The expression given in the image is:

$\ln \left( \frac{1}{4^k} \right)$

We are tasked with expanding this logarithm using properties of logarithms (sum, difference, or product). Let’s break it down:

Step 1: Use the logarithm of a fraction

The logarithm of a quotient can be written as:

$\ln \left( \frac{a}{b} \right) = \ln(a) - \ln(b)$

In this case, $a = 1$ and $b = 4^k$, so:

$\ln \left( \frac{1}{4^k} \right) = \ln(1) - \ln(4^k)$

Since $\ln(1) = 0$, we are left with:

$0 - \ln(4^k) = - \ln(4^k)$

Step 2: Use the power rule of logarithms

The power rule states:

$\ln(a^b) = b \ln(a)$

Here, we apply this to $\ln(4^k)$:

$- \ln(4^k) = - k \ln(4)$

Thus, the expanded expression is:

$\ln \left( \frac{1}{4^k} \right) = -k \ln(4)$

This is the fully expanded logarithmic expression.

Do you want more details or have any questions?

Here are 5 related questions to deepen your understanding:

1. How would you expand $\ln \left( \frac{1}{x^2} \right)$?
2. Can you simplify $\ln \left( \frac{1}{a^b} \right)$ using the same properties?
3. What is the expanded form of $\ln(a^b c^d)$?
4. How does the property $\ln(ab) = \ln(a) + \ln(b)$ differ from the quotient rule?
5. How does $\ln(1) = 0$ affect logarithmic expansions involving fractions?

Tip: Always look for ways to apply the basic logarithmic properties (product, quotient, and power rules) when simplifying complex logarithmic expressions.