Question

Show that the Quotient rule of Logarithms works for the base 4 expression log(base 4) 64 - log(base 4) 16.

Answer 1

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 \(\log_4 16\).

1. **Calculating \(\log_4 64\)**:
We know that \(64 = 4^3\), so:

\[
\log_4 64 = \log_4 (4^3) = 3.
\]

2. **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.