate \( \log_4 64 - \log_4 16 \):
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Calculate \( \log_4 64 \): We express 64 as a power of 4: \[ 64 = 4^3 \] Thus, \[ \log_4 64 = \log_4 (4^3) = 3 \]
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Calculate \( \log_4 16 \): We express 16 as a power of 4: \[ 16 = 4^2 \] Thus, \[ \log_4 16 = \log_4 (4^2) = 2 \]
Now, we can substitute these values into our original expression:
\[ \log_4 64 - \log_4 16 = 3 - 2 = 1 \]
- Calculate \( \log_4 \frac{64}{16} \): Now, according to the Quotient Rule, we can compute: \[ \frac{64}{16} = 4 \] Thus, \[ \log_4 \frac{64}{16} = \log_4 4 \] Since \( 4 = 4^1 \), we have: \[ \log_4 4 = 1 \]
Finally, we have shown that:
\[ \log_4 64 - \log_4 16 = 1 \]
and using the Quotient Rule we found:
\[ \log_4 \frac{64}{16} = 1 \]
Thus, we conclude that:
\[ \log_4 64 - \log_4 16 = \log_4 \frac{64}{16} \]
This verifies that the Quotient Rule of Logarithms indeed works for the expression \( \log_4 64 - \log_4 16 \). Therefore, the claim is proven.