Question

Show that the Quotient Rule of Logarithms works for the base 4 expression log^4 64−log^4 16 .

Answer 1

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 \( \log_4 64 - \log_4 16 \):

1. **Calculate \( \log_4 64 \)**:
We express 64 as a power of 4:
\[
64 = 4^3
\]
Thus,
\[
\log_4 64 = \log_4 (4^3) = 3
\]

2. **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:

\[
\log_4 64 - \log_4 16 = 3 - 2 = 1
\]

3. **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.