To show that the Quotient Rule of Logarithms holds for the expression \(\log_4 64 - \log_4 16\), we first recall the Quotient Rule of Logarithms:
\[ \log_b a - \log_b c = \log_b \left(\frac{a}{c}\right). \]
In this case, we have:
- \(a = 64\)
- \(c = 16\)
- \(b = 4\)
According to the Quotient Rule, we should show that:
\[ \log_4 64 - \log_4 16 = \log_4 \left(\frac{64}{16}\right). \]
Now, let's calculate the right-hand side:
\[ \frac{64}{16} = 4. \]
So we need to evaluate \(\log_4 4\):
\[ \log_4 4 = 1, \]
because the logarithm of a number to its own base is always 1.
Next, we calculate the left-hand side, which involves evaluating \(\log_4 64\) and \(\log_4 16\).
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Calculating \(\log_4 64\): We know that \(64 = 4^3\), so:
\[ \log_4 64 = \log_4 (4^3) = 3. \]
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Calculating \(\log_4 16\): Similarly, \(16 = 4^2\), so:
\[ \log_4 16 = \log_4 (4^2) = 2. \]
Now substituting these values back into the left-hand side:
\[ \log_4 64 - \log_4 16 = 3 - 2 = 1. \]
Now we have both sides:
- Left-hand side: \(3 - 2 = 1\),
- Right-hand side: \(\log_4 4 = 1\).
Since both sides are equal, we confirm that:
\[ \log_4 64 - \log_4 16 = \log_4 \left(\frac{64}{16}\right), \]
is indeed true and supports the Quotient Rule of Logarithms. Thus, the Quotient Rule works for the given base 4 expression.
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