Evaluate.
log Subscript 3 Baseline StartFraction 1 Over 243 EndFraction
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Part 1
log Subscript 3 Baseline StartFraction 1 Over 243 EndFraction
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To evaluate \( \log_3 \left( \frac{1}{243} \right) \), let's proceed step by step:

### Step 1: Express \( \frac{1}{243} \) as a power of 3.
We know that \( 243 = 3^5 \). Therefore:
\[
\frac{1}{243} = 3^{-5}
\]

### Step 2: Use the logarithmic property.
We apply the logarithmic rule \( \log_b (a^c) = c \cdot \log_b (a) \). In this case:
\[
\log_3 \left( 3^{-5} \right) = -5 \cdot \log_3 (3)
\]

### Step 3: Simplify using \( \log_3 (3) = 1 \).
Thus:
\[
-5 \cdot 1 = -5
\]

### Final answer:
\[
\log_3 \left( \frac{1}{243} \right) = -5
\]

Would you like further details or explanations?

Here are 5 related questions:
1. What is \( \log_3 (243) \)?
2. How do you simplify \( \log_b (1/a) \) for any base \( b \)?
3. What is \( \log_2 \left( \frac{1}{32} \right) \)?
4. Can you explain the change of base formula for logarithms?
5. How do logarithms behave when the base is a fraction?

**Tip:** When evaluating logarithms of fractions, express the fraction in terms of a power of the base to simplify.

Learn how to evaluate log_3 (1/243) with a detailed step-by-step solution. This problem involves logarithmic concepts such as powers of base 3 and key properties of logarithms. Suitable for students in Grades 9-11.

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