The equation \( \ln \left( \frac{1}{4} \right) = -2x \) being the same as \( \ln 4 \) involves an understanding of logarithmic properties. Here’s a step-by-step breakdown:

1. Start with the original equation:
   \[
   \ln \left( \frac{1}{4} \right) = -2x
   \]

2. Rewrite \( \ln \left( \frac{1}{4} \right) \) using the property of logarithms that allows us to express it as:
   \[
   \ln \left( \frac{1}{4} \right) = \ln(4^{-1}) = -\ln(4)
   \]
   So the equation becomes:
   \[
   -\ln(4) = -2x
   \]

3. Divide both sides by \(-1\) to simplify:
   \[
   \ln(4) = 2x
   \]

4. Now, divide by \(2\) to solve for \(x\):
   \[
   x = \frac{\ln(4)}{2}
   \]

This shows that the equation \( \ln \left( \frac{1}{4} \right) = -2x \) ultimately implies that \( x = \frac{\ln(4)}{2} \).

Would you like more details on any of these steps?

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Here are 5 questions related to this topic:

1. What properties of logarithms are used to solve equations like \( \ln \left( \frac{1}{4} \right) = -2x \)?
2. How would you solve an equation involving logarithms with different bases, like \( \log_2(8) = 3x \)?
3. What are some other ways to simplify expressions with fractions in logarithms?
4. How does solving logarithmic equations compare with solving exponential equations?
5. What is the general formula for changing the base of a logarithm?

**Tip:** Remember, \(\ln \left( \frac{1}{a} \right) = -\ln(a)\) is a useful property for simplifying expressions.

Explore the relationship between ln(1/4) = -2x and ln(4) through logarithmic properties. This guide breaks down the steps involved and explains key concepts suitable for high school students.

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