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 base 4 expression \( \log_4 64 - \log_4 16 \), we will first apply the Quotient Rule, which states that:

\[
\log_b \left( \frac{a}{c} \right) = \log_b a - \log_b c
\]

In our case, we need to verify that:

\[
\log_4 64 - \log_4 16 = \log_4 \left( \frac{64}{16} \right)
\]

First, let's compute \( \frac{64}{16} \):

\[
\frac{64}{16} = 4
\]

Now, we can rewrite the expression using our quotient:

\[
\log_4 64 - \log_4 16 = \log_4 4
\]

Next, we calculate \( \log_4 4 \). Since \( 4 \) is \( 4^1 \), we have:

\[
\log_4 4 = 1
\]

Now, we will compute \( \log_4 64 \) and \( \log_4 16 \) to ensure our calculations are consistent.

1. **Calculate \( \log_4 64 \)**:
- Since \( 64 = 4^3 \) (because \( 4^3 = 64 \)):
\[
\log_4 64 = 3
\]

2. **Calculate \( \log_4 16 \)**:
- Since \( 16 = 4^2 \) (because \( 4^2 = 16 \)):
\[
\log_4 16 = 2
\]

Now substituting these values back into our original expression:

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

This matches with our earlier result of \( \log_4 4 = 1 \).

Thus, we have shown that:

\[
\log_4 64 - \log_4 16 = \log_4 4 = 1
\]

Finally, we have verified that the Quotient Rule of Logarithms holds for the base 4 expression:

\[
\log_4 64 - \log_4 16 = \log_4 \left( \frac{64}{16} \right)
\]

Hence, the Quotient Rule is confirmed to be valid in this scenario.